Integrand size = 99, antiderivative size = 28 \[ \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx=-1+\frac {5 e^{-2+x-\frac {x^2}{2+3 e^x}}}{2 x} \]
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\[ \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx=\int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{2 \left (2+3 e^x\right )^2 x^2} \, dx \\ & = \frac {1}{2} \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{\left (2+3 e^x\right )^2 x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {5 e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} (-2+x)}{2+3 e^x}+\frac {5 e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} (-1+x)}{x^2}-\frac {10 e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} x}{\left (2+3 e^x\right )^2}\right ) \, dx \\ & = \frac {5}{2} \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} (-2+x)}{2+3 e^x} \, dx+\frac {5}{2} \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} (-1+x)}{x^2} \, dx-5 \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} x}{\left (2+3 e^x\right )^2} \, dx \\ & = \frac {5}{2} \int \left (-\frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}}}{x^2}+\frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}}}{x}\right ) \, dx+\frac {5}{2} \int \left (-\frac {2 e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}}}{2+3 e^x}+\frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} x}{2+3 e^x}\right ) \, dx-5 \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} x}{\left (2+3 e^x\right )^2} \, dx \\ & = -\left (\frac {5}{2} \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}}}{x^2} \, dx\right )+\frac {5}{2} \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}}}{x} \, dx+\frac {5}{2} \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} x}{2+3 e^x} \, dx-5 \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}}}{2+3 e^x} \, dx-5 \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} x}{\left (2+3 e^x\right )^2} \, dx \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx=\frac {5 e^{-2+x-\frac {x^2}{2+3 e^x}}}{2 x} \]
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Time = 0.88 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \(\frac {5 \,{\mathrm e}^{\frac {\left (-2+x \right ) {\mathrm e}^{\ln \left (3\right )+x}-x^{2}+2 x -4}{{\mathrm e}^{\ln \left (3\right )+x}+2}}}{2 x}\) | \(36\) |
norman | \(\frac {\frac {5 \,{\mathrm e}^{\ln \left (3\right )+x} {\mathrm e}^{\frac {\left (-2+x \right ) {\mathrm e}^{\ln \left (3\right )+x}-x^{2}+2 x -4}{{\mathrm e}^{\ln \left (3\right )+x}+2}}}{2}+5 \,{\mathrm e}^{\frac {\left (-2+x \right ) {\mathrm e}^{\ln \left (3\right )+x}-x^{2}+2 x -4}{{\mathrm e}^{\ln \left (3\right )+x}+2}}}{x \left ({\mathrm e}^{\ln \left (3\right )+x}+2\right )}\) | \(84\) |
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx=\frac {5 \, e^{\left (-\frac {x^{2} - {\left (x - 2\right )} e^{\left (x + \log \left (3\right )\right )} - 2 \, x + 4}{e^{\left (x + \log \left (3\right )\right )} + 2}\right )}}{2 \, x} \]
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx=\frac {5 e^{\frac {- x^{2} + 2 x + 3 \left (x - 2\right ) e^{x} - 4}{3 e^{x} + 2}}}{2 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (25) = 50\).
Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx=\frac {5 \, e^{\left (-\frac {x^{2}}{3 \, e^{x} + 2} + \frac {3 \, x e^{x}}{3 \, e^{x} + 2} + \frac {2 \, x}{3 \, e^{x} + 2} - \frac {6 \, e^{x}}{3 \, e^{x} + 2} - \frac {4}{3 \, e^{x} + 2}\right )}}{2 \, x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.81 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.50 \[ \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx=\frac {5 \, e^{\left (-\frac {1}{2} \, \log \left (3\right )^{2} + \frac {e^{\left (x + \log \left (3\right )\right )} \log \left (3\right )^{2} - 2 \, {\left (x + \log \left (3\right )\right )}^{2} + 2 \, {\left (x + \log \left (3\right )\right )} e^{\left (x + \log \left (3\right )\right )} + 4 \, {\left (x + \log \left (3\right )\right )} \log \left (3\right ) + 4 \, x + 4 \, \log \left (3\right )}{2 \, {\left (e^{\left (x + \log \left (3\right )\right )} + 2\right )}} - 2\right )}}{6 \, x} \]
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Time = 0.33 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.46 \[ \int \frac {e^{\frac {-4+3 e^x (-2+x)+2 x-x^2}{2+3 e^x}} \left (-20+20 x-20 x^2+9 e^{2 x} (-5+5 x)+3 e^x \left (-20+20 x-10 x^2+5 x^3\right )\right )}{8 x^2+24 e^x x^2+18 e^{2 x} x^2} \, dx=\frac {5\,{\mathrm {e}}^{\frac {3\,x\,{\mathrm {e}}^x}{3\,{\mathrm {e}}^x+2}}\,{\mathrm {e}}^{\frac {2\,x}{3\,{\mathrm {e}}^x+2}}\,{\mathrm {e}}^{-\frac {x^2}{3\,{\mathrm {e}}^x+2}}\,{\mathrm {e}}^{-\frac {6\,{\mathrm {e}}^x}{3\,{\mathrm {e}}^x+2}}\,{\mathrm {e}}^{-\frac {4}{3\,{\mathrm {e}}^x+2}}}{2\,x} \]
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