Integrand size = 83, antiderivative size = 25 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=5-e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \]
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\[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{\left (5-2 e^x-x\right )^2 x} \, dx \\ & = \int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} (-6+x) \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (1-x+x \log \left (\frac {4}{x}\right )\right )}{x \left (-5+2 e^x+x\right )}\right ) \, dx \\ & = -\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} (-6+x) \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (1-x+x \log \left (\frac {4}{x}\right )\right )}{x \left (-5+2 e^x+x\right )} \, dx \\ & = -\int \left (-\frac {6 e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx+\int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x}\right ) \, dx \\ & = 6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx \\ & = 6 \int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx-\int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx \\ & = -\left (6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{\left (-5+2 e^x+x\right )^2} \, dx\right )+6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x}{\left (-5+2 e^x+x\right )^2} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx \\ \end{align*}
\[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx \]
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Time = 3.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(-{\mathrm e}^{\frac {\ln \left (\frac {4}{x}\right )-1}{2 \,{\mathrm e}^{x}+x -5}}\) | \(22\) |
risch | \(-{\mathrm e}^{\frac {2 \ln \left (2\right )-\ln \left (x \right )-1}{2 \,{\mathrm e}^{x}+x -5}}\) | \(24\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 5}\right )} \]
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Time = 0.49 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=- e^{\frac {\log {\left (\frac {4}{x} \right )} - 1}{x + 2 e^{x} - 5}} \]
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Time = 0.38 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-e^{\left (\frac {2 \, \log \left (2\right )}{x + 2 \, e^{x} - 5} - \frac {\log \left (x\right )}{x + 2 \, e^{x} - 5} - \frac {1}{x + 2 \, e^{x} - 5}\right )} \]
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\[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=\int { -\frac {{\left (2 \, {\left (x - 1\right )} e^{x} - {\left (2 \, x e^{x} + x\right )} \log \left (\frac {4}{x}\right ) + 5\right )} e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 5}\right )}}{x^{3} - 10 \, x^{2} + 4 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} - 5 \, x\right )} e^{x} + 25 \, x} \,d x } \]
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Time = 9.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-{\mathrm {e}}^{-\frac {1}{x+2\,{\mathrm {e}}^x-5}}\,{\left (\frac {4}{x}\right )}^{\frac {1}{x+2\,{\mathrm {e}}^x-5}} \]
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