Integrand size = 69, antiderivative size = 27 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=e^{\frac {\left (-1+\frac {5 e^x (2-x)}{x}\right ) (1+x)}{\log (3)}}+x \]
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\[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=\int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2} \, dx}{\log (3)} \\ & = \frac {\int \left (-\frac {\exp \left (-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) \left (10 e^x-10 e^x x+x^2+5 e^x x^3\right )}{x^2}+\log (3)\right ) \, dx}{\log (3)} \\ & = x-\frac {\int \frac {\exp \left (-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) \left (10 e^x-10 e^x x+x^2+5 e^x x^3\right )}{x^2} \, dx}{\log (3)} \\ & = x-\frac {\int \frac {\exp \left (-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) \left (x^2+5 e^x \left (2-2 x+x^3\right )\right )}{x^2} \, dx}{\log (3)} \\ & = x-\frac {\int \left (\exp \left (-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right )+\frac {5 \exp \left (x-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) \left (2-2 x+x^3\right )}{x^2}\right ) \, dx}{\log (3)} \\ & = x-\frac {\int \exp \left (-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) \, dx}{\log (3)}-\frac {5 \int \frac {\exp \left (x-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) \left (2-2 x+x^3\right )}{x^2} \, dx}{\log (3)} \\ & = x-\frac {\int \exp \left (-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) \, dx}{\log (3)}-\frac {5 \int \left (\frac {2 \exp \left (x-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right )}{x^2}-\frac {2 \exp \left (x-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right )}{x}+\exp \left (x-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) x\right ) \, dx}{\log (3)} \\ & = x-\frac {\int \exp \left (-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) \, dx}{\log (3)}-\frac {5 \int \exp \left (x-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right ) x \, dx}{\log (3)}-\frac {10 \int \frac {\exp \left (x-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right )}{x^2} \, dx}{\log (3)}+\frac {10 \int \frac {\exp \left (x-\frac {(1+x) \left (5 e^x (-2+x)+x\right )}{x \log (3)}\right )}{x} \, dx}{\log (3)} \\ \end{align*}
Time = 2.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=e^{-\frac {1}{\log (3)}-\frac {x}{\log (3)}-\frac {5 e^x (-2+x) (1+x)}{x \log (3)}}+x \]
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Time = 0.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x +{\mathrm e}^{-\frac {\left (1+x \right ) \left (5 \,{\mathrm e}^{x} x -10 \,{\mathrm e}^{x}+x \right )}{\ln \left (3\right ) x}}\) | \(27\) |
norman | \(\frac {x^{2}+x \,{\mathrm e}^{\frac {\left (-5 x^{2}+5 x +10\right ) {\mathrm e}^{x}-x^{2}-x}{x \ln \left (3\right )}}}{x}\) | \(42\) |
parallelrisch | \(\frac {x \ln \left (3\right )+\ln \left (3\right ) {\mathrm e}^{\frac {\left (-5 x^{2}+5 x +10\right ) {\mathrm e}^{x}-x^{2}-x}{x \ln \left (3\right )}}}{\ln \left (3\right )}\) | \(45\) |
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=x + e^{\left (-\frac {x^{2} + 5 \, {\left (x^{2} - x - 2\right )} e^{x} + x}{x \log \left (3\right )}\right )} \]
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Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=x + e^{\frac {- x^{2} - x + \left (- 5 x^{2} + 5 x + 10\right ) e^{x}}{x \log {\left (3 \right )}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (24) = 48\).
Time = 0.38 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=\frac {x \log \left (3\right ) + e^{\left (-\frac {5 \, x e^{x}}{\log \left (3\right )} - \frac {x}{\log \left (3\right )} + \frac {5 \, e^{x}}{\log \left (3\right )} + \frac {10 \, e^{x}}{x \log \left (3\right )} - \frac {1}{\log \left (3\right )}\right )} \log \left (3\right )}{\log \left (3\right )} \]
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\[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=\int { \frac {x^{2} \log \left (3\right ) - {\left (x^{2} + 5 \, {\left (x^{3} - 2 \, x + 2\right )} e^{x}\right )} e^{\left (-\frac {x^{2} + 5 \, {\left (x^{2} - x - 2\right )} e^{x} + x}{x \log \left (3\right )}\right )}}{x^{2} \log \left (3\right )} \,d x } \]
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Time = 11.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=x+{\mathrm {e}}^{-\frac {x}{\ln \left (3\right )}}\,{\mathrm {e}}^{\frac {10\,{\mathrm {e}}^x}{x\,\ln \left (3\right )}}\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^x}{\ln \left (3\right )}}\,{\mathrm {e}}^{-\frac {1}{\ln \left (3\right )}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^x}{\ln \left (3\right )}} \]
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