Integrand size = 163, antiderivative size = 27 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {9}{5 \left (\frac {x^2}{4}+\log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )} \]
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\[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {72 \left (-2 x \left (1+2 e^x+x\right )-\left (e^x+x\right ) \left (2+x^2\right ) \log \left (e^x \left (e^x+x\right )\right )\right )}{5 x \left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx \\ & = \frac {72}{5} \int \frac {-2 x \left (1+2 e^x+x\right )-\left (e^x+x\right ) \left (2+x^2\right ) \log \left (e^x \left (e^x+x\right )\right )}{x \left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx \\ & = \frac {72}{5} \int \left (\frac {2 (-1+x)}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}+\frac {-4 x-2 \log \left (e^x \left (e^x+x\right )\right )-x^2 \log \left (e^x \left (e^x+x\right )\right )}{x \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}\right ) \, dx \\ & = \frac {72}{5} \int \frac {-4 x-2 \log \left (e^x \left (e^x+x\right )\right )-x^2 \log \left (e^x \left (e^x+x\right )\right )}{x \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx+\frac {144}{5} \int \frac {-1+x}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx \\ & = \frac {72}{5} \int \frac {-4 x-\left (2+x^2\right ) \log \left (e^x \left (e^x+x\right )\right )}{x \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx+\frac {144}{5} \int \left (-\frac {1}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}+\frac {x}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}\right ) \, dx \\ & = \frac {72}{5} \int \left (-\frac {2}{x \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}-\frac {x}{\left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}-\frac {4}{\log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}\right ) \, dx-\frac {144}{5} \int \frac {1}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx+\frac {144}{5} \int \frac {x}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx \\ & = -\left (\frac {72}{5} \int \frac {x}{\left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx\right )-\frac {144}{5} \int \frac {1}{x \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx-\frac {144}{5} \int \frac {1}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx+\frac {144}{5} \int \frac {x}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx-\frac {288}{5} \int \frac {1}{\log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx \\ \end{align*}
Time = 0.44 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )} \]
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Time = 24.60 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
parallelrisch | \(\frac {36}{5 \left (x^{2}+4 \ln \left (x \ln \left (\left ({\mathrm e}^{x}+x \right ) {\mathrm e}^{x}\right )\right )\right )}\) | \(22\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x \log \left (x e^{x} + e^{\left (2 \, x\right )}\right )\right )\right )}} \]
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Time = 0.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 x^{2} + 20 \log {\left (x \log {\left (x e^{x} + e^{2 x} \right )} \right )}} \]
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Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x + \log \left (x + e^{x}\right )\right ) + 4 \, \log \left (x\right )\right )}} \]
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Time = 1.94 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x \log \left (x e^{x} + e^{\left (2 \, x\right )}\right )\right )\right )}} \]
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Time = 15.64 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx=\frac {36}{5\,\left (4\,\ln \left (x\,\ln \left ({\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^x\right )\right )+x^2\right )} \]
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