More Elliptic Integration Examples

All Products Maple MapleSim

Home : Support : Online Help : Applications and Example Worksheets : Integration : More Elliptic Integration Examples

Elliptic Integration in Maple

$restart$

Introduction

Integrals of the form $\int_{a}^{b} \frac{a (x)}{b (x) \sqrt{y (x)}} &DifferentialD; x$ , with $a (x)$ , $b (x)$ , and $y (x)$ polynomials, often return answers in closed form. For example, when $y (x)$ has degree 1 or 2, the integral can always be returned in terms of elementary functions, such as logs, exponentials, trig, or arctrig functions. Thus, you have:

>	$\int_{0}^{1} \frac{1 - x}{(1 + x^{2}) \sqrt{1 + 3 x^{2}}} &DifferentialD; x = \int_{0}^{1} \frac{1 - x}{(1 + x^{2}) \sqrt{1 + 3 x^{2}}} &DifferentialD; x$

$\int_{0}^{1} \frac{1 - x}{(x^{2} + 1) \sqrt{3 x^{2} + 1}} &DifferentialD; x = \frac{\sqrt{2} \arctan (\frac{\sqrt{2}}{2})}{2} - \frac{\sqrt{2} \arctan (\sqrt{2})}{2} + \frac{\sqrt{2} arctanh (\frac{\sqrt{2}}{2})}{2}$

(1.1)

Note that the previous example evaluated the integral in an equation with the inert form on the left-hand side in order to see the integral "prettyprinted."

When y(x) has degree 3 or 4, the integral is called an elliptic integral. Only sometimes do these return answers in terms of elementary functions. In these cases, they are called pseudo-elliptic integrals.

>	$\int_{0}^{\frac{1}{2}} \frac{\sqrt{1 + x^{4}}}{1 - x^{4}} &DifferentialD; x = \int_{0}^{\frac{1}{2}} \frac{\sqrt{1 + x^{4}}}{1 - x^{4}} &DifferentialD; x$

$\int_{0}^{\frac{1}{2}} \frac{\sqrt{x^{4} + 1}}{- x^{4} + 1} &DifferentialD; x = - \frac{\sqrt{2} \ln (\sqrt{17} - 2 \sqrt{2})}{8} + \frac{\sqrt{2} \ln (\sqrt{17} + 2 \sqrt{2})}{8} - \frac{\sqrt{2} \arctan (\frac{\sqrt{17} \sqrt{2}}{4})}{4} + \frac{π \sqrt{2}}{8}$

(1.2)

However, in most cases, elliptic integrals return answers in terms of nonelementary functions, for example, the Legendre elliptic functions. A small variation of the previous problem gives:

>	$\int_{0}^{1} \frac{\sqrt{1 + x^{4}}}{4 - x^{4}} &DifferentialD; x = \int_{0}^{1} \frac{\sqrt{1 + x^{4}}}{4 - x^{4}} &DifferentialD; x$

$\int_{0}^{1} \frac{\sqrt{x^{4} + 1}}{- x^{4} + 4} &DifferentialD; x = - \frac{\sqrt{10} \arctan (\frac{\sqrt{5}}{2})}{16} + \frac{\sqrt{10} arccoth (\frac{(\sqrt{2} - 3) \sqrt{5} \sqrt{2}}{5})}{32} + \frac{\sqrt{10} arccoth (\frac{(\sqrt{2} + 3) \sqrt{5} \sqrt{2}}{5})}{32} - \frac{4 EllipticK (\frac{\sqrt{2}}{2})}{3} + \frac{15 EllipticPi (- \frac{1}{8}, \frac{\sqrt{2}}{2})}{32} + \frac{5 EllipticPi (\frac{9}{8}, \frac{\sqrt{2}}{2})}{96}$

(1.3)

Elliptic integrals appear in a variety of applications (some mentioned in the classical text by Byrd and Friedmann). For example, the length of an arc of a hyperbola $\frac{x^{2}}{α^{2}} - \frac{y^{2}}{β^{2}} = 1$ measured from the vertex to any point $(x, y)$ is determined by the integral

>	$len := \int_{0}^{y} \sqrt{\frac{(1 + \frac{α^{2}}{β^{2}}) t^{2} + β^{2}}{t^{2} + β^{2}}} &DifferentialD; t &colon;$

For example, when $α = 2$ and $β = 1$ , the length from the vertex to the point $(2 \sqrt{5}, 2)$ is given by

>	$\binom{len}{} \| \binom{}{α = 2, β = 1, y = 2}$

$\int_{0}^{2} \sqrt{\frac{5 t^{2} + 1}{t^{2} + 1}} &DifferentialD; t$

(1.4)

which is

>	$ans := value ()$

$ans ≔ \frac{\sqrt{5} EllipticPi (\frac{2 \sqrt{105}}{21}, 1, \frac{2 \sqrt{5}}{5})}{5}$

(1.5)

with the result given in terms of Elliptic functions. Numerically, this is approximated by

>	$evalf (ans)$

$3.267095734$

(1.6)

Definitions of F, E, Pi

The following equations were constructed to show Maple definitions of the Legendre F, E, and Pi functions.

>	$assume (0 < k, k < 1)$

>	$EllipticF (x, k) = \int_{0}^{x} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - k^{2} t^{2}}} &DifferentialD; t$

(2.1)

>	$EllipticE (x, k) = \int_{0}^{x} \frac{\sqrt{1 - k^{2} t^{2}}}{\sqrt{1 - t^{2}}} &DifferentialD; t$

(2.2)

>	$EllipticPi (x, a, k) = \int_{0}^{x} \frac{1}{(1 - a t^{2}) \sqrt{1 - t^{2}} \sqrt{1 - k^{2} t^{2}}} &DifferentialD; t$

(2.3)

with the variable $k$ in $(0, 1)$ . For $x = 1$ , you obtain the complete forms of these integrals. For example,

>	$EllipticK (k) = \int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}} \sqrt{1 - k^{2} t^{2}}} &DifferentialD; t$

(2.4)

For more information, see EllipticE, EllipticF, EllipticK, and EllipticPi.

Trigonometric Forms of Elliptic Integrals

Integrals of the form $\int_{0}^{θ} R (\sin (t), \cos (t)) \sqrt{y (\sin (t), \cos (t))} &DifferentialD; t$ where $R$ is a rational function of $\sin (t)$ and $\cos (t)$ and $y$ is a degree 2 polynomial in $\sin (t)$ and $\cos (t)$ are elliptic integrals in trigonometric form. Corresponding forms for hyperbolic trig expressions (as above) also exist. Such integrals can also be expressed in terms of Legendre's elliptic functions. For example,

>	$\int_{0}^{\frac{π}{2}} \sqrt{\sin (x)} &DifferentialD; x = \int_{0}^{\frac{π}{2}} \sqrt{\sin (x)} &DifferentialD; x$

$\int_{0}^{\frac{π}{2}} \sqrt{\sin (x)} &DifferentialD; x = - \sqrt{2} EllipticK (\frac{\sqrt{2}}{2}) + 2 \sqrt{2} EllipticE (\frac{\sqrt{2}}{2})$

(3.1)

and

>	$\int_{0}^{\infty} \frac{1}{\sqrt{\sinh (x)}} &DifferentialD; x = \int_{0}^{\infty} \frac{1}{\sqrt{\sinh (x)}} &DifferentialD; x$

$\int_{0}^{\infty} \frac{1}{\sqrt{\sinh (x)}} &DifferentialD; x = 2 EllipticK (\frac{\sqrt{2}}{2})$

(3.2)

In both cases you can compare the results numerically:

>	$evalf ()$

$1.198140235 = 1.198140235$

(3.3)

>	$evalf ()$

$3.708149355 = 3.708149354$

(3.4)

Why Reduce to Normal Form

Because of the possible presence of poles in the integrand, elliptic integrals are often very difficult to compute numerically with standard numerical integration methods. However, the Legendre functions can be numerically evaluated very efficiently using the AGM method. Therefore, reducing to normal form allows for easy numerical evaluation. For example,

>	$ques := \int_{0}^{1} \frac{1}{(x^{5} - x + 2) \sqrt{x^{4} - 3}} &DifferentialD; x &colon;$

>	$ans1 := value (ques)$

$ans1 ≔ \frac{I (\sum_{_α = RootOf ({_Z}^{5} -_Z + 2)} \frac{(- 40 {_α}^{4} - 100 {_α}^{3} - 250 {_α}^{2} - 625_α + 32) (- \ln (3) - \ln (- (\sqrt{- 3 {_α}^{4} + 9} + 3) (- \frac{1}{4} {_α}^{4} - \frac{1}{2} {_α}^{3} + \frac{1}{4})) + \ln (- (\sqrt{- 3 {_α}^{4} + 9} \sqrt{6} - 3 {_α}^{2} + 9) (- \frac{1}{2} {_α}^{4} - \frac{1}{2} {_α}^{3} - \frac{1}{2} {_α}^{2} - \frac{1}{2}_α) (- \frac{1}{2} {_α}^{4} + \frac{1}{2} {_α}^{3} - \frac{1}{2} {_α}^{2} + \frac{1}{2}_α)))}{\sqrt{- {_α}^{4} + 3}})}{12436} - \frac{I \sqrt{2} 3^{\frac{3}{4}} (\sum_{_α = RootOf ({_Z}^{5} -_Z + 2)} (- \frac{1361 {_α}^{2}}{24872} + \frac{1321_α \sqrt{3}}{24872} + \frac{1261_α}{24872} - \frac{87}{12436} - \frac{1337 {_α}^{4}}{24872} + \frac{1321 {_α}^{3}}{24872} - \frac{387 \sqrt{3}}{12436} - \frac{587 {_α}^{4} \sqrt{3}}{24872} + \frac{87 {_α}^{3} \sqrt{3}}{24872} - \frac{1337 {_α}^{2} \sqrt{3}}{24872}) EllipticPi (\frac{3 {_α}^{4}}{4} + \frac{{_α}^{4} \sqrt{3}}{4} - \frac{3 {_α}^{3}}{4} - \frac{{_α}^{3} \sqrt{3}}{4} + \frac{3 {_α}^{2}}{4} + \frac{3 {_α}^{2} \sqrt{3}}{4} - \frac{3_α}{4} - \frac{3_α \sqrt{3}}{4} + \frac{3}{2} + \frac{\sqrt{3}}{2}, \frac{\sqrt{2}}{2}))}{6} + \frac{I \sqrt{2} 3^{\frac{3}{4}} (\sum_{_α = RootOf ({_Z}^{5} -_Z + 2)} (- \frac{1361 {_α}^{2}}{24872} + \frac{1321_α \sqrt{3}}{24872} + \frac{1261_α}{24872} - \frac{87}{12436} - \frac{1337 {_α}^{4}}{24872} + \frac{1321 {_α}^{3}}{24872} - \frac{387 \sqrt{3}}{12436} - \frac{587 {_α}^{4} \sqrt{3}}{24872} + \frac{87 {_α}^{3} \sqrt{3}}{24872} - \frac{1337 {_α}^{2} \sqrt{3}}{24872}) EllipticPi (\frac{\sqrt{\sqrt{3} - 1} 3^{\frac{3}{4}}}{3}, \frac{3 {_α}^{4}}{4} + \frac{{_α}^{4} \sqrt{3}}{4} - \frac{3 {_α}^{3}}{4} - \frac{{_α}^{3} \sqrt{3}}{4} + \frac{3 {_α}^{2}}{4} + \frac{3 {_α}^{2} \sqrt{3}}{4} - \frac{3_α}{4} - \frac{3_α \sqrt{3}}{4} + \frac{3}{2} + \frac{\sqrt{3}}{2}, \frac{\sqrt{2}}{2}))}{6}$

(4.1)

which can be (quickly) numerically approximated by

>	$evalf ()$

$0. - 0.3633182165 I$

(4.2)

Of course, one can always compute the numerical approximation directly by using

>	$evalf (ques)$

$- 0.3633182165 I$

(4.3)

However, having a formula for an answer allows one to also compute to higher digits by using the same form:

>	$evalf (ans1, 30)$

$0. - 0.363318216528577575158770044176 I$

(4.4)

which can also be done directly with evalf. However, in this case, the numerical integrator takes considerable time to come up with a result:

>	$evalf (ques, 30)$

$- 0.363318216528577575158770044176 I$

(4.5)

Why Use RootOf

Let

>	$a := 14 x^{10} - 4 x^{8} + 149 x^{6} - 228 x^{4} + 102 x^{2} - 93 &colon;$

>	$b := x^{4} - 7 x^{2} - 1 &colon;$

and consider the problem of integrating

>	$\int_{0}^{1} \frac{a}{b^{2} \sqrt{x^{4} - 3 x^{2} + 2}} &DifferentialD; x &colon;$

and

>	$\int_{0}^{1} \frac{a}{b^{3} \sqrt{x^{4} - 3 x^{2} + 2}} &DifferentialD; x &colon;$

The classical technique of reducing such integrals to Legendre normal form involves working with the poles of the rational function, converting the rational function to partial fractions, and reducing each "smaller" part. However, the poles of $\frac{a}{b^{2}}$ and $\frac{a}{b^{3}}$ are seen to be

>	$poles := solve (x^{4} + x^{2} - 1)$

$poles ≔ \frac{\sqrt{- 2 + 2 \sqrt{5}}}{2}, - \frac{\sqrt{- 2 + 2 \sqrt{5}}}{2}, \frac{I}{2} \sqrt{2 \sqrt{5} + 2}, - \frac{I}{2} \sqrt{2 \sqrt{5} + 2}$

(5.1)

so all the poles involve nested radicals. Numerically, at the current setting of Digits, these are given by:

>	$evalf ([poles])$

$[0.7861513775, −0.7861513775, 1.272019650 I, - 1.272019650 I]$

(5.2)

Converting say $\frac{a}{b^{3}}$ to a complete partial fraction decomposition can be accomplished by

>	$new_b := mul (x - i, i = poles)$

$new_b ≔ (x - \frac{\sqrt{- 2 + 2 \sqrt{5}}}{2}) (x + \frac{\sqrt{- 2 + 2 \sqrt{5}}}{2}) (x - \frac{I \sqrt{2 \sqrt{5} + 2}}{2}) (x + \frac{I \sqrt{2 \sqrt{5} + 2}}{2})$

(5.3)

>	$convert (\frac{a}{{new_b}^{3}}, parfrac, x)$

$\frac{- 92798976000 + 40475033600 \sqrt{5}}{(2 x + \sqrt{- 2 + 2 \sqrt{5}}) {(\sqrt{5} - 1)}^{5} {(I \sqrt{5} + 2 + I)}^{5} {(- 2 + 2 \sqrt{5})}^{\frac{9}{2}} {(I \sqrt{5} - 2 + I)}^{5}} + \frac{11632640 \sqrt{5} - 28672000}{{(2 x + \sqrt{- 2 + 2 \sqrt{5}})}^{3} {(- 2 + 2 \sqrt{5})}^{\frac{9}{2}} {(I \sqrt{5} + 2 + I)}^{3} {(I \sqrt{5} - 2 + I)}^{3}} + \frac{10485760 I (175 + 71 \sqrt{5})}{{(I \sqrt{- 2 + 2 \sqrt{5}} \sqrt{5} + I \sqrt{- 2 + 2 \sqrt{5}} + 4 x)}^{3} {(- 2 + 2 \sqrt{5})}^{\frac{9}{2}} {(\sqrt{5} + 1)}^{3} {(I \sqrt{5} + 2 + I)}^{3} {(I \sqrt{5} - 2 + I)}^{3}} + \frac{- 6658457600 + 1520435200 \sqrt{5}}{{(I \sqrt{- 2 + 2 \sqrt{5}} \sqrt{5} + I \sqrt{- 2 + 2 \sqrt{5}} + 4 x)}^{2} {(\sqrt{5} - 1)}^{9} {(\sqrt{5} + 1)}^{4} {(I \sqrt{5} + 2 + I)}^{4} {(I \sqrt{5} - 2 + I)}^{4}} + \frac{3355443200 I (405 + 47 \sqrt{5})}{(- I \sqrt{- 2 + 2 \sqrt{5}} \sqrt{5} - I \sqrt{- 2 + 2 \sqrt{5}} + 4 x) {(\sqrt{5} - 1)}^{5} {(\sqrt{5} + 1)}^{5} {(I \sqrt{5} + 2 + I)}^{5} {(- 2 + 2 \sqrt{5})}^{\frac{9}{2}} {(I \sqrt{5} - 2 + I)}^{5}} + \frac{- 6658457600 + 1520435200 \sqrt{5}}{{(- I \sqrt{- 2 + 2 \sqrt{5}} \sqrt{5} - I \sqrt{- 2 + 2 \sqrt{5}} + 4 x)}^{2} {(\sqrt{5} - 1)}^{9} {(\sqrt{5} + 1)}^{4} {(I \sqrt{5} + 2 + I)}^{4} {(I \sqrt{5} - 2 + I)}^{4}} + \frac{- 11632640 \sqrt{5} + 28672000}{{(2 x - \sqrt{- 2 + 2 \sqrt{5}})}^{3} {(- 2 + 2 \sqrt{5})}^{\frac{9}{2}} {(I \sqrt{5} + 2 + I)}^{3} {(I \sqrt{5} - 2 + I)}^{3}} + \frac{461209600 - 202342400 \sqrt{5}}{{(2 x - \sqrt{- 2 + 2 \sqrt{5}})}^{2} {(\sqrt{5} - 1)}^{9} {(I \sqrt{5} + 2 + I)}^{4} {(I \sqrt{5} - 2 + I)}^{4}} - \frac{10485760 I (175 + 71 \sqrt{5})}{{(- I \sqrt{- 2 + 2 \sqrt{5}} \sqrt{5} - I \sqrt{- 2 + 2 \sqrt{5}} + 4 x)}^{3} {(- 2 + 2 \sqrt{5})}^{\frac{9}{2}} {(\sqrt{5} + 1)}^{3} {(I \sqrt{5} + 2 + I)}^{3} {(I \sqrt{5} - 2 + I)}^{3}} + \frac{92798976000 - 40475033600 \sqrt{5}}{(2 x - \sqrt{- 2 + 2 \sqrt{5}}) {(\sqrt{5} - 1)}^{5} {(I \sqrt{5} + 2 + I)}^{5} {(- 2 + 2 \sqrt{5})}^{\frac{9}{2}} {(I \sqrt{5} - 2 + I)}^{5}} - \frac{3355443200 I (405 + 47 \sqrt{5})}{(I \sqrt{- 2 + 2 \sqrt{5}} \sqrt{5} + I \sqrt{- 2 + 2 \sqrt{5}} + 4 x) {(\sqrt{5} - 1)}^{5} {(\sqrt{5} + 1)}^{5} {(I \sqrt{5} + 2 + I)}^{5} {(- 2 + 2 \sqrt{5})}^{\frac{9}{2}} {(I \sqrt{5} - 2 + I)}^{5}} + \frac{461209600 - 202342400 \sqrt{5}}{{(2 x + \sqrt{- 2 + 2 \sqrt{5}})}^{2} {(\sqrt{5} - 1)}^{9} {(I \sqrt{5} + 2 + I)}^{4} {(I \sqrt{5} - 2 + I)}^{4}}$

(5.4)

again an expression that involves complicated nested radicals. The elliptic integration algorithm in Maple takes special care to avoid working with unnecessary radicals whenever possible. Integrals are first reduced to ones involving no repeated poles by using a classical reduction technique discovered by Hermite in the last century which introduces no new radicals during the reduction. Thus, you obtain

>	$\int_{0}^{1} \frac{a}{b^{2} \sqrt{x^{4} - 3 x^{2} + 2}} &DifferentialD; x = \int_{0}^{1} \frac{a}{b^{2} \sqrt{x^{4} - 3 x^{2} + 2}} &DifferentialD; x$

$\int_{0}^{1} \frac{14 x^{10} - 4 x^{8} + 149 x^{6} - 228 x^{4} + 102 x^{2} - 93}{{(x^{4} - 7 x^{2} - 1)}^{2} \sqrt{x^{4} - 3 x^{2} + 2}} &DifferentialD; x = \frac{60715 \sqrt{2} EllipticK (\frac{\sqrt{2}}{2})}{742} - \frac{119772 \sqrt{2} EllipticE (\frac{\sqrt{2}}{2})}{4081} + \frac{\sqrt{2} (\sum_{_α = RootOf ({_Z}^{4} - 7 {_Z}^{2} - 1)} (- \frac{1006000 {_α}^{2}}{216293} - \frac{4977338}{216293}) EllipticPi ({_α}^{2} - 7, \frac{\sqrt{2}}{2}))}{2}$

(5.5)

In addition, whenever possible, the elliptic integration algorithm in Maple works with an implicit rather than with an explicit representation of such roots. After Hermite reduction to eliminate multiple poles, a final answer is reduced to normal form by using the sum over roots functionality available in Maple. Thus, you obtain

>	$\int_{0}^{1} \frac{a}{b^{3} \sqrt{x^{4} - 3 x^{2} + 2}} &DifferentialD; x = \int_{0}^{1} \frac{a}{b^{3} \sqrt{x^{4} - 3 x^{2} + 2}} &DifferentialD; x$

$\int_{0}^{1} \frac{14 x^{10} - 4 x^{8} + 149 x^{6} - 228 x^{4} + 102 x^{2} - 93}{{(x^{4} - 7 x^{2} - 1)}^{3} \sqrt{x^{4} - 3 x^{2} + 2}} &DifferentialD; x = - \frac{26481451 \sqrt{2} EllipticK (\frac{\sqrt{2}}{2})}{12112408} + \frac{160910307 \sqrt{2} EllipticE (\frac{\sqrt{2}}{2})}{66618244} + \frac{\sqrt{2} (\sum_{_α = RootOf ({_Z}^{4} - 7 {_Z}^{2} - 1)} (- \frac{19624487705 {_α}^{2}}{7061533864} + \frac{137971253003}{7061533864}) EllipticPi ({_α}^{2} - 7, \frac{\sqrt{2}}{2}))}{2}$

(5.6)

You can evaluate this numerically by

>	$ans := rhs ()$

$ans ≔ - \frac{26481451 \sqrt{2} EllipticK (\frac{\sqrt{2}}{2})}{12112408} + \frac{160910307 \sqrt{2} EllipticE (\frac{\sqrt{2}}{2})}{66618244} + \frac{\sqrt{2} (\sum_{_α = RootOf ({_Z}^{4} - 7 {_Z}^{2} - 1)} (- \frac{19624487705 {_α}^{2}}{7061533864} + \frac{137971253003}{7061533864}) EllipticPi ({_α}^{2} - 7, \frac{\sqrt{2}}{2}))}{2}$

(5.7)

>	$evalf (ans)$

$14.90025848 + 8.420766585 \times 10^{−19} I$

(5.8)

If you want to evaluate the integral numerically to higher digits, you would simply do

>	$evalf (ans, 50)$

$14.900258483139842257241717964774763836672323405342 + 5.6084898176386662799736895664138375463221347382190 \times 10^{−58} I$

(5.9)

Sometimes one also uses the roots of the polynomial under the radical in an implicit rather than an explicit form. For example, you obtain (after some waiting and using alias to obtain a simpler form of answer):

>	$y := 97 x^{4} + 50 x^{3} + 79 x^{2} + 56 x + 49 &colon;$

>	$rts := fsolve (y, x, complex)$

$rts ≔ −0.545608960370471 - 0.548081748438361 I, −0.545608960370471 + 0.548081748438361 I, 0.287877001607584 - 0.872783001252362 I, 0.287877001607584 + 0.872783001252362 I$

(5.10)

>	$alias (seq (cat (R, i) = RootOf (subs (x =_Z, y), index = i), i = 1 .. 4))$

$R1, R2, R3, R4$

(5.11)

>	$\int_{0}^{1} \frac{1}{\sqrt{y}} &DifferentialD; x = \int_{0}^{1} \frac{1}{\sqrt{y}} &DifferentialD; x$

$\int_{0}^{1} \frac{1}{\sqrt{97 x^{4} + 50 x^{3} + 79 x^{2} + 56 x + 49}} &DifferentialD; x = - \frac{2 (R4 - R1) \sqrt{97} (\sqrt{\frac{(R4 - R2) R1}{(R4 - R1) R2}} {R2}^{2} \sqrt{- \frac{(R2 - R1) R3}{(R1 - R3) R2}} \sqrt{\frac{R4 (R2 - R1)}{(R4 - R1) R2}} EllipticF (\sqrt{\frac{(R4 - R2) R1}{(R4 - R1) R2}}, \sqrt{- \frac{(R4 - R1) (R2 - R3)}{(R1 - R3) (R2 - R4)}}) \sqrt{(- 1 + R1) (- 1 + R2) (- 1 + R3) (- 1 + R4)} - {R2}^{2} \sqrt{R1 R2 R3 R4} \sqrt{\frac{(R2 - R4) (- 1 + R1)}{(R1 - R4) (- 1 + R2)}} \sqrt{- \frac{(R2 - R1) (- 1 + R3)}{(- 1 + R2) (R1 - R3)}} \sqrt{\frac{(R2 - R1) (- 1 + R4)}{(R4 - R1) (- 1 + R2)}} EllipticF (\sqrt{- \frac{(R2 - R4) (- 1 + R1)}{(R4 - R1) (- 1 + R2)}}, \sqrt{- \frac{(R4 - R1) (R2 - R3)}{(R1 - R3) (R2 - R4)}}) + 2 R2 \sqrt{R1 R2 R3 R4} \sqrt{\frac{(R2 - R4) (- 1 + R1)}{(R1 - R4) (- 1 + R2)}} \sqrt{- \frac{(R2 - R1) (- 1 + R3)}{(- 1 + R2) (R1 - R3)}} \sqrt{\frac{(R2 - R1) (- 1 + R4)}{(R4 - R1) (- 1 + R2)}} EllipticF (\sqrt{- \frac{(R2 - R4) (- 1 + R1)}{(R4 - R1) (- 1 + R2)}}, \sqrt{- \frac{(R4 - R1) (R2 - R3)}{(R1 - R3) (R2 - R4)}}) - \sqrt{\frac{(R2 - R4) (- 1 + R1)}{(R1 - R4) (- 1 + R2)}} \sqrt{- \frac{(R2 - R1) (- 1 + R3)}{(- 1 + R2) (R1 - R3)}} \sqrt{\frac{(R2 - R1) (- 1 + R4)}{(R4 - R1) (- 1 + R2)}} EllipticF (\sqrt{- \frac{(R2 - R4) (- 1 + R1)}{(R4 - R1) (- 1 + R2)}}, \sqrt{- \frac{(R4 - R1) (R2 - R3)}{(R1 - R3) (R2 - R4)}}) \sqrt{R1 R2 R3 R4})}{97 (R2 - R4) (R2 - R1) \sqrt{R1 R2 R3 R4} \sqrt{(- 1 + R1) (- 1 + R2) (- 1 + R3) (- 1 + R4)}}$

(5.12)

Here the answer is expressed in a (complicated) expression involving implicit roots indexed by their proximity to a numerical root. In this case, you can evaluate such an integral numerically to any number of digits by using:

>	$ans := rhs ()$

$ans ≔ - \frac{2 (R4 - R1) \sqrt{97} (\sqrt{\frac{(R4 - R2) R1}{(R4 - R1) R2}} {R2}^{2} \sqrt{- \frac{(R2 - R1) R3}{(R1 - R3) R2}} \sqrt{\frac{R4 (R2 - R1)}{(R4 - R1) R2}} EllipticF (\sqrt{\frac{(R4 - R2) R1}{(R4 - R1) R2}}, \sqrt{- \frac{(R4 - R1) (R2 - R3)}{(R1 - R3) (R2 - R4)}}) \sqrt{(- 1 + R1) (- 1 + R2) (- 1 + R3) (- 1 + R4)} - {R2}^{2} \sqrt{R1 R2 R3 R4} \sqrt{\frac{(R2 - R4) (- 1 + R1)}{(R1 - R4) (- 1 + R2)}} \sqrt{- \frac{(R2 - R1) (- 1 + R3)}{(- 1 + R2) (R1 - R3)}} \sqrt{\frac{(R2 - R1) (- 1 + R4)}{(R4 - R1) (- 1 + R2)}} EllipticF (\sqrt{- \frac{(R2 - R4) (- 1 + R1)}{(R4 - R1) (- 1 + R2)}}, \sqrt{- \frac{(R4 - R1) (R2 - R3)}{(R1 - R3) (R2 - R4)}}) + 2 R2 \sqrt{R1 R2 R3 R4} \sqrt{\frac{(R2 - R4) (- 1 + R1)}{(R1 - R4) (- 1 + R2)}} \sqrt{- \frac{(R2 - R1) (- 1 + R3)}{(- 1 + R2) (R1 - R3)}} \sqrt{\frac{(R2 - R1) (- 1 + R4)}{(R4 - R1) (- 1 + R2)}} EllipticF (\sqrt{- \frac{(R2 - R4) (- 1 + R1)}{(R4 - R1) (- 1 + R2)}}, \sqrt{- \frac{(R4 - R1) (R2 - R3)}{(R1 - R3) (R2 - R4)}}) - \sqrt{\frac{(R2 - R4) (- 1 + R1)}{(R1 - R4) (- 1 + R2)}} \sqrt{- \frac{(R2 - R1) (- 1 + R3)}{(- 1 + R2) (R1 - R3)}} \sqrt{\frac{(R2 - R1) (- 1 + R4)}{(R4 - R1) (- 1 + R2)}} EllipticF (\sqrt{- \frac{(R2 - R4) (- 1 + R1)}{(R4 - R1) (- 1 + R2)}}, \sqrt{- \frac{(R4 - R1) (R2 - R3)}{(R1 - R3) (R2 - R4)}}) \sqrt{R1 R2 R3 R4})}{97 (R2 - R4) (R2 - R1) \sqrt{R1 R2 R3 R4} \sqrt{(- 1 + R1) (- 1 + R2) (- 1 + R3) (- 1 + R4)}}$

(5.13)

>	$evalf (ans)$

$0.09691069513 - 2.589567301 \times 10^{−10} I$

(5.14)

>	$evalf (ans, 50)$

$0.096910695142820439269099009374730407267346394794458 + 1.3044894546755286314153240166438825889924180855545 \times 10^{−50} I$

(5.15)

Indefinite Elliptic Integrals

You can also use a reduction procedure to return closed form solutions of Elliptic integrals in the indefinite case. In such cases, one does not return a form with elliptic functions having a k between 0 and 1; instead, an arbitrary complex number is used. For example, you have

>	$\int \frac{\sqrt{2 - x^{4}}}{1 - x^{4}} &DifferentialD; x = \int \frac{\sqrt{2 - x^{4}}}{1 - x^{4}} &DifferentialD; x$

$\int \frac{\sqrt{- x^{4} + 2}}{- x^{4} + 1} &DifferentialD; x = \frac{2^{\frac{1}{4}} \sqrt{4 - 2 \sqrt{2} x^{2}} \sqrt{4 + 2 \sqrt{2} x^{2}} EllipticF (\frac{x 2^{\frac{3}{4}}}{2}, I)}{4 \sqrt{- x^{4} + 2}} - \frac{2^{\frac{3}{4}} \sqrt{4 - 2 \sqrt{2} x^{2}} \sqrt{4 + 2 \sqrt{2} x^{2}} (EllipticF (\frac{x 2^{\frac{3}{4}}}{2}, I) - EllipticE (\frac{x 2^{\frac{3}{4}}}{2}, I))}{8 \sqrt{- x^{4} + 2}} + \frac{2^{\frac{1}{4}} \sqrt{1 - \frac{\sqrt{2} x^{2}}{2}} \sqrt{1 + \frac{\sqrt{2} x^{2}}{2}} EllipticPi (\frac{x 2^{\frac{3}{4}}}{2}, \sqrt{2}, \sqrt{- \frac{\sqrt{2}}{2}} 2^{\frac{1}{4}})}{2 \sqrt{- x^{4} + 2}} + \frac{2^{\frac{3}{4}} \sqrt{4 - 2 \sqrt{2} x^{2}} \sqrt{4 + 2 \sqrt{2} x^{2}} EllipticF (\frac{x 2^{\frac{3}{4}}}{2}, I)}{8 \sqrt{- x^{4} + 2}} - \frac{2^{\frac{3}{4}} \sqrt{4 - 2 \sqrt{2} x^{2}} \sqrt{4 + 2 \sqrt{2} x^{2}} EllipticE (\frac{x 2^{\frac{3}{4}}}{2}, I)}{8 \sqrt{- x^{4} + 2}} + \frac{2^{\frac{1}{4}} \sqrt{1 - \frac{\sqrt{2} x^{2}}{2}} \sqrt{1 + \frac{\sqrt{2} x^{2}}{2}} EllipticPi (\frac{x 2^{\frac{3}{4}}}{2}, - \sqrt{2}, \sqrt{- \frac{\sqrt{2}}{2}} 2^{\frac{1}{4}})}{2 \sqrt{- x^{4} + 2}}$

(6.1)

a result which is verified by

>	$ans := &colon;$

>	$normal (\frac{&DifferentialD;}{&DifferentialD; x} (lhs (ans) - rhs (ans)))$

$0$

(6.2)

Similarly, in the case of the trig form of elliptic integrals, you have

>	$\int \sqrt{1 + 2 \cos (x)} &DifferentialD; x = \int \sqrt{1 + 2 \cos (x)} &DifferentialD; x$

$\int \sqrt{1 + 2 \cos (x)} &DifferentialD; x = \frac{2 \sqrt{(4 {\cos (\frac{x}{2})}^{2} - 1) {\sin (\frac{x}{2})}^{2}} \sqrt{{\sin (\frac{x}{2})}^{2}} \sqrt{- 4 {\cos (\frac{x}{2})}^{2} + 1} EllipticE (\cos (\frac{x}{2}), 2)}{\sqrt{- 4 {\sin (\frac{x}{2})}^{4} + 3 {\sin (\frac{x}{2})}^{2}} \sin (\frac{x}{2}) \sqrt{4 {\cos (\frac{x}{2})}^{2} - 1}}$

(6.3)

which again can be verified by

>	$ans := &colon;$

>	$simplify (\frac{&DifferentialD;}{&DifferentialD; x} (lhs (ans) - rhs (ans)))$

$0$

(6.4)

	See Also
	diff, EllipticE, EllipticF, EllipticK, EllipticPi, evalf, Int, simplify

Return to Index for Example Worksheets

Download Help Document

Maple

Erweiterungen für Maple

Math Success Platform

Verbesserung der Studienerfolgsquote

Maple Flow

MapleSim

Beratende Dienstleistungen

Produkte zur Online-Ausbildung

Ausbildung

Industrie

Automobil und Luftfahrt

Robotik

Maschinendesign & industrielle Automation

Andere

Lösungen für die Industrie

Kaufen: Einzelheiten und Preise

Kaufen

Institutionelle Studentenlizenzierung

Elite-Wartungsprogramm

Support

Produktschulung

Online Hilfe zu Produkten

Webinare und Veranstaltungen

Veröffentlichungen

Content Hubs

Anwendungsbeispiele

Community

Über Maplesoft

Pressezentrum

User Community

Kontakt

Online Help

All Products Maple MapleSim

Maple

Leistungsfähige intuitive Mathematiksoftware

Erweiterungen für Maple

Math Success Platform

Verbesserung der Studienerfolgsquote

Maple Flow

Engineering calculations & documentation

MapleSim

Modellierung auf Systemebene

Beratende Dienstleistungen

Produkte zur Online-Ausbildung

Ausbildung

Industrie

Automobil und Luftfahrt

Robotik

Maschinendesign & industrielle Automation

Andere

Lösungen für die Industrie

Kaufen: Einzelheiten und Preise

Kaufen

Institutionelle Studentenlizenzierung

Elite-Wartungsprogramm

Support

Produktschulung

Online Hilfe zu Produkten

Webinare und Veranstaltungen

Veröffentlichungen

Content Hubs

Anwendungsbeispiele

Community

Über Maplesoft

Pressezentrum

User Community

Kontakt

Online Help

All Products Maple MapleSim