Integrand size = 92, antiderivative size = 32 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {2+\frac {x}{2}+\log (5)}{4 \left (-4+e^{1-x}+\frac {\log \left (x^2\right )}{4}\right )} \]
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\[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (4 e x (5+x+\log (25))-2 e^x (4+9 x+\log (25))+e^x x \log \left (x^2\right )\right )}{2 x \left (4 \left (e-4 e^x\right )+e^x \log \left (x^2\right )\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {e^x \left (4 e x (5+x+\log (25))-2 e^x (4+9 x+\log (25))+e^x x \log \left (x^2\right )\right )}{x \left (4 \left (e-4 e^x\right )+e^x \log \left (x^2\right )\right )^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {e^x \left (18 x+8 \left (1+\frac {\log (5)}{2}\right )-x \log \left (x^2\right )\right )}{x \left (16-\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )}+\frac {4 e^{1+x} (4+x+\log (25)) \left (2-16 x+x \log \left (x^2\right )\right )}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^x \left (18 x+8 \left (1+\frac {\log (5)}{2}\right )-x \log \left (x^2\right )\right )}{x \left (16-\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+2 \int \frac {e^{1+x} (4+x+\log (25)) \left (2-16 x+x \log \left (x^2\right )\right )}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx \\ & = \frac {1}{2} \int \left (-\frac {18 e^x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )}-\frac {4 e^x (2+\log (5))}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )}+\frac {e^x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )}\right ) \, dx+2 \int \left (\frac {e^{1+x} \left (2-16 x+x \log \left (x^2\right )\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}+\frac {e^{1+x} (4+\log (25)) \left (2-16 x+x \log \left (x^2\right )\right )}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+2 \int \frac {e^{1+x} \left (2-16 x+x \log \left (x^2\right )\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx-9 \int \frac {e^x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx-(2 (2+\log (5))) \int \frac {e^x}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+(2 (4+\log (25))) \int \frac {e^{1+x} \left (2-16 x+x \log \left (x^2\right )\right )}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {e^x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+2 \int \left (\frac {2 e^{1+x}}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}-\frac {16 e^{1+x} x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}+\frac {e^{1+x} x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}\right ) \, dx-9 \int \frac {e^x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx-(2 (2+\log (5))) \int \frac {e^x}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+(2 (4+\log (25))) \int \left (-\frac {16 e^{1+x}}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}+\frac {2 e^{1+x}}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}+\frac {e^{1+x} \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+2 \int \frac {e^{1+x} x \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx+4 \int \frac {e^{1+x}}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx-9 \int \frac {e^x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx-32 \int \frac {e^{1+x} x}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx-(2 (2+\log (5))) \int \frac {e^x}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \, dx+(2 (4+\log (25))) \int \frac {e^{1+x} \log \left (x^2\right )}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx+(4 (4+\log (25))) \int \frac {e^{1+x}}{x \left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx-(32 (4+\log (25))) \int \frac {e^{1+x}}{\left (-16+\log \left (x^2\right )\right ) \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )^2} \, dx \\ \end{align*}
Time = 5.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {e^x (4+x+\log (25))}{2 \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \]
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Time = 0.41 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {16+8 \ln \left (5\right )+4 x}{8 \ln \left (x^{2}\right )+32 \,{\mathrm e}^{1-x}-128}\) | \(28\) |
risch | \(\frac {i \left (2 \ln \left (5\right )+x +4\right )}{\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+8 i {\mathrm e}^{1-x}+4 i \ln \left (x \right )-32 i}\) | \(74\) |
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Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {x + 2 \, \log \left (5\right ) + 4}{2 \, {\left (4 \, e^{\left (-x + 1\right )} + \log \left (x^{2}\right ) - 16\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {x + 2 \log {\left (5 \right )} + 4}{8 e^{1 - x} + 2 \log {\left (x^{2} \right )} - 32} \]
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Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {{\left (x + 2 \, \log \left (5\right ) + 4\right )} e^{x}}{4 \, {\left ({\left (\log \left (x\right ) - 8\right )} e^{x} + 2 \, e\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {x + 2 \, \log \left (5\right ) + 4}{2 \, {\left (4 \, e^{\left (-x + 1\right )} + \log \left (x^{2}\right ) - 16\right )}} \]
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Timed out. \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=-\int \frac {18\,x+4\,\ln \left (5\right )-x\,\ln \left (x^2\right )-{\mathrm {e}}^{1-x}\,\left (20\,x+8\,x\,\ln \left (5\right )+4\,x^2\right )+8}{2\,x\,{\ln \left (x^2\right )}^2+\left (16\,x\,{\mathrm {e}}^{1-x}-64\,x\right )\,\ln \left (x^2\right )+512\,x-256\,x\,{\mathrm {e}}^{1-x}+32\,x\,{\mathrm {e}}^{2-2\,x}} \,d x \]
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