Integrand size = 54, antiderivative size = 26 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=x^2-\frac {1}{10} e^{2-2 e^x} x \left (-4+\log ^2(6 x)\right ) \]
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\[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=\int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{10} \int e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx \\ & = \frac {1}{10} \int e^{-2 \left (-1+e^x\right )} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx \\ & = \frac {1}{10} \int \left (4 e^{-2 \left (-1+e^x\right )}+20 x-8 e^{-2 \left (-1+e^x\right )+x} x-2 e^{-2 \left (-1+e^x\right )} \log (6 x)+e^{-2 \left (-1+e^x\right )} \left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx \\ & = x^2+\frac {1}{10} \int e^{-2 \left (-1+e^x\right )} \left (-1+2 e^x x\right ) \log ^2(6 x) \, dx-\frac {1}{5} \int e^{-2 \left (-1+e^x\right )} \log (6 x) \, dx+\frac {2}{5} \int e^{-2 \left (-1+e^x\right )} \, dx-\frac {4}{5} \int e^{-2 \left (-1+e^x\right )+x} x \, dx \\ & = x^2-\frac {1}{5} e^2 \operatorname {ExpIntegralEi}\left (-2 e^x\right ) \log (6 x)+\frac {1}{10} \int \left (-e^{-2 \left (-1+e^x\right )} \log ^2(6 x)+2 e^{-2 \left (-1+e^x\right )+x} x \log ^2(6 x)\right ) \, dx+\frac {1}{5} \int \frac {e^2 \operatorname {ExpIntegralEi}\left (-2 e^x\right )}{x} \, dx+\frac {2}{5} \text {Subst}\left (\int \frac {e^{2-2 x}}{x} \, dx,x,e^x\right )-\frac {4}{5} \int e^{-2 \left (-1+e^x\right )+x} x \, dx \\ & = x^2+\frac {2}{5} e^2 \operatorname {ExpIntegralEi}\left (-2 e^x\right )-\frac {1}{5} e^2 \operatorname {ExpIntegralEi}\left (-2 e^x\right ) \log (6 x)-\frac {1}{10} \int e^{-2 \left (-1+e^x\right )} \log ^2(6 x) \, dx+\frac {1}{5} \int e^{-2 \left (-1+e^x\right )+x} x \log ^2(6 x) \, dx-\frac {4}{5} \int e^{-2 \left (-1+e^x\right )+x} x \, dx+\frac {1}{5} e^2 \int \frac {\operatorname {ExpIntegralEi}\left (-2 e^x\right )}{x} \, dx \\ \end{align*}
Time = 1.99 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=\frac {1}{10} \left (4 e^{2-2 e^x} x+10 x^2-e^{2-2 e^x} x \log ^2(6 x)\right ) \]
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Time = 0.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
method | result | size |
risch | \(x^{2}+\frac {\left (-x \ln \left (6 x \right )^{2}+4 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}+2}}{10}\) | \(27\) |
parallelrisch | \(-\frac {\left (-10 \,{\mathrm e}^{2 \,{\mathrm e}^{x}-2} x^{2}+x \ln \left (6 x \right )^{2}-4 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}+2}}{10}\) | \(34\) |
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=\frac {1}{10} \, {\left (10 \, x^{2} e^{\left (2 \, e^{x} - 2\right )} - x \log \left (6 \, x\right )^{2} + 4 \, x\right )} e^{\left (-2 \, e^{x} + 2\right )} \]
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Time = 6.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=x^{2} + \frac {\left (- x \log {\left (6 x \right )}^{2} + 4 x\right ) e^{2 - 2 e^{x}}}{10} \]
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\[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=\int { \frac {1}{10} \, {\left ({\left (2 \, x e^{x} - 1\right )} \log \left (6 \, x\right )^{2} - 8 \, x e^{x} + 20 \, x e^{\left (2 \, e^{x} - 2\right )} - 2 \, \log \left (6 \, x\right ) + 4\right )} e^{\left (-2 \, e^{x} + 2\right )} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.62 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx=-\frac {1}{10} \, {\left (x e^{\left (x - 2 \, e^{x} + 2\right )} \log \left (6\right )^{2} + 2 \, x e^{\left (x - 2 \, e^{x} + 2\right )} \log \left (6\right ) \log \left (x\right ) + x e^{\left (x - 2 \, e^{x} + 2\right )} \log \left (x\right )^{2} - 10 \, x^{2} e^{x} - 4 \, x e^{\left (x - 2 \, e^{x} + 2\right )}\right )} e^{\left (-x\right )} \]
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Time = 14.65 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1}{10} e^{2-2 e^x} \left (4+20 e^{-2+2 e^x} x-8 e^x x-2 \log (6 x)+\left (-1+2 e^x x\right ) \log ^2(6 x)\right ) \, dx={\mathrm {e}}^{2-2\,{\mathrm {e}}^x}\,\left (\frac {2\,x}{5}-\frac {x\,{\ln \left (6\,x\right )}^2}{10}\right )+x^2 \]
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