Integrand size = 107, antiderivative size = 29 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=6 x-\frac {x (2 x-\log (\log (2)))}{5 \log \left (\log \left (1+\frac {5}{x}\right )\right )} \]
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\[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=\int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\frac {5 (-2 x+\log (\log (2)))}{(5+x) \log \left (\frac {5+x}{x}\right )}+\log \left (\log \left (\frac {5+x}{x}\right )\right ) \left (-4 x+\log (\log (2))+30 \log \left (\log \left (\frac {5+x}{x}\right )\right )\right )}{5 \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx \\ & = \frac {1}{5} \int \frac {\frac {5 (-2 x+\log (\log (2)))}{(5+x) \log \left (\frac {5+x}{x}\right )}+\log \left (\log \left (\frac {5+x}{x}\right )\right ) \left (-4 x+\log (\log (2))+30 \log \left (\log \left (\frac {5+x}{x}\right )\right )\right )}{\log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx \\ & = \frac {1}{5} \int \left (30+\frac {5 (-2 x+\log (\log (2)))}{(5+x) \log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )}+\frac {-4 x+\log (\log (2))}{\log \left (\log \left (1+\frac {5}{x}\right )\right )}\right ) \, dx \\ & = 6 x+\frac {1}{5} \int \frac {-4 x+\log (\log (2))}{\log \left (\log \left (1+\frac {5}{x}\right )\right )} \, dx+\int \frac {-2 x+\log (\log (2))}{(5+x) \log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx \\ & = 6 x+\frac {1}{5} \int \left (-\frac {4 x}{\log \left (\log \left (1+\frac {5}{x}\right )\right )}+\frac {\log (\log (2))}{\log \left (\log \left (1+\frac {5}{x}\right )\right )}\right ) \, dx+\int \left (-\frac {2}{\log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )}+\frac {10+\log (\log (2))}{(5+x) \log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )}\right ) \, dx \\ & = 6 x-\frac {4}{5} \int \frac {x}{\log \left (\log \left (1+\frac {5}{x}\right )\right )} \, dx-2 \int \frac {1}{\log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx+\frac {1}{5} \log (\log (2)) \int \frac {1}{\log \left (\log \left (1+\frac {5}{x}\right )\right )} \, dx+(10+\log (\log (2))) \int \frac {1}{(5+x) \log \left (1+\frac {5}{x}\right ) \log ^2\left (\log \left (1+\frac {5}{x}\right )\right )} \, dx \\ \end{align*}
Time = 0.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=\frac {1}{5} x \left (30+\frac {-2 x+\log (\log (2))}{\log \left (\log \left (\frac {5+x}{x}\right )\right )}\right ) \]
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Time = 1.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66
method | result | size |
parallelrisch | \(\frac {x \ln \left (\ln \left (2\right )\right )-2 x^{2}+30 \ln \left (\ln \left (\frac {5+x}{x}\right )\right ) x -300 \ln \left (\ln \left (\frac {5+x}{x}\right )\right )}{5 \ln \left (\ln \left (\frac {5+x}{x}\right )\right )}\) | \(48\) |
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Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=-\frac {2 \, x^{2} - x \log \left (\log \left (2\right )\right ) - 30 \, x \log \left (\log \left (\frac {x + 5}{x}\right )\right )}{5 \, \log \left (\log \left (\frac {x + 5}{x}\right )\right )} \]
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Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=6 x + \frac {- 2 x^{2} + x \log {\left (\log {\left (2 \right )} \right )}}{5 \log {\left (\log {\left (\frac {x + 5}{x} \right )} \right )}} \]
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Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=-\frac {2 \, x^{2} - 30 \, x \log \left (\log \left (x + 5\right ) - \log \left (x\right )\right ) - x \log \left (\log \left (2\right )\right )}{5 \, \log \left (\log \left (x + 5\right ) - \log \left (x\right )\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (27) = 54\).
Time = 0.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=6 \, x - \frac {2 \, x^{2} \log \left (\frac {x + 5}{x}\right ) - x \log \left (\frac {x + 5}{x}\right ) \log \left (\log \left (2\right )\right )}{5 \, {\left (\log \left (x + 5\right ) \log \left (\log \left (\frac {x + 5}{x}\right )\right ) - \log \left (x\right ) \log \left (\log \left (\frac {x + 5}{x}\right )\right )\right )}} \]
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Time = 12.65 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-10 x+5 \log (\log (2))+\left (\left (-20 x-4 x^2\right ) \log \left (\frac {5+x}{x}\right )+(5+x) \log \left (\frac {5+x}{x}\right ) \log (\log (2))\right ) \log \left (\log \left (\frac {5+x}{x}\right )\right )+(150+30 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )}{(25+5 x) \log \left (\frac {5+x}{x}\right ) \log ^2\left (\log \left (\frac {5+x}{x}\right )\right )} \, dx=\frac {x\,\left (\ln \left (\ln \left (2\right )\right )-2\,x+30\,\ln \left (\ln \left (\frac {x+5}{x}\right )\right )\right )}{5\,\ln \left (\ln \left (\frac {x+5}{x}\right )\right )} \]
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