Integrand size = 161, antiderivative size = 27 \[ \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx=\frac {1+\frac {\log \left (e^5+x\right )}{20 x}}{1-x+\log (4 x)} \]
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\[ \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx=\int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\log (4 x) \left (x-\left (e^5+x\right ) \log \left (e^5+x\right )\right )+(-1+x) \left (x \left (-1+20 e^5+20 x\right )+2 \left (e^5+x\right ) \log \left (e^5+x\right )\right )}{20 x^2 \left (e^5+x\right ) (1-x+\log (4 x))^2} \, dx \\ & = \frac {1}{20} \int \frac {\log (4 x) \left (x-\left (e^5+x\right ) \log \left (e^5+x\right )\right )+(-1+x) \left (x \left (-1+20 e^5+20 x\right )+2 \left (e^5+x\right ) \log \left (e^5+x\right )\right )}{x^2 \left (e^5+x\right ) (1-x+\log (4 x))^2} \, dx \\ & = \frac {1}{20} \int \left (\frac {1-20 e^5-21 \left (1-\frac {20 e^5}{21}\right ) x+20 x^2+\log (4 x)}{x \left (e^5+x\right ) (1-x+\log (4 x))^2}+\frac {(-2+2 x-\log (4 x)) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2}\right ) \, dx \\ & = \frac {1}{20} \int \frac {1-20 e^5-21 \left (1-\frac {20 e^5}{21}\right ) x+20 x^2+\log (4 x)}{x \left (e^5+x\right ) (1-x+\log (4 x))^2} \, dx+\frac {1}{20} \int \frac {(-2+2 x-\log (4 x)) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx \\ & = \frac {1}{20} \int \frac {(-1+x) \left (-1+20 e^5+20 x\right )+\log (4 x)}{x \left (e^5+x\right ) (1-x+\log (4 x))^2} \, dx+\frac {1}{20} \int \left (-\frac {2 \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2}+\frac {2 \log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2}-\frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2}\right ) \, dx \\ & = \frac {1}{20} \int \left (\frac {20 (-1+x)}{x (-1+x-\log (4 x))^2}-\frac {1}{x \left (e^5+x\right ) (-1+x-\log (4 x))}\right ) \, dx-\frac {1}{20} \int \frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx-\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx+\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2} \, dx \\ & = -\left (\frac {1}{20} \int \frac {1}{x \left (e^5+x\right ) (-1+x-\log (4 x))} \, dx\right )-\frac {1}{20} \int \frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx-\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx+\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2} \, dx+\int \frac {-1+x}{x (-1+x-\log (4 x))^2} \, dx \\ & = \frac {1}{1-x+\log (4 x)}-\frac {1}{20} \int \left (\frac {1}{e^5 x (-1+x-\log (4 x))}-\frac {1}{e^5 \left (e^5+x\right ) (-1+x-\log (4 x))}\right ) \, dx-\frac {1}{20} \int \frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx-\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx+\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2} \, dx \\ & = \frac {1}{1-x+\log (4 x)}-\frac {1}{20} \int \frac {\log (4 x) \log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx-\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x^2 (-1+x-\log (4 x))^2} \, dx+\frac {1}{10} \int \frac {\log \left (e^5+x\right )}{x (-1+x-\log (4 x))^2} \, dx-\frac {\int \frac {1}{x (-1+x-\log (4 x))} \, dx}{20 e^5}+\frac {\int \frac {1}{\left (e^5+x\right ) (-1+x-\log (4 x))} \, dx}{20 e^5} \\ \end{align*}
Time = 0.68 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx=\frac {20 x+\log \left (e^5+x\right )}{20 \left (x-x^2+x \log (4 x)\right )} \]
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Time = 11.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(-\frac {\ln \left ({\mathrm e}^{5}+x \right )+20 x}{20 x \left (x -\ln \left (4 x \right )-1\right )}\) | \(26\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{5}+x \right )}{20 x \left (x -\ln \left (4 x \right )-1\right )}-\frac {1}{x -\ln \left (4 x \right )-1}\) | \(36\) |
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx=-\frac {20 \, x + \log \left (x + e^{5}\right )}{20 \, {\left (x^{2} - x \log \left (4 \, x\right ) - x\right )}} \]
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Exception generated. \[ \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx=\text {Exception raised: TypeError} \]
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Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx=-\frac {20 \, x + \log \left (x + e^{5}\right )}{20 \, {\left (x^{2} - x {\left (2 \, \log \left (2\right ) + 1\right )} - x \log \left (x\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx=-\frac {20 \, x + \log \left (x + e^{5}\right )}{20 \, {\left ({\left (x + e^{5}\right )}^{2} - 2 \, {\left (x + e^{5}\right )} e^{5} - {\left (x + e^{5}\right )} \log \left (4 \, x\right ) + e^{5} \log \left (4 \, x\right ) - x + e^{10}\right )}} \]
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Time = 8.87 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {x-21 x^2+20 x^3+e^5 \left (-20 x+20 x^2\right )+x \log (4 x)+\left (-2 x+2 x^2+e^5 (-2+2 x)+\left (-e^5-x\right ) \log (4 x)\right ) \log \left (e^5+x\right )}{20 x^3-40 x^4+20 x^5+e^5 \left (20 x^2-40 x^3+20 x^4\right )+\left (40 x^3-40 x^4+e^5 \left (40 x^2-40 x^3\right )\right ) \log (4 x)+\left (20 e^5 x^2+20 x^3\right ) \log ^2(4 x)} \, dx=\frac {20\,x+\ln \left (x+{\mathrm {e}}^5\right )}{20\,x\,\left (\ln \left (4\,x\right )-x+1\right )} \]
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