Integrand size = 74, antiderivative size = 29 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{7+x+\frac {1}{3} x \left (x-\left (\frac {3}{5}+x\right ) \log \left (e^{-e^x} x\right )\right )} \]
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\[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\int \frac {1}{15} \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{15} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx \\ & = \frac {1}{15} \int \left (12 \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right )+5 \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x+\exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x (3+5 x)-\exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) (3+10 x) \log \left (e^{-e^x} x\right )\right ) \, dx \\ & = \frac {1}{15} \int \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x (3+5 x) \, dx-\frac {1}{15} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) (3+10 x) \log \left (e^{-e^x} x\right ) \, dx+\frac {1}{3} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \, dx+\frac {4}{5} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \, dx \\ & = \frac {1}{15} \int \left (3 \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x+5 \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x^2\right ) \, dx-\frac {1}{15} \int \left (3 \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \log \left (e^{-e^x} x\right )+10 \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \log \left (e^{-e^x} x\right )\right ) \, dx+\frac {1}{3} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \, dx+\frac {4}{5} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \, dx \\ & = \frac {1}{5} \int \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \, dx-\frac {1}{5} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \log \left (e^{-e^x} x\right ) \, dx+\frac {1}{3} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \, dx+\frac {1}{3} \int \exp \left (x+\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x^2 \, dx-\frac {2}{3} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) x \log \left (e^{-e^x} x\right ) \, dx+\frac {4}{5} \int \exp \left (\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )\right ) \, dx \\ \end{align*}
Time = 3.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{7+x+\frac {x^2}{3}} \left (e^{-e^x} x\right )^{-\frac {1}{15} x (3+5 x)} \]
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Time = 1.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (-5 x^{2}-3 x \right ) \ln \left (x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )}{15}+\frac {x^{2}}{3}+x +7}\) | \(29\) |
risch | \(x^{-\frac {x^{2}}{3}} x^{-\frac {x}{5}} \left ({\mathrm e}^{{\mathrm e}^{x}}\right )^{\frac {x^{2}}{3}} \left ({\mathrm e}^{{\mathrm e}^{x}}\right )^{\frac {x}{5}} {\mathrm e}^{7+\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{3} x^{2}}{6}+\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{3} x}{10}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i x \right ) x^{2}}{6}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i x \right ) x}{10}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x^{2}}{6}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x}{10}+\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x^{2}}{6}+\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x}{10}+\frac {x^{2}}{3}+x}\) | \(235\) |
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} - \frac {1}{15} \, {\left (5 \, x^{2} + 3 \, x\right )} \log \left (x e^{\left (-e^{x}\right )}\right ) + x + 7\right )} \]
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Timed out. \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\text {Timed out} \]
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Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} e^{x} - \frac {1}{3} \, x^{2} \log \left (x\right ) + \frac {1}{3} \, x^{2} + \frac {1}{5} \, x e^{x} - \frac {1}{5} \, x \log \left (x\right ) + x + 7\right )} \]
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Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (-\frac {1}{3} \, x^{2} \log \left (x e^{\left (-e^{x}\right )}\right ) + \frac {1}{3} \, x^{2} - \frac {1}{5} \, x \log \left (x e^{\left (-e^{x}\right )}\right ) + x + 7\right )} \]
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Time = 11.97 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^7\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^x}{3}}\,{\mathrm {e}}^{\frac {x^2}{3}}\,{\mathrm {e}}^x}{x^{\frac {x^2}{3}+\frac {x}{5}}} \]
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