Integrand size = 50, antiderivative size = 19 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\frac {2 x}{(2+\log (6)) \log \left (\frac {1}{1+x^2}\right )} \]
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\[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+\log (6)+x^2 (2+\log (6))\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx \\ & = \int \left (\frac {4 x^2}{\left (1+x^2\right ) (2+\log (6)) \log ^2\left (\frac {1}{1+x^2}\right )}+\frac {2}{(2+\log (6)) \log \left (\frac {1}{1+x^2}\right )}\right ) \, dx \\ & = \frac {2 \int \frac {1}{\log \left (\frac {1}{1+x^2}\right )} \, dx}{2+\log (6)}+\frac {4 \int \frac {x^2}{\left (1+x^2\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx}{2+\log (6)} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\frac {2 x}{(2+\log (6)) \log \left (\frac {1}{1+x^2}\right )} \]
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Time = 0.47 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05
method | result | size |
norman | \(\frac {2 x}{\left (\ln \left (6\right )+2\right ) \ln \left (\frac {1}{x^{2}+1}\right )}\) | \(20\) |
parallelrisch | \(\frac {2 x}{\left (\ln \left (6\right )+2\right ) \ln \left (\frac {1}{x^{2}+1}\right )}\) | \(20\) |
risch | \(\frac {2 x}{\left (\ln \left (2\right )+\ln \left (3\right )+2\right ) \ln \left (\frac {1}{x^{2}+1}\right )}\) | \(22\) |
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Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\frac {2 \, x}{{\left (\log \left (6\right ) + 2\right )} \log \left (\frac {1}{x^{2} + 1}\right )} \]
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Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=\frac {2 x}{\left (\log {\left (6 \right )} + 2\right ) \log {\left (\frac {1}{x^{2} + 1} \right )}} \]
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Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=-\frac {2 \, x}{{\left (\log \left (3\right ) + \log \left (2\right ) + 2\right )} \log \left (x^{2} + 1\right )} \]
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Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=-\frac {2 \, x}{\log \left (6\right ) \log \left (x^{2} + 1\right ) + 2 \, \log \left (x^{2} + 1\right )} \]
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Time = 0.36 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {4 x^2+\left (2+2 x^2\right ) \log \left (\frac {1}{1+x^2}\right )}{\left (2+2 x^2+\left (1+x^2\right ) \log (6)\right ) \log ^2\left (\frac {1}{1+x^2}\right )} \, dx=-\frac {2\,x}{\ln \left (x^2+1\right )\,\left (\ln \left (6\right )+2\right )} \]
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