Integrand size = 347, antiderivative size = 31 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{-5+\log \left (\log \left (\frac {e^{2-e^4}}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )} \]
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\[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-e^2-\left (e^2+4 x\right ) \log \left (-4-\frac {e^2}{x}\right ) \left (-2+e^4-\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right ) \left (-5+\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )\right )}{\left (e^2+4 x\right ) \log \left (-4-\frac {e^2}{x}\right ) \left (2 \left (1-\frac {e^4}{2}\right )+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right ) \left (5-\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )\right )^2} \, dx \\ & = \int \frac {-5+\frac {e^2}{\left (e^2+4 x\right ) \log \left (-4-\frac {e^2}{x}\right ) \left (-2+e^4-\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )}+\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )}{\left (5-\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )\right )^2} \, dx \\ & = \int \left (\frac {e^2}{\left (-e^2-4 x\right ) \log \left (-4-\frac {e^2}{x}\right ) \left (2 \left (1-\frac {e^4}{2}\right )+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right ) \left (5-\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )\right )^2}+\frac {1}{-5+\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )}\right ) \, dx \\ & = e^2 \int \frac {1}{\left (-e^2-4 x\right ) \log \left (-4-\frac {e^2}{x}\right ) \left (2 \left (1-\frac {e^4}{2}\right )+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right ) \left (5-\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )\right )^2} \, dx+\int \frac {1}{-5+\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )} \, dx \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{-5+\log \left (2-e^4+\log \left (\frac {1}{\log \left (-4-\frac {e^2}{x}\right )}\right )\right )} \]
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Time = 20.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {x}{\ln \left (\ln \left (\frac {{\mathrm e}^{2-{\mathrm e}^{4}}}{\ln \left (-\frac {{\mathrm e}^{2}+4 x}{x}\right )}\right )\right )-5}\) | \(31\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{\log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right ) - 5} \]
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Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{\log {\left (\log {\left (\frac {1}{e^{-2 + e^{4}} \log {\left (\frac {- 4 x - e^{2}}{x} \right )}} \right )} \right )} - 5} \]
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Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{\log \left (-e^{4} - \log \left (-\log \left (x\right ) + \log \left (-4 \, x - e^{2}\right )\right ) + 2\right ) - 5} \]
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\[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\int { \frac {{\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right ) \log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right ) - 5 \, {\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right ) - e^{2}}{{\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right ) \log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right )^{2} - 10 \, {\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right ) \log \left (\log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )\right ) + 25 \, {\left (4 \, x + e^{2}\right )} \log \left (-\frac {4 \, x + e^{2}}{x}\right ) \log \left (\frac {e^{\left (-e^{4} + 2\right )}}{\log \left (-\frac {4 \, x + e^{2}}{x}\right )}\right )} \,d x } \]
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Time = 11.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {-e^2+\left (-5 e^2-20 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )}{\left (25 e^2+100 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )+\left (-10 e^2-40 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log \left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )+\left (e^2+4 x\right ) \log \left (\frac {-e^2-4 x}{x}\right ) \log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right ) \log ^2\left (\log \left (\frac {e^{2-e^4}}{\log \left (\frac {-e^2-4 x}{x}\right )}\right )\right )} \, dx=\frac {x}{\ln \left (\ln \left (\frac {{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^2}{\ln \left (-\frac {4\,x+{\mathrm {e}}^2}{x}\right )}\right )\right )-5} \]
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