Integrand size = 78, antiderivative size = 19 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{2 x-\frac {361}{8} e^{-e^x x} x} \]
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\[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=\int \frac {\exp \left (-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}\right ) \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \frac {\exp \left (-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}\right ) \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{x} \, dx \\ & = \frac {1}{8} \int e^{2 x-e^x x-\frac {361}{8} e^{-e^x x} x} \left (-361+16 e^{e^x x}+361 e^x x (1+x)\right ) \, dx \\ & = \frac {1}{8} \int \left (16 e^{2 x-\frac {361}{8} e^{-e^x x} x}-361 e^{2 x-e^x x-\frac {361}{8} e^{-e^x x} x}+361 e^{3 x-e^x x-\frac {361}{8} e^{-e^x x} x} x (1+x)\right ) \, dx \\ & = 2 \int e^{2 x-\frac {361}{8} e^{-e^x x} x} \, dx-\frac {361}{8} \int e^{2 x-e^x x-\frac {361}{8} e^{-e^x x} x} \, dx+\frac {361}{8} \int e^{3 x-e^x x-\frac {361}{8} e^{-e^x x} x} x (1+x) \, dx \\ & = 2 \int e^{2 x-\frac {361}{8} e^{-e^x x} x} \, dx-\frac {361}{8} \int e^{2 x-e^x x-\frac {361}{8} e^{-e^x x} x} \, dx+\frac {361}{8} \int \left (e^{3 x-e^x x-\frac {361}{8} e^{-e^x x} x} x+e^{3 x-e^x x-\frac {361}{8} e^{-e^x x} x} x^2\right ) \, dx \\ & = 2 \int e^{2 x-\frac {361}{8} e^{-e^x x} x} \, dx-\frac {361}{8} \int e^{2 x-e^x x-\frac {361}{8} e^{-e^x x} x} \, dx+\frac {361}{8} \int e^{3 x-e^x x-\frac {361}{8} e^{-e^x x} x} x \, dx+\frac {361}{8} \int e^{3 x-e^x x-\frac {361}{8} e^{-e^x x} x} x^2 \, dx \\ \end{align*}
Time = 2.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{2 x-\frac {361}{8} e^{-e^x x} x} \]
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Time = 1.55 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05
method | result | size |
risch | \({\mathrm e}^{\frac {x \left (16 \,{\mathrm e}^{{\mathrm e}^{x} x}-361\right ) {\mathrm e}^{-{\mathrm e}^{x} x}}{8}}\) | \(20\) |
parallelrisch | \({\mathrm e}^{\frac {\left (16 x \,{\mathrm e}^{{\mathrm e}^{x} x}-361 x \right ) {\mathrm e}^{-{\mathrm e}^{x} x}}{8}}\) | \(31\) |
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.47 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{\left (-\frac {1}{8} \, {\left (361 \, x^{2} + 8 \, {\left (x e^{x} - 2 \, x + \log \left (x\right )\right )} e^{\left (x e^{x} + \log \left (x\right )\right )}\right )} e^{\left (-x e^{x} - \log \left (x\right )\right )} + x e^{x} + \log \left (x\right )\right )} \]
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Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{\frac {2 \left (x^{2} e^{x e^{x}} - \frac {361 x^{2}}{16}\right ) e^{- x e^{x}}}{x}} \]
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Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=e^{\left (-\frac {361}{8} \, x e^{\left (-x e^{x}\right )} + 2 \, x\right )} \]
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\[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx=\int { \frac {1}{8} \, {\left (361 \, {\left (x^{3} + x^{2}\right )} e^{x} - 361 \, x + 16 \, e^{\left (x e^{x} + \log \left (x\right )\right )}\right )} e^{\left (-\frac {1}{8} \, {\left (361 \, x^{2} - 16 \, x e^{\left (x e^{x} + \log \left (x\right )\right )}\right )} e^{\left (-x e^{x} - \log \left (x\right )\right )} - x e^{x} - \log \left (x\right )\right )} \,d x } \]
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Time = 11.62 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-e^x x+\frac {e^{-e^x x} \left (-361 x^2+16 e^{e^x x} x^2\right )}{8 x}} \left (-361 x+16 e^{e^x x} x+e^x \left (361 x^2+361 x^3\right )\right )}{8 x} \, dx={\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-\frac {361\,x\,{\mathrm {e}}^{-x\,{\mathrm {e}}^x}}{8}} \]
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