Integrand size = 106, antiderivative size = 38 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\frac {x}{-i \pi +\frac {16-\frac {x^2}{e^4}}{4 x}-\log (-\log (2 \log (\log (5))))} \]
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Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.50, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6, 1607, 1694, 12, 790} \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=-\frac {16 e^8 (4-x (\log (-\log (2 \log (\log (5))))+i \pi ))}{x^2+4 e^4 x (\log (-\log (2 \log (\log (5))))+i \pi )-16 e^4} \]
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Rule 6
Rule 12
Rule 790
Rule 1607
Rule 1694
Rubi steps \begin{align*} \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 \\ & = \text {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 ) \text {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{align*}
Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\frac {16 e^8 (-4+i \pi x+x \log (-\log (2 \log (\log (5)))))}{x^2+4 e^4 (-4+i \pi x+x \log (-\log (2 \log (\log (5)))))} \]
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Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16
method | result | size |
gosper | \(\frac {16 \left (x \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )-4\right ) {\mathrm e}^{8}}{4 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} x -16 \,{\mathrm e}^{4}+x^{2}}\) | \(44\) |
risch | \(\frac {4 x \,{\mathrm e}^{8} \ln \left (\ln \left (2\right )+\ln \left (\ln \left (\ln \left (5\right )\right )\right )\right )-16 \,{\mathrm e}^{8}}{\ln \left (\ln \left (2\right )+\ln \left (\ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} x -4 \,{\mathrm e}^{4}+\frac {x^{2}}{4}}\) | \(44\) |
parallelrisch | \(\frac {16 x \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )-64 \,{\mathrm e}^{8}}{4 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} x -16 \,{\mathrm e}^{4}+x^{2}}\) | \(49\) |
norman | \(\frac {16 x \,{\mathrm e}^{8} \ln \left (\ln \left (2\right )+\ln \left (\ln \left (\ln \left (5\right )\right )\right )\right )-64 \,{\mathrm e}^{8}}{4 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} x -16 \,{\mathrm e}^{4}+x^{2}}\) | \(50\) |
default | \(-4 \,{\mathrm e}^{8} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+8 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} \textit {\_Z}^{3}+\left (16 \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )^{2}-32 \,{\mathrm e}^{4}\right ) \textit {\_Z}^{2}-128 \textit {\_Z} \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )+256 \,{\mathrm e}^{8}\right )}{\sum }\frac {\left (\ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) \textit {\_R}^{2}-8 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{8 \textit {\_R} \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )^{2}-32 \,{\mathrm e}^{8} \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right )+6 \ln \left (\ln \left (2 \ln \left (\ln \left (5\right )\right )\right )\right ) {\mathrm e}^{4} \textit {\_R}^{2}-16 \textit {\_R} \,{\mathrm e}^{4}+\textit {\_R}^{3}}\right )\) | \(139\) |
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Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\frac {16 \, {\left (x e^{8} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) - 4 \, e^{8}\right )}}{4 \, x e^{4} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) + x^{2} - 16 \, e^{4}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (27) = 54\).
Time = 1.00 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.92 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=- \frac {x \left (- 16 e^{8} \log {\left (- \log {\left (2 \right )} - \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )} \right )} - 16 i \pi e^{8}\right ) + 64 e^{8}}{x^{2} + x \left (4 e^{4} \log {\left (- \log {\left (2 \right )} - \log {\left (\log {\left (\log {\left (5 \right )} \right )} \right )} \right )} + 4 i \pi e^{4}\right ) - 16 e^{4}} \]
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Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\frac {16 \, {\left (x e^{8} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) - 4 \, e^{8}\right )}}{4 \, x e^{4} \log \left (\log \left (2 \, \log \left (\log \left (5\right )\right )\right )\right ) + x^{2} - 16 \, e^{4}} \]
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Timed out. \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=\text {Timed out} \]
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Time = 0.83 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53 \[ \int \frac {128 e^8 x-16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))}{256 e^8-32 e^4 x^2+x^4+\left (-128 e^8 x+8 e^4 x^3\right ) (i \pi +\log (-\log (2 \log (\log (5)))))+16 e^8 x^2 (i \pi +\log (-\log (2 \log (\log (5)))))^2} \, dx=-\frac {64\,{\mathrm {e}}^8-16\,x\,{\mathrm {e}}^8\,\left (\ln \left (-\ln \left (2\,\ln \left (\ln \left (5\right )\right )\right )\right )+\pi \,1{}\mathrm {i}\right )}{x^2+\left (4\,{\mathrm {e}}^4\,\ln \left (-\ln \left (2\,\ln \left (\ln \left (5\right )\right )\right )\right )+\pi \,{\mathrm {e}}^4\,4{}\mathrm {i}\right )\,x-16\,{\mathrm {e}}^4} \]
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