Integrand size = 60, antiderivative size = 27 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=3+9 x-e^{5-e^2-e^{-x} x} x^2 \]
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\[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=\int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (9+e^{5 \left (1-\frac {e^2}{5}\right )-x-e^{-x} x} x \left (-2 e^x+x-x^2\right )\right ) \, dx \\ & = 9 x+\int e^{5 \left (1-\frac {e^2}{5}\right )-x-e^{-x} x} x \left (-2 e^x+x-x^2\right ) \, dx \\ & = 9 x+\int \exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x \left (-2 e^x+x-x^2\right ) \, dx \\ & = 9 x+\int \left (-2 \exp \left (x-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x-\exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) (-1+x) x^2\right ) \, dx \\ & = 9 x-2 \int \exp \left (x-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x \, dx-\int \exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) (-1+x) x^2 \, dx \\ & = 9 x-2 \int e^{5 \left (1-\frac {e^2}{5}\right )-e^{-x} x} x \, dx-\int \left (-\exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x^2+\exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x^3\right ) \, dx \\ & = 9 x-2 \int e^{-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x\right )} x \, dx+\int \exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x^2 \, dx-\int \exp \left (-e^{-x} \left (-5 e^x \left (1-\frac {e^2}{5}\right )+x+e^x x\right )\right ) x^3 \, dx \\ \end{align*}
Time = 0.69 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=9 x-e^{5-e^2-e^{-x} x} x^2 \]
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Time = 2.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30
method | result | size |
parallelrisch | \(9 x -{\mathrm e}^{-\left ({\mathrm e}^{2} {\mathrm e}^{x}-{\mathrm e}^{x} \ln \left (x^{2}\right )-5 \,{\mathrm e}^{x}+x \right ) {\mathrm e}^{-x}}\) | \(35\) |
default | \(\left (9 \,{\mathrm e}^{x} x -{\mathrm e}^{x} {\mathrm e}^{\left ({\mathrm e}^{x} \ln \left (x^{2}\right )+\left (5-{\mathrm e}^{2}\right ) {\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}}\right ) {\mathrm e}^{-x}\) | \(44\) |
norman | \(\left (9 \,{\mathrm e}^{x} x -{\mathrm e}^{x} {\mathrm e}^{\left ({\mathrm e}^{x} \ln \left (x^{2}\right )+\left (5-{\mathrm e}^{2}\right ) {\mathrm e}^{x}-x \right ) {\mathrm e}^{-x}}\right ) {\mathrm e}^{-x}\) | \(44\) |
risch | \(9 x -{\mathrm e}^{\frac {\left (-i {\mathrm e}^{x} \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i {\mathrm e}^{x} \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i {\mathrm e}^{x} \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+4 \,{\mathrm e}^{x} \ln \left (x \right )+10 \,{\mathrm e}^{x}-2 \,{\mathrm e}^{2+x}-2 x \right ) {\mathrm e}^{-x}}{2}}\) | \(89\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=9 \, x - e^{\left (-{\left ({\left (e^{2} - 5\right )} e^{x} - e^{x} \log \left (x^{2}\right ) + x\right )} e^{\left (-x\right )}\right )} \]
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Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=9 x - e^{\left (- x + e^{x} \log {\left (x^{2} \right )} + \left (5 - e^{2}\right ) e^{x}\right ) e^{- x}} \]
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\[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=\int { -\frac {{\left ({\left (x^{2} - x + 2 \, e^{x}\right )} e^{\left (-{\left ({\left (e^{2} - 5\right )} e^{x} - e^{x} \log \left (x^{2}\right ) + x\right )} e^{\left (-x\right )}\right )} - 9 \, x e^{x}\right )} e^{\left (-x\right )}}{x} \,d x } \]
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Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=-{\left (x^{2} e^{\left (-x e^{\left (-x\right )} - x - e^{2} + 5\right )} - 9 \, x e^{\left (-x\right )}\right )} e^{x} \]
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Time = 9.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-x} \left (9 e^x x+e^{e^{-x} \left (e^x \left (5-e^2\right )-x+e^x \log \left (x^2\right )\right )} \left (-2 e^x+x-x^2\right )\right )}{x} \, dx=9\,x-x^2\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{-x}} \]
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