The int command has seen improvements to various methods. The following new results come from the asymptotic method:${\int }_0^{\mathrm{\infty }}\frac{\sin \left(x\right)}{x^3}\ⅆx$

${\int }_0^{\mathrm{\infty }}\frac{\cos \left(x\right)\cos \left(\frac{x}{3}\right)}{x^2}\ⅆx$

This new result results from an improvement in the elliptic integration code:

${\int }_0^1{\int }_0^1\left|{\ⅇ}^{2\mathrm{I}\mathrm{\pi }x}\+{\ⅇ}^{2\mathrm{I}\mathrm{\pi } y}\right|\ⅆx\ⅆy$

The next two examples are due to improvements to the Risch algorithm:

$\mathrm{normal}\left(\int \frac{\sin \left(x\arctan \left(a\right)\right){\left(a^2\+1\right)}^{-x}}{1\+{\ⅇ}^a}\ⅆx\right)$

$\int \left(\left(\frac{83{\ln \left(u\right)}^2}{30}-\frac{20162\ln \left(u\right)}{2025}+\frac{584972}{30375}\right)u^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\left(\frac{4723{\ln \left(u\right)}^2}{1260}-\frac{3961877\ln \left(u\right)}{297675}+\frac{802885334}{31255875}\right)u^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\ⅆu$

This previously unsolved example is due to an improvement in the hyperexponential algorithm:
$\int \frac{\sqrt{\sinh \left(x\right)\+\cosh \left(x\right)}}{\sinh \left(x\right)\+1}\ⅆx$

Two new commands, Homogenize and IsHomogeneous, for performing and testing (weighted) homogenization were added to the PolynomialTools package.

$\mathrm{with}\left(\mathrm{PolynomialTools}\right)\:$

$f≔w^5\+w^{2}x\+w y\+y^3$

$\mathrm{IsHomogeneous}\left(f\right)$

$g≔\mathrm{Homogenize}\left(f\,z\right)$

$\mathrm{IsHomogeneous}\left(g\right)$

$\mathrm{Homogenize}\left(f\,z\,\left[w\,x\right]\,\left[1\,2\right]\right)$

The simplify command has been enhanced with respect to logarithms whose arguments are constant radical expressions. Some examples:

$\mathrm{simplify}\left(\ln \left(1-\sqrt{2}\right)\+\ln \left(1\+\sqrt{2}\right)\right)$

$\mathrm{simplify}\left(\ln \left(\sqrt{2\+\sqrt{2}}\+\sqrt{-2\+\sqrt{2}}\right)\right)$

$\mathrm{simplify}\left(-7 \mathrm{\π} \ln \left(2-\sqrt{3}\right)+11\mathrm{\pi }\ln \left(2+\sqrt{3}\right)+38\ln \left(\sqrt{3}-1\right)\mathrm{\pi }+2\mathrm{\pi }\ln \left(1+\sqrt{3}\right)+4\ln \left(2\right)\mathrm{\pi }-2\sqrt{3}\ln {\left(2\right)}^2+\sqrt{3}{\ln \left(2-\sqrt{3}\right)}^2+\sqrt{3}{\ln \left(2+\sqrt{3}\right)}^2+4\sqrt{3}\ln \left(2\right)\ln \left(\sqrt{3}-1\right)+4\sqrt{3}\ln \left(2\right)\ln \left(1+\sqrt{3}\right)-4\sqrt{3}{\ln \left(\sqrt{3}-1\right)}^2-4\sqrt{3}{\ln \left(1+\sqrt{3}\right)}^2\right)$

simplify can now recognize more trig and exp simplifications:
$\mathrm{simplify}\left(-\cos \left(\frac{1}{12}\mathrm{Pi}\right)\+\sin \left(\frac{1}{12}\mathrm{Pi}\right)\+\frac{1}{2}\sqrt{2}\right)$

$\mathrm{simplify}\left(16\sin \left(\frac{4}{9}\mathrm{Pi}\right)-24\sin \left(\frac{1}{9}\mathrm{Pi}\right)-32\sin \left(\frac{2}{9}\mathrm{Pi}\right)\+8\sqrt{3}\cos \left(\frac{4}{9}\mathrm{Pi}\right)\+8\sqrt{3}\cos \left(\frac{2}{9}\mathrm{Pi}\right)\right)$

$\mathrm{simplify}\left({\ⅇ}^{\frac{\mathrm{I}}{5}\mathrm{\pi }}\left(\frac{\sqrt{5}}{4}+\frac{1}{4}-\frac{\mathrm{I}\sqrt{2}\sqrt{5-\sqrt{5}}}{4}\right)\right)$$$

simplify now tries harder to simplify constants that appear as subexpressions:
$\mathrm{simplify}\left(\left({\left(-1\right)}^{1\/5}-{\left(-1\right)}^{2\/5}\+{\left(-1\right)}^{3\/5}-{\left(-1\right)}^{4\/5}-1\right)x\right)$

The sum command with the parametric option has been improved for the case of hypergeometric sums with more than one parameter. The following sums used to return unevaluated in Maple 2017.

$\mathrm{sum}\left(\frac{x^n}{n\+b}\, n \= 0 .. \mathrm{infinity}\,\mathrm{parametric}\right)$

$\mathrm{sum}\left(\left(\genfrac{}{}{0ex}{}{i\+k-1}{i}\right) z^i\, i \= 0 .. \mathrm{infinity}\, \mathrm{parametric}\right)$

$\mathrm{sum}\left(\left(k+a\right)z^k\,k=0..\mathrm{\infty }\,\mathrm{parametric}\right)$

The commands sturm and sturmseq have been extended to support polynomials with real algebraic number coefficients.

$f≔\mathrm{expand}\left(\left(x-\sqrt{2}\right)\cdot \left(x\+1\right)\cdot \left(x\+2 \sqrt{2}\right)\right)$

$s≔\mathrm{sturmseq}\left(f\,x\right)$

$\mathrm{sturm}\left(s\,x\=-\infty ..0\right)$

$\mathrm{sturm}\left(s\,x\=0..5\right)$

The signum command has had various improvements made. The following examples previously returned with unevaluated signum calls:

$\mathrm{signum}~\left(\left[{\left(-1\right)}^x\,{\mathrm{I}}^x\right]\right) assuming x\colon\colon \mathrm{real}$

$\mathrm{signum}~\left(\left[{\mathrm{I}}^{2x}\,{\mathrm{I}}^{2x\+1}\right]\right) assuming x\colon\colon \mathrm{even}$

$\mathrm{signum}~\left(\left[{\mathrm{I}}^{2x}\,{\mathrm{I}}^{2x\+1}\right]\right) assuming x\colon\colon \mathrm{odd}$

$\mathrm{signum}\left(\sqrt{-\frac{6\left(\sqrt{\frac{{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{2\/3}\+48}{{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{1\/3}}}{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{2\/3}\+12\sqrt{6}{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{1\/3}\+48\sqrt{\frac{{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{2\/3}\+48}{{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{1\/3}}}\right)}{{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{1\/3}\sqrt{\frac{{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{2\/3}\+48}{{\left(108\+12\mathrm{I}\sqrt{687}\right)}^{1\/3}}}}}\right)$

The is and coulditbe commands have had various improvements made. is performs more simplification than it did previously:

$\mathrm{is}\left(\ⅇ{\ⅇ}^{-1}\,\mathrm{integer}\right)$

$\mathrm{is}\left(-\mathrm{I}{\mathrm{I}}^{2n\+1}\>0\right) assuming n\colon\colon \mathrm{even}$

and takes into account more function properties:

$\mathrm{is}\left( \mathrm{arccosh}\left(x\right)-\mathrm{I}\mathrm{\pi } \ge 0\right) assuming x<-1$

and operation properties:

$\mathrm{is}\left(z^n\&comma;\mathrm{imaginary}\right)\mathbf{assuming} z\colon\colon \mathrm{imaginary}\&comma;n\colon\colon \mathrm{integer}\&comma;\left(z^n\right)\colon\colon \mathrm{Non}\left(\mathrm{real}\right)$

The internal solver for systems of inequalities in is and coulditbe has been improved:

$\mathrm{map2}\left(\mathrm{coulditbe}\&comma; p\&comma; \left[-1\&comma; 1\right]\right) assuming -1 < p\&comma; -1 < q\&comma; -1 < r\&comma; -1 < p\&plus;q\&comma; -1 < p\&plus;r\&comma; -1 < r\&plus;q\&comma; -1 < p\&plus;q\&plus;r$

The Re and Im commands have had various improvements made. The following examples previously returned with unevaluated Re and Im calls:

$\mathrm{\Im }\left(\mathrm{z1}\right) assuming z\&gt;0\&comma; z<1\&comma; \mathrm{z1}\&equals;\mathrm{arccosh}\left(z\right)$

$\left[\mathrm{\Re }\&comma;\mathrm{\Im }\right]\left(\mathrm{I}\Gamma \left(\frac{3}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\right)\Gamma \left(\frac{3}{2}\&plus;\frac{1}{2}\mathrm{I}\sqrt{3}\right)\right)$

In some cases, the max and min commands can now recognize numbers as real even though they are composed from nonreal ingredients:

$\max \left(1\&comma;\frac{1}{2}{\left(-4\&plus;4\mathrm{I}\sqrt{3}\right)}^{1\&sol;3}\&plus;\frac{2}{{\left(-4\&plus;4\mathrm{I}\sqrt{3}\right)}^{1\&sol;3}}\right)$

$\mathrm{evalc}\left(\%\right)$

$\mathrm{evalf}\left(\%\right)\&semi;$

The arctan command now performs some more automatic simplifications:

$\arctan \left(\frac{b\&plus;\sqrt{a^2\&plus;b^2}}{a}\right) assuming a \&gt; 0\&comma; b\&gt;0$

$\arctan \left(t\&plus;\sqrt{t^2\&plus;1}\right) assuming t \&gt; 0$

$\arctan \left(\frac{-1\&plus;\sqrt{z^2\&plus;1}}{z}\right)$

Furthermore, combine performs more simplifications on arctan functions:
$\mathrm{combine}\left(\arctan \left(\frac{1}{z}\right)\&plus;\arctan \left(z\right)\right)$

$\mathrm{combine}\left(\mathrm{expand}\left(\frac{\arctan \left(b\right)}{\mathrm{Pi}}-\frac{2\arctan \left(b\&plus;\sqrt{b^2\&plus;1}\right)}{\mathrm{Pi}}\right)\right) assuming 0<b$

The SMTLIB package has been extended to support satisfiability queries on Boolean combinations of polynomial equations and inequalities.

Consider the following description of a set:

$\mathrm{sys}≔\left(0\le y^2+x\wedge \frac{2x}{3}-3<2y\wedge y<\frac{x}{3}+1\right)\vee \left(1<y\wedge 0<x+y-1\wedge x^3\le y\right)\vee \left(x<-2\wedge y<-2\right)\&colon;$We can use first use SMTLIB[Satisfiable] to verify that a solution exists:
$\mathrm{SMTLIB}\left[\mathrm{Satisfiable}\right]\left(\mathrm{sys}\right)$

In this simple two-dimensional case, we can use plots[inequal] to visualize the solution space:

$\mathrm{with}\left(\mathrm{plots}\right)\&colon;$$\mathrm{inequal}\left(\mathrm{sys}\&comma; x\&equals;-5..5\&comma; y\&equals;-5..5\right)$

The new command SMTLIB[Satisfy] offers an efficient method of finding a concrete example for a point in the solution space:
$\mathrm{SMTLIB}\left[\mathrm{Satisfy}\right]\left(\mathrm{sys}\right)$

To produce a satisfying point within the visual bounds of the plot above, we can simply augment our system with a bounding rectangle:
$\mathrm{pt} ≔ \mathrm{SMTLIB}\left[\mathrm{Satisfy}\right]\left(\mathrm{sys} \mathbf{and} x\&gt;-5 \mathbf{and} x<5 \mathbf{and} y\&gt;-5 \mathbf{and} y<5\right)$

$\mathrm{display}\left( \mathrm{inequal}\left(\mathrm{sys}\&comma; x\&equals;-5..5\&comma; y\&equals;-5..5\right)\&comma; \mathrm{pointplot}\left( \left[\mathrm{eval}\left(\left[x\&comma;y\right]\&comma;\mathrm{pt}\right)\right]\&comma; \mathrm{symbol}\&equals;\mathrm{solidcircle}\&comma; \mathrm{symbolsize}\&equals;25\&comma; \mathrm{color}\&equals;\mathrm{red} \right) \right)\&semi;$