There have been various improvements made to the int command for Maple 2019.

New results from int:

${\int }_0^{\mathrm{\pi }}\left|\sin \left(x^2\right)\right|\ⅆx$

$-\frac{\mathrm{FresnelS}\left(\sqrt{2}\sqrt{\mathrm{\pi }}\right)\sqrt{2}\sqrt{\mathrm{\pi }}}{2}+\mathrm{FresnelS}\left(\sqrt{2}\right)\sqrt{2}\sqrt{\mathrm{\pi }}-\mathrm{FresnelS}\left(2\right)\sqrt{2}\sqrt{\mathrm{\pi }}+\mathrm{FresnelS}\left(\sqrt{3}\sqrt{2}\right)\sqrt{2}\sqrt{\mathrm{\pi }}$

${\int }_0^{\mathrm{\pi }}{\ⅇ}^{\mathrm{I}\sin \left(x\right)}\ⅆx$

$\mathrm{\pi }\mathrm{BesselJ}\left(0\,1\right)+\mathrm{I}\mathrm{\pi }\mathrm{StruveH}\left(0\,1\right)$

${\int }_0^{\frac{\mathrm{\pi }}{2}}{\sec \left(u\right)}^2{\ⅇ}^{-\sec \left(u\right)}\ⅆu$

$\mathrm{BesselK}\left(1\,1\right)$

${\int }_0^1\mathrm{signum}\left(\arccos \left(\frac{1}{z}\right)\right)\ⅆz$

$\mathrm{I}$

${\int }_0^{\mathrm{\pi }}\left|\cos \left(\left(p-1\right)x\right)-\cos \left(\left(p+1\right)x\right)\right|\ⅆx\mathbf{assuming} p>1$

$-\frac{2\left(\sin \left(\mathrm{\pi }p\right)\mathrm{signum}\left(\sin \left(\mathrm{\pi }p\right)\right)\cos \left(\frac{\mathrm{\pi }}{p}\right)-\sin \left(\mathrm{\pi }p\right)\mathrm{signum}\left(\sin \left(\mathrm{\pi }p\right)\right)-\sin \left(\frac{\mathrm{\pi }\left(⌊p⌋+1\right)}{p}\right)p+\sin \left(\frac{\mathrm{\pi }}{p}\right)p+p\sin \left(\frac{\mathrm{\pi }⌊p⌋}{p}\right)\right)}{\cos \left(\frac{\mathrm{\pi }}{p}\right)p^2-p^2-\cos \left(\frac{\mathrm{\pi }}{p}\right)+1}$

${\int }_0^{\frac{\mathrm{\pi }}{2}}{\left(\csc \left(x\right)-1\right)}^{\mathrm{\nu }}\cos \left(x\right)\ⅆx$

$\frac{\mathrm{\pi }\mathrm{\nu }}{\sin \left(\mathrm{\nu }\mathrm{\pi }\right)}$

${\int }_0^2\left(x^2-\frac{1}{2}+\frac{\mathrm{I}\ln \left(-{\ⅇ}^{2\mathrm{I}\mathrm{\pi }{x}^2}\right)}{2\mathrm{\pi }}\right)\ⅆx$

$5-\sqrt{2}-\sqrt{3}$

Improved answers for definite integrals when the AllSolutions option is given:

$\mathrm{int}\left(\mathrm{round}\left(t\right)\,t\=0..x\,\mathrm{AllSolutions}\right) \#\mathrm{Similar\; improvements\; with\; ceil,\; floor,\; frac,\; trunc\; instead\; of\; round.}$

$\frac{\left(\left\{\begin{array}{cc}-{⌈x-\frac{1}{2}⌉}^2+x\left(\left\{\begin{array}{cc}2x-1& \left(x+\frac{1}{2}\right)::\mathbb{Z}\\ 2⌊x⌉& \mathrm{otherwise}\end{array}\right.\right)& 0<x\\ -{⌊x+\frac{1}{2}⌋}^2+x\left(\left\{\begin{array}{cc}2x+1& \left(x-\frac{1}{2}\right)::\mathbb{Z}\\ 2⌊x⌉& \mathrm{otherwise}\end{array}\right.\right)& x\le 0\end{array}\right.\right)}{2}$

$\mathrm{int}\left(\frac{1}{2\&plus;\cos \left(t\right)}\&comma;t\&equals;0..x\&comma;\mathrm{AllSolutions}\right)$

$\frac{\left(\left\{\begin{array}{cc}2\mathrm{\pi }⌈-\frac{\mathrm{\pi }-x}{2\mathrm{\pi }}⌉+\left(\left\{\begin{array}{cc}\mathrm{\pi }& \left(-\frac{\mathrm{\pi }-x}{2\mathrm{\pi }}\right)::\mathbb{Z}\\ 2\arctan \left(\frac{\tan \left(\frac{x}{2}\right)\sqrt{3}}{3}\right)& \mathrm{otherwise}\end{array}\right.\right)& 0<x\\ 2\mathrm{\pi }⌊-\frac{\mathrm{\pi }-x}{2\mathrm{\pi }}⌋+2\mathrm{\pi }-\left(\left\{\begin{array}{cc}\mathrm{\pi }& \left(-\frac{\mathrm{\pi }-x}{2\mathrm{\pi }}\right)::\mathbb{Z}\\ -2\arctan \left(\frac{\tan \left(\frac{x}{2}\right)\sqrt{3}}{3}\right)& \mathrm{otherwise}\end{array}\right.\right)& x\le 0\end{array}\right.\right)\sqrt{3}}{3}$

$\mathrm{int}\left(\frac{1}{\ln \left(t\right)}\&comma;t\&equals;0..x\&comma;\mathrm{AllSolutions}\&comma;\mathrm{CauchyPrincipalValue}\right)$

$\left\{\begin{array}{cc}-\mathrm{I}\mathrm{\pi }+\mathrm{Ei}\left(\ln \left(-x\right)+\mathrm{I}\mathrm{\pi }\right)& x<0\\ 0& x=0\\ \mathrm{Ei}\left(\ln \left(x\right)\right)& x<1\\ -\mathrm{\infty }& x=1\\ \mathrm{Ei}\left(\ln \left(x\right)\right)& 1<x\end{array}\right.$

The inttrans package in Maple 2019 has had several transforms, specifically laplace, invlaplace, fourier and invfourier, extended to handle a larger class of problems, and in some cases already handled classes of problems faster. This has been accomplished via an integration by differentiation approach described in the following:
- A. Kempf, D.M. Jackson and A.H. Morales, "New Dirac delta function based methods with applications to perturbative expansions in quantum field theory", J. Phys. A:47, 2014
- D. Jia, E. Tang, and A. Kempf, "Integration by differentiation: new proofs, methods and examples", J. Phys. A:50, 2017

One can view this approach, in simplest possible terms, as a product rule.

Here are a few examples which failed to transform in prior versions of Maple, but now transform quite rapidly:

$\mathrm{ex1}≔\frac{\frac{{\&ExponentialE;}^{-\frac{t}{2}}}{1+{\&ExponentialE;}^{-t}}\sin \left(t\right)}{t}$

$\mathrm{ex1}≔\frac{{\&ExponentialE;}^{-\frac{t}{2}}\sin \left(t\right)}{\left(1+{\&ExponentialE;}^{-t}\right)t}$

$\mathrm{inttrans}\left[\mathrm{fourier}\right]\left(\mathrm{ex1}\&comma;t\&comma;s\right)$

$-\frac{\mathrm{I}}{2}\ln \left(\frac{\csc \left(\mathrm{I}\mathrm{\pi }s-\left(\frac{1}{2}+\mathrm{I}\right)\mathrm{\pi }\right)+\cot \left(\mathrm{I}\mathrm{\pi }s-\left(\frac{1}{2}+\mathrm{I}\right)\mathrm{\pi }\right)}{\csc \left(\mathrm{I}\mathrm{\pi }s+\left(-\frac{1}{2}+\mathrm{I}\right)\mathrm{\pi }\right)+\cot \left(\mathrm{I}\mathrm{\pi }s+\left(-\frac{1}{2}+\mathrm{I}\right)\mathrm{\pi }\right)}\right)$

$\mathrm{ex2}≔{\&ExponentialE;}^{\frac{t}{2}}\ln \left(1+{\&ExponentialE;}^{-t}\right)\cos \left(t\right)$

$\mathrm{ex2}≔{\&ExponentialE;}^{\frac{t}{2}}\ln \left(1+{\&ExponentialE;}^{-t}\right)\cos \left(t\right)$

$\mathrm{inttrans}\left[\mathrm{fourier}\right]\left(\mathrm{ex2}\&comma;t\&comma;s\right)$

$\mathrm{\pi }\left(\frac{\csc \left(\mathrm{\pi }\left(-\frac{1}{2}+\mathrm{I}\left(s-1\right)\right)\right)}{2\mathrm{I}s-1-2\mathrm{I}}+\frac{\csc \left(\mathrm{\pi }\left(-\frac{1}{2}+\mathrm{I}\left(s+1\right)\right)\right)}{2\mathrm{I}s-1+2\mathrm{I}}\right)$

$\mathrm{ex3}≔\frac{{\&ExponentialE;}^{-t}\mathrm{Heaviside}\left(t\right)+{\&ExponentialE;}^t\mathrm{Heaviside}\left(-t\right)}{2\left(t+\mathrm{I}\right)}$

$\mathrm{ex3}≔\frac{{\&ExponentialE;}^{-t}\mathrm{Heaviside}\left(t\right)+{\&ExponentialE;}^t\mathrm{Heaviside}\left(-t\right)}{2t+2\mathrm{I}}$

$\mathrm{inttrans}\left[\mathrm{fourier}\right]\left(\mathrm{ex3}\&comma;t\&comma;s\right)$

$\frac{{\&ExponentialE;}^{-s}\left(-{\mathrm{Ei}}_1\left(-\mathrm{I}-s\right){\&ExponentialE;}^{\mathrm{-I}}+{\mathrm{Ei}}_1\left(\mathrm{I}-s\right){\&ExponentialE;}^{\mathrm{I}}\right)}{2}$

$\mathrm{ex4}≔\frac{\mathrm{csch}\left(t\right)\sin \left(t\right)}{t}$

$\mathrm{ex4}≔\frac{\mathrm{csch}\left(t\right)\sin \left(t\right)}{t}$

$\mathrm{inttrans}\left[\mathrm{fourier}\right]\left(\mathrm{ex4}\&comma;t\&comma;s\right)$

$\mathrm{-I}\left(-2\mathrm{\pi }+\ln \left(\frac{{\&ExponentialE;}^{\mathrm{\pi }\left(s+1\right)}+1}{{\&ExponentialE;}^{\mathrm{\pi }\left(s-1\right)}+1}\right)\right)$

The simplify command in Maple 2019 has undergone several improvements, especially with regard to expressions containing piecewise functions.

Simplification of expressions containing piecewise functions has been improved.

$\mathrm{simplify}\left(\left\{\begin{array}{cc}5& x=1\\ 5& x=2\\ 6& \mathrm{otherwise}\end{array}\right.\right)$

$\left\{\begin{array}{cc}5& x=1\vee x=2\\ 6& \mathrm{otherwise}\end{array}\right.$

$\mathrm{simplify}\left(\left\{\begin{array}{cc}0& \frac{\mathrm{\pi }}{2}=T\\ \cos \left(T\right)& \left(\frac{2T-\mathrm{\pi }}{2\mathrm{\pi }}\right)\colon\colon \mathrm{integer}\\ 1& \mathrm{otherwise}\end{array}\right.\right)$

$\left\{\begin{array}{cc}\cos \left(T\right)& \left(-\frac{-2T+\mathrm{\pi }}{2\mathrm{\pi }}\right)::\mathbb{Z}\\ 1& \mathrm{otherwise}\end{array}\right.$

$\mathrm{simplify}\left(\left\{\begin{array}{cc}0& a^2+b^2=0\\ \sqrt{\frac{a^2+b^2}{a^2}}& \mathrm{otherwise}\end{array}\right.\right)$

$\sqrt{\frac{a^2+b^2}{a^2}}$

$\mathrm{simplify}\left(\left\{\begin{array}{cc}⌊\frac{m}{2}⌋& m < 2\\ 0& \mathrm{otherwise}\end{array}\right.\right)\mathbf{assuming} m\colon\colon \mathrm{posint}$

$0$

$\mathrm{simplify}\&lpar;\&lcub;\begin{array}{cc}f\left(x\right)& \mathrm{round}\left(x\right)\&gt;\frac{1}{2}\\ g\left(x\right)& \mathrm{otherwise}\end{array}\&rpar;$

$\left\{\begin{array}{cc}g\left(x\right)& x<\frac{1}{2}\\ f\left(x\right)& \frac{1}{2}\le x\end{array}\right.$

$\mathrm{simplify}\left(\left\{\begin{array}{cc}⌊x⌋& x\colon\colon \mathrm{integer}\\ x^2& \mathrm{otherwise}\end{array}\right.\right)$

$\left\{\begin{array}{cc}x& x::\mathbb{Z}\\ x^2& \mathrm{otherwise}\end{array}\right.$

$\mathrm{simplify}\left(\left\{\begin{array}{cc}{\left(x^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}& x\colon\colon \mathrm{posint}\\ x& \mathrm{otherwise}\end{array}\right.\right)$

$x$

$\mathrm{simplify}\left(\left\{\begin{array}{cc}0& a=1\vee \left(a=1\wedge b\colon\colon \mathrm{integer}\right)\\ 1& \mathrm{otherwise}\end{array}\right.\right)$

$\left\{\begin{array}{cc}0& a=1\\ 1& \mathrm{otherwise}\end{array}\right.$

$\mathrm{simplify}\left(\left\{\begin{array}{cc}f\left(x\right)& x^2=y^2\vee x^2<y^2\\ g\left(x\right)& \mathrm{otherwise}\end{array}\right.\right)$

$\left\{\begin{array}{cc}f\left(x\right)& x^2\le y^2\\ g\left(x\right)& \mathrm{otherwise}\end{array}\right.$

$\mathrm{simplify}\left(\left(\left\{\begin{array}{cc}4& a=b\\ 5& a\ne b\end{array}\right.\right)-\left(\left\{\begin{array}{cc}5& a\ne b\\ 4& a=b\end{array}\right.\right)\right)$

$0$

$\mathrm{simplify}\left(\left(\left\{\begin{array}{cc}f\&lpar;x\&rpar;& x\ne \frac{t}{2}\\ 0& \mathrm{otherwise}\end{array}\right.\right)-\left(\left\{\begin{array}{cc}f\&lpar;x\&rpar;& 2 x-t\ne 0\\ 0& \mathrm{otherwise}\end{array}\right.\right)\right)$

$0$

$\mathrm{simplify}\left(\left(\left\{\begin{array}{cc}y& z-y<\frac{1}{2}\\ y+1& \mathrm{otherwise}\end{array}\right.\right)-\left(\left\{\begin{array}{cc}2y& z-y<\frac{1}{2}\\ 2y+1& \mathrm{otherwise}\end{array}\right.\right)\right)$

$-y$

$\mathrm{simplify}\left(\left\{\begin{array}{cc}-4\mathrm{\pi }{\mathrm{\_C8}}_{\mathrm{\_Z1},\mathrm{\_Z2}}{\mathrm{BesselJ}\left(\mathrm{\_Z1}\&comma;\mathrm{BesselJZeros}\left(\mathrm{\_Z1}\&comma;\mathrm{\_Z2}\right)r\right)}^{2}r& x=0\\ -2\mathrm{\pi }{\mathrm{\_C8}}_{\mathrm{\_Z1},\mathrm{\_Z2}}{\mathrm{BesselJ}\left(\mathrm{\_Z1}\&comma;\mathrm{BesselJZeros}\left(\mathrm{\_Z1}\&comma;\mathrm{\_Z2}\right)r\right)}^{2}r& \mathrm{otherwise}\end{array}\right.\right)$

$\left(\left\{\begin{array}{cc}\mathrm{-4}& \mathrm{\_Z1}=0\\ \mathrm{-2}& \mathrm{otherwise}\end{array}\right.\right)\mathrm{\pi }{\mathrm{\_C8}}_{\mathrm{\_Z1},\mathrm{\_Z2}}{\mathrm{BesselJ}\left(\mathrm{\_Z1}\&comma;\mathrm{BesselJZeros}\left(\mathrm{\_Z1}\&comma;\mathrm{\_Z2}\right)r\right)}^{2}r$

Nonpiecewise-related improvements made to simplify:

$\mathrm{simplify}\left(\frac{1}{\mathrm{\Gamma }\left(1-n\right)\mathrm{\Gamma }\left(n\right)}\right) assuming n\colon\colon \mathrm{integer}$

$0$

$\mathrm{simplify}\left(\sec \left(z\right)+\mathrm{I}\left(\mathrm{I}\tan \left(\frac{\mathrm{\pi }}{4}+\frac{z}{2}\right)-\mathrm{I}\tan \left(z\right)\right)\right)$

$0$

$\mathrm{simplify}\left(\sin \left(\frac{\arctan \left(\frac{\sqrt{111}}{9}\right)}{3}+\frac{\mathrm{\pi }}{6}\right)-\cos \left(\frac{\arctan \left(\frac{3\sqrt{111}}{5}\right)}{6}+\frac{\mathrm{\pi }}{6}\right)\&comma;\mathrm{arctrig}\right)$

$0$

$\mathrm{simplify}\left(-\sqrt{y^2}\mathrm{csgn}\left(y\right)+y\&comma;\mathrm{csgn}\right)$

$0$

$\mathrm{simplify}\left(\cos \left(x\right)-\frac{{\&ExponentialE;}^{\mathrm{I}x}}{2}\right)$

$\frac{{\&ExponentialE;}^{\mathrm{-I}x}}{2}$

$\mathrm{simplify}\left(\sinh \left(x\right)-\cosh \left(x\right)+{\&ExponentialE;}^{-x}\right)$

$0$

$\mathrm{simplify}\left(\left(-3\sin \left(x\right)+9\mathrm{I}\right){\&ExponentialE;}^{-\frac{\mathrm{I}}{3}\sin \left(x\right)}+\left(-9\mathrm{I}-3\sin \left(x\right)\right){\&ExponentialE;}^{\frac{\mathrm{I}}{3}\sin \left(x\right)}+6\sin \left(x\right)\cos \left(\frac{\sin \left(x\right)}{3}\right)-18\sin \left(\frac{\sin \left(x\right)}{3}\right)\right)$

$0$

$\mathrm{simplify}\left(-{\left(z+1\right)}^{\frac{b}{2}+\frac{1}{2}}{\&ExponentialE;}^{\left(b+1\right)\mathrm{\pi }\mathrm{I}}\left(z^{-a-b-1}-z^{-a-b+1}\right){\left(z-1\right)}^{\frac{b}{2}+\frac{1}{2}}+{\left(z-1\right)}^{\frac{3}{2}+\frac{b}{2}}{\left(z+1\right)}^{\frac{3}{2}+\frac{b}{2}}{\&ExponentialE;}^{\left(b+2\right)\mathrm{\pi }\mathrm{I}}{z}^{-a-b-1}\right)$

$0$

$\mathrm{simplify}\left({\left(\mathrm{-1}\right)}^{2k}\left(4^k{25}^k-{100}^k\right)\right)$

$0$

$\mathrm{simplify}\left({\left(\mathrm{-1}\right)}^{\mathrm{\_k2}}{4}^{\mathrm{\_k2}}{2}^{-\mathrm{\nu }}{16}^{-\mathrm{\_k2}}\right)$

$2^{-2\mathrm{\_k2}-\mathrm{\nu }}{\left(\mathrm{-1}\right)}^{\mathrm{\_k2}}$

$\mathrm{simplify}\left(\sqrt{3}\sqrt{2}\right)$

$\sqrt{6}$

$\mathrm{simplify}\left(⌊z+\frac{1}{2}⌋+⌊z⌋\right)\mathbf{assuming} z\colon\colon \mathrm{real}$

$⌊2z⌋$

$\mathrm{simplify}\left(⌊z⌉-⌊2z⌋\right)\mathbf{assuming}\left(z-\frac{1}{2}\right)::\left(\mathrm{real}\wedge \neg \mathrm{negint}\right)$

$-⌊z⌋$

$\mathrm{simplify}\left(\mathrm{EllipticE}\left({\left(\mathrm{-1}\right)}^n\&comma;x\right)\right)\mathbf{assuming} n\colon\colon \mathrm{integer}$

${\left(\mathrm{-1}\right)}^n\mathrm{EllipticE}\left(x\right)$

The solve command in Maple 2019 has undergone several improvements.

$\mathrm{solve}\left(\sqrt{x}\le 0\right)$

$0$

$\mathrm{solve}\left(-\frac{1}{\sqrt{x}}\le 2\right)$

$\left(0\&comma;\mathrm{\infty }\right)$

$\left[\mathrm{solve}\left(\sqrt{x}\le -y^2\&comma;\left\{x\&comma;y\right\}\right)\right] \&semi;$

$\left[\left\{x=0\&comma;y=0\right\}\right]$

$\mathrm{solve}\left(\left\{\frac{x\sqrt{x+5}}{\sqrt{x+6}}<0\right\}\&comma;\left\{x\right\}\right)\&semi;$

$\left\{x<\mathrm{-6}\right\},\left\{\mathrm{-5}<x\&comma;x<0\right\}$

There are other commands which have improved.

$\mathrm{minimize}\left(\frac{x^4+y^4+z^4}{x^2+y^2+z^2}\&comma;x=\mathrm{-1}..1\&comma;y=\mathrm{-1}..1\&comma;z=\mathrm{-1}..1\right)\&semi;$

$0$

$\mathrm{expand}\left({\left(\frac{c}{c-1}\right)}^m{\left(\frac{c-1}{c}\right)}^m{c}^m-{\left(\frac{c-1}{c}\right)}^m{c}^m+{\left(c-1\right)}^m-c^m\right)\mathbf{assuming}m\colon\colon \&apos;\mathrm{integer}\&apos;\&semi;$

$0$

$⌊w⌋\mathbf{assuming}w\ge 0,w<\frac{1}{2}$

$0$

$\mathrm{rationalize}\left(\frac{1}{{\left(\frac{y}{x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)\&semi;$

$\frac{{\left(\frac{y}{x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{x}^{\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{y^3}$

$\mathrm{\Re }\left({\mathrm{\Re }\left(x\right)}^2+\mathrm{\Im }\left(x^2\right)\right)\&semi;$

${\mathrm{\Re }\left(x\right)}^2+\mathrm{\Im }\left(x^2\right)$