Integrand size = 183, antiderivative size = 34 \[ \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx=\frac {e^{-e^x} x}{-\frac {x^2}{4}+\frac {\log (-2+x-x (3+\log (5)))}{x}} \]
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\[ \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx=\int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+x^7 (2+\log (5))+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx \\ & = \int \frac {4 e^{-e^x} x \left (2 x^3-4 x (2+\log (5))+e^x x^5 (2+\log (5))+x^4 \left (2+2 e^x+\log (5)\right )-4 \left (-2+e^x x\right ) (2+x (2+\log (5))) \log (-2-x (2+\log (5)))\right )}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx \\ & = 4 \int \frac {e^{-e^x} x \left (2 x^3-4 x (2+\log (5))+e^x x^5 (2+\log (5))+x^4 \left (2+2 e^x+\log (5)\right )-4 \left (-2+e^x x\right ) (2+x (2+\log (5))) \log (-2-x (2+\log (5)))\right )}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx \\ & = 4 \int \left (\frac {2 e^{-e^x} x^4}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {4 e^{-e^x} x^2 (-2-\log (5))}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {2 e^{-e^x} x^5 \left (1+\frac {\log (5)}{2}\right )}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {e^{-e^x+x} x^2}{x^3-4 \log (-2-x (2+\log (5)))}+\frac {8 e^{-e^x} x \log (-2-x (2+\log (5)))}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}\right ) \, dx \\ & = 4 \int \frac {e^{-e^x+x} x^2}{x^3-4 \log (-2-x (2+\log (5)))} \, dx+8 \int \frac {e^{-e^x} x^4}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx+32 \int \frac {e^{-e^x} x \log (-2-x (2+\log (5)))}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx+(4 (2+\log (5))) \int \frac {e^{-e^x} x^5}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx-(16 (2+\log (5))) \int \frac {e^{-e^x} x^2}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx \\ & = 4 \int \frac {e^{-e^x+x} x^2}{x^3-4 \log (-2-x (2+\log (5)))} \, dx+8 \int \left (-\frac {8 e^{-e^x}}{(2+\log (5))^4 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {4 e^{-e^x} x}{(2+\log (5))^3 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}-\frac {2 e^{-e^x} x^2}{(2+\log (5))^2 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {e^{-e^x} x^3}{(2+\log (5)) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {16 e^{-e^x}}{(2+\log (5))^4 (2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}\right ) \, dx+32 \int \left (\frac {e^{-e^x} x^4}{4 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}-\frac {e^{-e^x} x}{4 \left (x^3-4 \log (-2-x (2+\log (5)))\right )}\right ) \, dx+(4 (2+\log (5))) \int \left (\frac {16 e^{-e^x}}{(2+\log (5))^5 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}-\frac {8 e^{-e^x} x}{(2+\log (5))^4 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {4 e^{-e^x} x^2}{(2+\log (5))^3 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}-\frac {2 e^{-e^x} x^3}{(2+\log (5))^2 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {e^{-e^x} x^4}{(2+\log (5)) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {32 e^{-e^x}}{(-2-\log (5))^5 (2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}\right ) \, dx-(16 (2+\log (5))) \int \left (-\frac {2 e^{-e^x}}{(2+\log (5))^2 \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {e^{-e^x} x}{(2+\log (5)) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}+\frac {4 e^{-e^x}}{(2+\log (5))^2 (2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2}\right ) \, dx \\ & = 4 \int \frac {e^{-e^x} x^4}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx+4 \int \frac {e^{-e^x+x} x^2}{x^3-4 \log (-2-x (2+\log (5)))} \, dx+8 \int \frac {e^{-e^x} x^4}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx-8 \int \frac {e^{-e^x} x}{x^3-4 \log (-2-x (2+\log (5)))} \, dx-16 \int \frac {e^{-e^x} x}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx+\frac {32 \int \frac {e^{-e^x}}{\left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx}{2+\log (5)}-\frac {64 \int \frac {e^{-e^x}}{(2+x (2+\log (5))) \left (x^3-4 \log (-2-x (2+\log (5)))\right )^2} \, dx}{2+\log (5)} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx=\frac {4 e^{-e^x} x^2}{-x^3+4 \log (-2-x (2+\log (5)))} \]
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Time = 33.84 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(-\frac {4 x^{2} {\mathrm e}^{-{\mathrm e}^{x}}}{x^{3}-4 \ln \left (-x \ln \left (5\right )-2 x -2\right )}\) | \(30\) |
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx=-\frac {4 \, x^{2} e^{\left (-e^{x}\right )}}{x^{3} - 4 \, \log \left (-x \log \left (5\right ) - 2 \, x - 2\right )} \]
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx=- \frac {4 x^{2} e^{- e^{x}}}{x^{3} - 4 \log {\left (- 2 x - x \log {\left (5 \right )} - 2 \right )}} \]
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Time = 0.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx=-\frac {4 \, x^{2}}{x^{3} e^{\left (e^{x}\right )} - 4 \, e^{\left (e^{x}\right )} \log \left (-x {\left (\log \left (5\right ) + 2\right )} - 2\right )} \]
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Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx=-\frac {4 \, x^{2} e^{\left (-e^{x}\right )}}{x^{3} - 4 \, \log \left (-x \log \left (5\right ) - 2 \, x - 2\right )} \]
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Time = 15.85 (sec) , antiderivative size = 254, normalized size of antiderivative = 7.47 \[ \int \frac {e^{-e^x} \left (-32 x^2+8 x^4+8 x^5+\left (-16 x^2+4 x^5\right ) \log (5)+e^x \left (8 x^5+8 x^6+4 x^6 \log (5)\right )+\left (64 x+64 x^2+32 x^2 \log (5)+e^x \left (-32 x^2-32 x^3-16 x^3 \log (5)\right )\right ) \log (-2-2 x-x \log (5))\right )}{2 x^6+2 x^7+x^7 \log (5)+\left (-16 x^3-16 x^4-8 x^4 \log (5)\right ) \log (-2-2 x-x \log (5))+(32+32 x+16 x \log (5)) \log ^2(-2-2 x-x \log (5))} \, dx=\frac {\frac {4\,x^2\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,\left (2\,x^3\,{\mathrm {e}}^x-4\,\ln \left (5\right )+2\,x^4\,{\mathrm {e}}^x+x^3\,\ln \left (5\right )+2\,x^2+2\,x^3+x^4\,{\mathrm {e}}^x\,\ln \left (5\right )-8\right )}{3\,x^3\,\ln \left (5\right )-4\,\ln \left (5\right )+6\,x^2+6\,x^3-8}-\frac {16\,x\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,\ln \left (-2\,x-x\,\ln \left (5\right )-2\right )\,\left (x\,{\mathrm {e}}^x-2\right )\,\left (2\,x+x\,\ln \left (5\right )+2\right )}{3\,x^3\,\ln \left (5\right )-4\,\ln \left (5\right )+6\,x^2+6\,x^3-8}}{4\,\ln \left (-2\,x-x\,\ln \left (5\right )-2\right )-x^3}-\frac {{\mathrm {e}}^{-{\mathrm {e}}^x}\,\left (\frac {16\,x}{\ln \left (125\right )+6}-\frac {8\,x^2\,{\mathrm {e}}^x}{\ln \left (125\right )+6}+\frac {x^2\,\left (8\,\ln \left (5\right )+16\right )}{\ln \left (125\right )+6}-\frac {x^3\,{\mathrm {e}}^x\,\left (\ln \left (625\right )+8\right )}{\ln \left (125\right )+6}\right )}{x^3+\frac {6\,x^2}{\ln \left (125\right )+6}-\frac {\ln \left (625\right )+8}{\ln \left (125\right )+6}} \]
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