Optimal. Leaf size=38 \[ \frac {x}{-i \pi +\frac {16-\frac {x^2}{e^4}}{4 x}-\log (-\log (2 \log (\log (5))))} \]
________________________________________________________________________________________
Rubi [A] time = 0.24, antiderivative size = 57, normalized size of antiderivative = 1.50, number of steps used = 5, number of rules used = 5, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6, 1593, 1680, 12, 776} \begin {gather*} -\frac {16 e^8 (4-x (\log (-\log (2 \log (\log (5))))+i \pi ))}{x^2+4 e^4 x (\log (-\log (2 \log (\log (5))))+i \pi )-16 e^4} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 6
Rule 12
Rule 776
Rule 1593
Rule 1680
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+x^2 \left (-32 e^4+16 e^8 (i \pi +\log (-\log (2 \log (\log (5)))))^2\right )} \, dx\\ &=\int \frac {x \left (128 e^8-16 e^8 x (i \pi +\log (-\log (2 \log (\log (5)))))\right )}{256 e^8+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+x^2 \left (-32 e^4+16 e^8 (i \pi +\log (-\log (2 \log (\log (5)))))^2\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {16 e^8 \left (x-2 e^4 (i \pi +\log (-\log (2 \log (\log (5)))))\right ) \left (2 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )-x (i \pi +\log (-\log (2 \log (\log (5)))))\right )}{\left (16 e^4-x^2-4 e^8 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )^2} \, dx,x,x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )\\ &=\left (16 e^8\right ) \operatorname {Subst}\left (\int \frac {\left (x-2 e^4 (i \pi +\log (-\log (2 \log (\log (5)))))\right ) \left (2 \left (4-e^4 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )-x (i \pi +\log (-\log (2 \log (\log (5)))))\right )}{\left (16 e^4-x^2-4 e^8 (\pi -i \log (-\log (2 \log (\log (5)))))^2\right )^2} \, dx,x,x+\frac {1}{4} \left (8 i e^4 \pi +8 e^4 \log (-\log (2 \log (\log (5))))\right )\right )\\ &=\frac {16 e^8 (4-x (i \pi +\log (-\log (2 \log (\log (5))))))}{16 e^4-x^2-4 e^4 x (i \pi +\log (-\log (2 \log (\log (5)))))}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.05, size = 54, normalized size = 1.42 \begin {gather*} \frac {16 e^8 (-4+i \pi x+x \log (-\log (2 \log (\log (5)))))}{x^2+4 e^4 (-4+i \pi x+x \log (-\log (2 \log (\log (5)))))} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.51, size = 40, normalized size = 1.05 \begin {gather*} \frac {16 \, {\left (x e^{8} \log \left (\log \left (2 \, \log \left (\log \relax (5)\right )\right )\right ) - 4 \, e^{8}\right )}}{4 \, x e^{4} \log \left (\log \left (2 \, \log \left (\log \relax (5)\right )\right )\right ) + x^{2} - 16 \, e^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.34, size = 44, normalized size = 1.16
method | result | size |
gosper | \(\frac {16 \left (\ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) x -4\right ) {\mathrm e}^{8}}{4 \,{\mathrm e}^{4} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) x -16 \,{\mathrm e}^{4}+x^{2}}\) | \(44\) |
risch | \(\frac {4 x \,{\mathrm e}^{8} \ln \left (\ln \relax (2)+\ln \left (\ln \left (\ln \relax (5)\right )\right )\right )-16 \,{\mathrm e}^{8}}{{\mathrm e}^{4} \ln \left (\ln \relax (2)+\ln \left (\ln \left (\ln \relax (5)\right )\right )\right ) x -4 \,{\mathrm e}^{4}+\frac {x^{2}}{4}}\) | \(44\) |
norman | \(\frac {16 x \,{\mathrm e}^{8} \ln \left (\ln \relax (2)+\ln \left (\ln \left (\ln \relax (5)\right )\right )\right )-64 \,{\mathrm e}^{8}}{4 \,{\mathrm e}^{4} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) x -16 \,{\mathrm e}^{4}+x^{2}}\) | \(50\) |
default | \(-4 \,{\mathrm e}^{8} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+8 \,{\mathrm e}^{4} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) \textit {\_Z}^{3}+\left (16 \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right )^{2}-32 \,{\mathrm e}^{4}\right ) \textit {\_Z}^{2}-128 \textit {\_Z} \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right )+256 \,{\mathrm e}^{8}\right )}{\sum }\frac {\left (\ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) \textit {\_R}^{2}-8 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{8 \textit {\_R} \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right )^{2}-32 \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right )+6 \,{\mathrm e}^{4} \ln \left (\ln \left (2 \ln \left (\ln \relax (5)\right )\right )\right ) \textit {\_R}^{2}-16 \textit {\_R} \,{\mathrm e}^{4}+\textit {\_R}^{3}}\right )\) | \(139\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.37, size = 40, normalized size = 1.05 \begin {gather*} \frac {16 \, {\left (x e^{8} \log \left (\log \left (2 \, \log \left (\log \relax (5)\right )\right )\right ) - 4 \, e^{8}\right )}}{4 \, x e^{4} \log \left (\log \left (2 \, \log \left (\log \relax (5)\right )\right )\right ) + x^{2} - 16 \, e^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 3.85, size = 58, normalized size = 1.53 \begin {gather*} -\frac {64\,{\mathrm {e}}^8-16\,x\,{\mathrm {e}}^8\,\left (\ln \left (-\ln \left (2\,\ln \left (\ln \relax (5)\right )\right )\right )+\pi \,1{}\mathrm {i}\right )}{x^2+\left (4\,{\mathrm {e}}^4\,\ln \left (-\ln \left (2\,\ln \left (\ln \relax (5)\right )\right )\right )+\pi \,{\mathrm {e}}^4\,4{}\mathrm {i}\right )\,x-16\,{\mathrm {e}}^4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [B] time = 1.46, size = 73, normalized size = 1.92 \begin {gather*} - \frac {x \left (16 e^{8} \log {\left (- \log {\relax (2 )} - \log {\left (\log {\left (\log {\relax (5 )} \right )} \right )} \right )} + 16 i \pi e^{8}\right ) - 64 e^{8}}{- x^{2} + x \left (- 4 e^{4} \log {\left (- \log {\relax (2 )} - \log {\left (\log {\left (\log {\relax (5 )} \right )} \right )} \right )} - 4 i \pi e^{4}\right ) + 16 e^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________