Integrand size = 112, antiderivative size = 33 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {\left (x-(x-\log (2))^2 \log ^2(2)\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{10 x} \]
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\[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-x^2 \log ^2(2)-\log ^4(2)+x \left (1+2 \log ^3(2)\right )+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx \\ & = \int \frac {-x^2 \log ^2(2)-\log ^4(2)+x \left (1+2 \log ^3(2)\right )+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{x^2 (10+10 x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx \\ & = \int \left (\frac {-x^2 \log ^2(2)-\log ^4(2)+x \left (1+2 \log ^3(2)\right )}{10 x^2 (1+x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )}-\frac {(x-\log (2)) \log ^2(2) (x+\log (2)) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{10 x^2}\right ) \, dx \\ & = \frac {1}{10} \int \frac {-x^2 \log ^2(2)-\log ^4(2)+x \left (1+2 \log ^3(2)\right )}{x^2 (1+x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx-\frac {1}{10} \log ^2(2) \int \frac {(x-\log (2)) (x+\log (2)) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{x^2} \, dx \\ & = \frac {1}{10} \int \left (-\frac {\log ^4(2)}{x^2 \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )}+\frac {-1-\log ^2(2)-2 \log ^3(2)-\log ^4(2)}{(1+x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )}+\frac {1+2 \log ^3(2)+\log ^4(2)}{x \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )}\right ) \, dx-\frac {1}{10} \log ^2(2) \int \left (\log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )-\frac {\log ^2(2) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{10} \log ^2(2) \int \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right ) \, dx\right )-\frac {1}{10} \log ^4(2) \int \frac {1}{x^2 \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx+\frac {1}{10} \log ^4(2) \int \frac {\log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{x^2} \, dx+\frac {1}{10} \left (-1-\log ^2(2)-2 \log ^3(2)-\log ^4(2)\right ) \int \frac {1}{(1+x) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx+\frac {1}{10} \left (1+2 \log ^3(2)+\log ^4(2)\right ) \int \frac {1}{x \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=-\frac {\left (x^2 \log ^2(2)+\log ^4(2)-x \left (1+2 \log ^3(2)\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{10 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(69\) vs. \(2(31)=62\).
Time = 46.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.12
method | result | size |
parallelrisch | \(-\frac {\ln \left (2\right )^{4} \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )-2 \ln \left (2\right )^{3} x \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )+\ln \left (2\right )^{2} x^{2} \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )-x \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )}{10 x}\) | \(70\) |
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Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=-\frac {{\left (x^{2} \log \left (2\right )^{2} - 2 \, x \log \left (2\right )^{3} + \log \left (2\right )^{4} - x\right )} \log \left (\log \left (\log \left (\frac {x}{x + 1}\right )\right )\right )}{10 \, x} \]
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Time = 0.50 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {\left (2 \log {\left (2 \right )}^{3} + 1\right ) \log {\left (\log {\left (\log {\left (\frac {x}{x + 1} \right )} \right )} \right )}}{10} + \frac {\left (- x^{2} \log {\left (2 \right )}^{2} - \log {\left (2 \right )}^{4}\right ) \log {\left (\log {\left (\log {\left (\frac {x}{x + 1} \right )} \right )} \right )}}{10 x} \]
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Time = 0.33 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=-\frac {{\left (x^{2} \log \left (2\right )^{2} + \log \left (2\right )^{4} - {\left (2 \, \log \left (2\right )^{3} + 1\right )} x\right )} \log \left (\log \left (-\log \left (x + 1\right ) + \log \left (x\right )\right )\right )}{10 \, x} \]
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Time = 0.51 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {1}{10} \, {\left (2 \, \log \left (2\right )^{3} + 1\right )} \log \left (\log \left (-\log \left (x + 1\right ) + \log \left (x\right )\right )\right ) - \frac {1}{10} \, {\left (x \log \left (2\right )^{2} + \frac {\log \left (2\right )^{4}}{x}\right )} \log \left (\log \left (\log \left (\frac {x}{x + 1}\right )\right )\right ) \]
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Time = 16.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {\ln \left (\ln \left (\ln \left (\frac {x}{x+1}\right )\right )\right )\,\left (x-x^2\,{\ln \left (2\right )}^2+2\,x\,{\ln \left (2\right )}^3-{\ln \left (2\right )}^4\right )}{10\,x} \]
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