Integrand size = 74, antiderivative size = 34 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=25 e^{-2 x}+\frac {e^{e^x+x}-\frac {5 e^{-e^x}}{x}}{x}+\log (3) \]
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\[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=\int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-50 e^{-2 x}+\frac {10 e^{-e^x}}{x^3}+\frac {e^{-e^x-2 x+2 \left (e^x+2 x\right )}}{x}+\frac {e^{-e^x+x} \left (5-e^{2 e^x}+e^{2 e^x} x\right )}{x^2}\right ) \, dx \\ & = 10 \int \frac {e^{-e^x}}{x^3} \, dx-50 \int e^{-2 x} \, dx+\int \frac {e^{-e^x-2 x+2 \left (e^x+2 x\right )}}{x} \, dx+\int \frac {e^{-e^x+x} \left (5-e^{2 e^x}+e^{2 e^x} x\right )}{x^2} \, dx \\ & = 25 e^{-2 x}+10 \int \frac {e^{-e^x}}{x^3} \, dx+\int \frac {e^{-e^x+x} \left (5+e^{2 e^x} (-1+x)\right )}{x^2} \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx \\ & = 25 e^{-2 x}+10 \int \frac {e^{-e^x}}{x^3} \, dx+\int \left (\frac {5 e^{-e^x+x}}{x^2}+\frac {e^{e^x+x} (-1+x)}{x^2}\right ) \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx \\ & = 25 e^{-2 x}+5 \int \frac {e^{-e^x+x}}{x^2} \, dx+10 \int \frac {e^{-e^x}}{x^3} \, dx+\int \frac {e^{e^x+x} (-1+x)}{x^2} \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx \\ & = 25 e^{-2 x}+5 \int \frac {e^{-e^x+x}}{x^2} \, dx+10 \int \frac {e^{-e^x}}{x^3} \, dx+\int \left (-\frac {e^{e^x+x}}{x^2}+\frac {e^{e^x+x}}{x}\right ) \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx \\ & = 25 e^{-2 x}+5 \int \frac {e^{-e^x+x}}{x^2} \, dx+10 \int \frac {e^{-e^x}}{x^3} \, dx-\int \frac {e^{e^x+x}}{x^2} \, dx+\int \frac {e^{e^x+x}}{x} \, dx+\int \frac {e^{e^x+2 x}}{x} \, dx \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=25 e^{-2 x}-\frac {5 e^{-e^x}}{x^2}+\frac {e^{e^x+x}}{x} \]
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Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79
method | result | size |
risch | \(25 \,{\mathrm e}^{-2 x}+\frac {{\mathrm e}^{{\mathrm e}^{x}+x}}{x}-\frac {5 \,{\mathrm e}^{-{\mathrm e}^{x}}}{x^{2}}\) | \(27\) |
parallelrisch | \(\frac {\left ({\mathrm e}^{2 x} {\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{{\mathrm e}^{x}+x} x +25 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-5 \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-{\mathrm e}^{x}} {\mathrm e}^{-2 x}}{x^{2}}\) | \(43\) |
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=\frac {{\left (25 \, x^{2} e^{\left (e^{x}\right )} + x e^{\left (3 \, x + 2 \, e^{x}\right )} - 5 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x - e^{x}\right )}}{x^{2}} \]
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Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=25 e^{- 2 x} + \frac {x^{2} e^{x} e^{e^{x}} - 5 x e^{- e^{x}}}{x^{3}} \]
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=\frac {x e^{\left (x + e^{x}\right )} - 5 \, e^{\left (-e^{x}\right )}}{x^{2}} + 25 \, e^{\left (-2 \, x\right )} \]
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\[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=\int { -\frac {{\left (50 \, x^{3} e^{\left (e^{x}\right )} - 5 \, x e^{\left (3 \, x\right )} - {\left (x^{2} e^{\left (3 \, x\right )} + {\left (x^{2} - x\right )} e^{\left (2 \, x\right )}\right )} e^{\left (x + 2 \, e^{x}\right )} - 10 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x - e^{x}\right )}}{x^{3}} \,d x } \]
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Time = 12.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-e^x-2 x} \left (10 e^{2 x}+5 e^{3 x} x-50 e^{e^x} x^3+e^{2 e^x+x} \left (e^{3 x} x^2+e^{2 x} \left (-x+x^2\right )\right )\right )}{x^3} \, dx=25\,{\mathrm {e}}^{-2\,x}+\frac {{\mathrm {e}}^{x+{\mathrm {e}}^x}}{x}-\frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^x}}{x^2} \]
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