Integrand size = 86, antiderivative size = 26 \[ \int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx=\frac {x (2+x) \left (1-x^4\right )}{2 (-2+x) (1+\log (x))} \]
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\[ \int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx=\int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 \left (x-4 x^4-x^5+x^6\right )+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{2 (2-x)^2 (1+\log (x))^2} \, dx \\ & = \frac {1}{2} \int \frac {-4 \left (x-4 x^4-x^5+x^6\right )+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{(2-x)^2 (1+\log (x))^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {-2-x+2 x^4+x^5}{(-2+x) (1+\log (x))^2}+\frac {-4-4 x+x^2+20 x^4+4 x^5-5 x^6}{(-2+x)^2 (1+\log (x))}\right ) \, dx \\ & = \frac {1}{2} \int \frac {-2-x+2 x^4+x^5}{(-2+x) (1+\log (x))^2} \, dx+\frac {1}{2} \int \frac {-4-4 x+x^2+20 x^4+4 x^5-5 x^6}{(-2+x)^2 (1+\log (x))} \, dx \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx=-\frac {x \left (-2-x+2 x^4+x^5\right )}{2 (-2+x) (1+\log (x))} \]
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Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
risch | \(-\frac {x \left (x^{5}+2 x^{4}-x -2\right )}{2 \left (-2+x \right ) \left (\ln \left (x \right )+1\right )}\) | \(28\) |
norman | \(\frac {x +\frac {1}{2} x^{2}-x^{5}-\frac {1}{2} x^{6}}{\left (\ln \left (x \right )+1\right ) \left (-2+x \right )}\) | \(30\) |
default | \(-\frac {x^{6}+2 x^{5}-x^{2}-2 x}{2 \left (-2+x \right ) \left (\ln \left (x \right )+1\right )}\) | \(31\) |
parallelrisch | \(\frac {-x^{6}-2 x^{5}+x^{2}+2 x}{2 \left (\ln \left (x \right )+1\right ) \left (-2+x \right )}\) | \(31\) |
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Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx=-\frac {x^{6} + 2 \, x^{5} - x^{2} - 2 \, x}{2 \, {\left ({\left (x - 2\right )} \log \left (x\right ) + x - 2\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx=\frac {- x^{6} - 2 x^{5} + x^{2} + 2 x}{2 x + \left (2 x - 4\right ) \log {\left (x \right )} - 4} \]
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Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx=-\frac {x^{6} + 2 \, x^{5} - x^{2} - 2 \, x}{2 \, {\left ({\left (x - 2\right )} \log \left (x\right ) + x - 2\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx=-\frac {x^{6} + 2 \, x^{5} - x^{2} - 2 \, x}{2 \, {\left (x \log \left (x\right ) + x - 2 \, \log \left (x\right ) - 2\right )}} \]
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Time = 8.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 4.08 \[ \int \frac {-4 x+16 x^4+4 x^5-4 x^6+\left (-4-4 x+x^2+20 x^4+4 x^5-5 x^6\right ) \log (x)}{8-8 x+2 x^2+\left (16-16 x+4 x^2\right ) \log (x)+\left (8-8 x+2 x^2\right ) \log ^2(x)} \, dx=\frac {60\,x}{x^2-4\,x+4}-\frac {31\,x}{2}-\frac {\frac {2\,x^2\,\left (-x^5+x^4+4\,x^3-1\right )}{{\left (x-2\right )}^2}-\frac {x\,\ln \left (x\right )\,\left (5\,x^6-4\,x^5-20\,x^4-x^2+4\,x+4\right )}{2\,{\left (x-2\right )}^2}}{\ln \left (x\right )+1}-16\,x^2-12\,x^3-8\,x^4-\frac {5\,x^5}{2} \]
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