Integrand size = 96, antiderivative size = 32 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 e^{\frac {e^x+\frac {1}{4} e^{-e^x x^2} x^2}{x}}}{x} \]
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\[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\int \frac {\exp \left (-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}\right ) \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {\exp \left (-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}\right ) \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{x^3} \, dx \\ & = \frac {1}{4} \int \left (\frac {20 \exp \left (\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}\right ) \left (-e^x-x+e^x x\right )}{x^3}-\frac {5 \exp \left (-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}\right ) \left (-1+2 e^x x^2+e^x x^3\right )}{x}\right ) \, dx \\ & = -\left (\frac {5}{4} \int \frac {\exp \left (-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}\right ) \left (-1+2 e^x x^2+e^x x^3\right )}{x} \, dx\right )+5 \int \frac {\exp \left (\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}\right ) \left (-e^x-x+e^x x\right )}{x^3} \, dx \\ & = -\left (\frac {5}{4} \int \frac {e^{\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )} \left (-1+e^x x^2 (2+x)\right )}{x} \, dx\right )+5 \int \frac {e^{\frac {e^x}{x}+\frac {1}{4} e^{-e^x x^2} x} \left (e^x (-1+x)-x\right )}{x^3} \, dx \\ & = -\left (\frac {5}{4} \int \left (-\frac {e^{\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )}}{x}+e^{x+\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )} x (2+x)\right ) \, dx\right )+5 \int \left (\frac {e^{\frac {e^x}{x}+x+\frac {1}{4} e^{-e^x x^2} x} (-1+x)}{x^3}-\frac {e^{\frac {e^x}{x}+\frac {1}{4} e^{-e^x x^2} x}}{x^2}\right ) \, dx \\ & = \frac {5}{4} \int \frac {e^{\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )}}{x} \, dx-\frac {5}{4} \int e^{x+\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )} x (2+x) \, dx+5 \int \frac {e^{\frac {e^x}{x}+x+\frac {1}{4} e^{-e^x x^2} x} (-1+x)}{x^3} \, dx-5 \int \frac {e^{\frac {e^x}{x}+\frac {1}{4} e^{-e^x x^2} x}}{x^2} \, dx \\ & = \frac {5}{4} \int \frac {e^{\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )}}{x} \, dx-\frac {5}{4} \int \left (2 e^{x+\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )} x+e^{x+\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )} x^2\right ) \, dx+5 \int \left (-\frac {e^{\frac {e^x}{x}+x+\frac {1}{4} e^{-e^x x^2} x}}{x^3}+\frac {e^{\frac {e^x}{x}+x+\frac {1}{4} e^{-e^x x^2} x}}{x^2}\right ) \, dx-5 \int \frac {e^{\frac {e^x}{x}+\frac {1}{4} e^{-e^x x^2} x}}{x^2} \, dx \\ & = \frac {5}{4} \int \frac {e^{\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )}}{x} \, dx-\frac {5}{4} \int e^{x+\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )} x^2 \, dx-\frac {5}{2} \int e^{x+\frac {1}{4} e^{-e^x x^2} x+e^x \left (\frac {1}{x}-x^2\right )} x \, dx-5 \int \frac {e^{\frac {e^x}{x}+x+\frac {1}{4} e^{-e^x x^2} x}}{x^3} \, dx-5 \int \frac {e^{\frac {e^x}{x}+\frac {1}{4} e^{-e^x x^2} x}}{x^2} \, dx+5 \int \frac {e^{\frac {e^x}{x}+x+\frac {1}{4} e^{-e^x x^2} x}}{x^2} \, dx \\ \end{align*}
Time = 5.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 e^{\frac {e^x}{x}+\frac {1}{4} e^{-e^x x^2} x}}{x} \]
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Time = 1.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
risch | \(\frac {5 \,{\mathrm e}^{\frac {\left (4 \,{\mathrm e}^{x \left ({\mathrm e}^{x} x +1\right )}+x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{x} x^{2}}}{4 x}}}{x}\) | \(35\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{\frac {\left (4 \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x} x^{2}}+x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{x} x^{2}}}{4 x}}}{x}\) | \(36\) |
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Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 \, e^{\left (x^{2} e^{x} + \frac {{\left (x^{2} e^{x} - 4 \, {\left (x^{3} - 1\right )} e^{\left (x^{2} e^{x} + 2 \, x\right )}\right )} e^{\left (-x^{2} e^{x} - x\right )}}{4 \, x}\right )}}{x} \]
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Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 e^{\frac {\left (\frac {x^{2}}{4} + e^{x} e^{x^{2} e^{x}}\right ) e^{- x^{2} e^{x}}}{x}}}{x} \]
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Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 \, e^{\left (\frac {1}{4} \, x e^{\left (-x^{2} e^{x}\right )} + \frac {e^{x}}{x}\right )}}{x} \]
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\[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\int { \frac {5 \, {\left (x^{2} + 4 \, {\left ({\left (x - 1\right )} e^{x} - x\right )} e^{\left (x^{2} e^{x}\right )} - {\left (x^{5} + 2 \, x^{4}\right )} e^{x}\right )} e^{\left (-x^{2} e^{x} + \frac {{\left (x^{2} + 4 \, e^{\left (x^{2} e^{x} + x\right )}\right )} e^{\left (-x^{2} e^{x}\right )}}{4 \, x}\right )}}{4 \, x^{3}} \,d x } \]
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Time = 12.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x}+\frac {x\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^x}}{4}}}{x} \]
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