Integrand size = 127, antiderivative size = 34 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=e^{4+\frac {x+\frac {-4+\frac {x}{2}-\log (x)}{\log (4-2 x-\log (5))}}{x}} \]
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\[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (2 x^3+x^2 (-4+\log (5))\right ) \log ^2(4-2 x-\log (5))} \, dx \\ & = \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{x^2 (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx \\ & = \int \left (\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (3+\log (x))}{x^2 \log (4-2 x-\log (5))}\right ) \, dx \\ & = \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (3+\log (x))}{x^2 \log (4-2 x-\log (5))} \, dx \\ & = \int \left (\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (-8+x-2 \log (x))}{(-4+\log (5)) (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx+\int \left (\frac {3 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))}\right ) \, dx \\ & = 3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (-8+x-2 \log (x))}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x \log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx \\ & = 3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \left (-\frac {8 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) x}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}-\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx}{4-\log (5)}+\frac {\int \left (-\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))}+\frac {8 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))}+\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))}\right ) \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx \\ & = 3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) x}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx \\ & = 3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \left (\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{2 \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (4-\log (5))}{2 (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx \\ & = 3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}-\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=e^{5+\frac {-8+x-2 \log (x)}{2 x \log (4-2 x-\log (5))}} \]
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Time = 42.62 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {-2 \ln \left (x \right )+10 x \ln \left (-\ln \left (5\right )+4-2 x \right )-8+x}{2 x \ln \left (-\ln \left (5\right )+4-2 x \right )}}\) | \(39\) |
risch | \({\mathrm e}^{-\frac {2 \ln \left (x \right )-10 x \ln \left (-\ln \left (5\right )+4-2 x \right )+8-x}{2 x \ln \left (-\ln \left (5\right )+4-2 x \right )}}\) | \(41\) |
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Time = 0.48 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=e^{\left (\frac {10 \, x \log \left (-2 \, x - \log \left (5\right ) + 4\right ) + x - 2 \, \log \left (x\right ) - 8}{2 \, x \log \left (-2 \, x - \log \left (5\right ) + 4\right )}\right )} \]
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Exception generated. \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=\text {Exception raised: TypeError} \]
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Exception generated. \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.76 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=e^{\left (-\frac {\log \left (x\right )}{x \log \left (-2 \, x - \log \left (5\right ) + 4\right )} + \frac {1}{2 \, \log \left (-2 \, x - \log \left (5\right ) + 4\right )} - \frac {4}{x \log \left (-2 \, x - \log \left (5\right ) + 4\right )} + 5\right )} \]
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Time = 11.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=\frac {{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {4}{x\,\ln \left (4-\ln \left (5\right )-2\,x\right )}}\,{\mathrm {e}}^{\frac {1}{2\,\ln \left (4-\ln \left (5\right )-2\,x\right )}}}{x^{\frac {1}{x\,\ln \left (4-\ln \left (5\right )-2\,x\right )}}} \]
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