Integrand size = 90, antiderivative size = 26 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log \left (\log \left (\frac {(-5+x) x}{4-\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )\right ) \]
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\[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{(5-x) x \left (4-\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {(5-x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx \\ & = \int \left (\frac {-35+16 x-x^2+5 \log \left (\frac {1}{100} e^{-x} x^3\right )-2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{5 x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {35-16 x+x^2-5 \log \left (\frac {1}{100} e^{-x} x^3\right )+2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{5 (-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx \\ & = \frac {1}{5} \int \frac {-35+16 x-x^2+5 \log \left (\frac {1}{100} e^{-x} x^3\right )-2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\frac {1}{5} \int \frac {35-16 x+x^2-5 \log \left (\frac {1}{100} e^{-x} x^3\right )+2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx \\ & = \frac {1}{5} \int \left (\frac {16}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {35}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {x}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {2 \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {5 \log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx+\frac {1}{5} \int \left (\frac {35}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {16 x}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {x^2}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}-\frac {5 \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {2 x \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx \\ & = -\left (\frac {1}{5} \int \frac {x}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx\right )+\frac {1}{5} \int \frac {x^2}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\frac {2}{5} \int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\frac {2}{5} \int \frac {x \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\frac {16}{5} \int \frac {1}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\frac {16}{5} \int \frac {x}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+7 \int \frac {1}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-7 \int \frac {1}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx \\ & = \frac {1}{5} \int \left (\frac {5}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {25}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {x}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx-\frac {1}{5} \int \frac {x}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\frac {2}{5} \int \left (\frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {5 \log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx-\frac {2}{5} \int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\frac {16}{5} \int \left (\frac {1}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}+\frac {5}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )}\right ) \, dx+\frac {16}{5} \int \frac {1}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+7 \int \frac {1}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-7 \int \frac {1}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx \\ & = 2 \int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+5 \int \frac {1}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+7 \int \frac {1}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-7 \int \frac {1}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-16 \int \frac {1}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\int \frac {1}{\left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx-\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{(-5+x) \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx+\int \frac {\log \left (\frac {1}{100} e^{-x} x^3\right )}{x \left (-4+\log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log \left (\log \left (-\frac {(-5+x) x}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )\right ) \]
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Time = 1.81 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
default | \(\ln \left (\ln \left (\frac {\left (-5+x \right ) x}{4-\ln \left (\frac {x^{3} {\mathrm e}^{-x}}{100}\right )}\right )\right )\) | \(24\) |
parallelrisch | \(\ln \left (\ln \left (\frac {-x^{2}+5 x}{\ln \left (\frac {x^{3} {\mathrm e}^{-x}}{100}\right )-4}\right )\right )\) | \(27\) |
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Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log \left (\log \left (-\frac {x^{2} - 5 \, x}{\log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 4}\right )\right ) \]
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Time = 0.48 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log {\left (\log {\left (\frac {- x^{2} + 5 x}{\log {\left (\frac {x^{3} e^{- x}}{100} \right )} - 4} \right )} \right )} \]
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Time = 0.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\log \left (\log \left (x + 2 \, \log \left (5\right ) + 2 \, \log \left (2\right ) - 3 \, \log \left (x\right ) + 4\right ) - \log \left (x - 5\right ) - \log \left (x\right )\right ) \]
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\[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\int { -\frac {x^{2} + {\left (2 \, x - 5\right )} \log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 16 \, x + 35}{{\left (4 \, x^{2} - {\left (x^{2} - 5 \, x\right )} \log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 20 \, x\right )} \log \left (-\frac {x^{2} - 5 \, x}{\log \left (\frac {1}{100} \, x^{3} e^{\left (-x\right )}\right ) - 4}\right )} \,d x } \]
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Time = 14.67 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {35-16 x+x^2+(-5+2 x) \log \left (\frac {1}{100} e^{-x} x^3\right )}{\left (20 x-4 x^2+\left (-5 x+x^2\right ) \log \left (\frac {1}{100} e^{-x} x^3\right )\right ) \log \left (\frac {5 x-x^2}{-4+\log \left (\frac {1}{100} e^{-x} x^3\right )}\right )} \, dx=\ln \left (\ln \left (-\frac {5\,x-x^2}{x-\ln \left (x^3\right )+2\,\ln \left (10\right )+4}\right )\right ) \]
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