Integrand size = 47, antiderivative size = 24 \[ \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx=\log \left (2+\frac {1}{2 x}+2 x+\log (4)+\frac {1}{8} x \log \left (x^2\right )\right ) \]
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\[ \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx=\int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^3+x^2 (16+8 \log (4))+x^3 \log \left (x^2\right )} \, dx \\ & = \int \left (\frac {1}{x}+\frac {2 \left (-4+x^2-4 x (2+\log (4))\right )}{x \left (4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )\right )}\right ) \, dx \\ & = \log (x)+2 \int \frac {-4+x^2-4 x (2+\log (4))}{x \left (4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )\right )} \, dx \\ & = \log (x)+2 \int \left (\frac {4}{x \left (-4-16 x^2-16 x (1+\log (2))-x^2 \log \left (x^2\right )\right )}+\frac {x}{4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )}+\frac {4 (-2-\log (4))}{4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )}\right ) \, dx \\ & = \log (x)+2 \int \frac {x}{4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )} \, dx+8 \int \frac {1}{x \left (-4-16 x^2-16 x (1+\log (2))-x^2 \log \left (x^2\right )\right )} \, dx-(8 (2+\log (4))) \int \frac {1}{4+16 x^2+16 x (1+\log (2))+x^2 \log \left (x^2\right )} \, dx \\ \end{align*}
Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx=-\log (x)+\log \left (4+16 x+16 x^2+8 x \log (4)+x^2 \log \left (x^2\right )\right ) \]
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Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25
method | result | size |
risch | \(\ln \left (x \right )+\ln \left (\ln \left (x^{2}\right )+\frac {16 x \ln \left (2\right )+16 x^{2}+16 x +4}{x^{2}}\right )\) | \(30\) |
norman | \(-\frac {\ln \left (x^{2}\right )}{2}+\ln \left (x^{2} \ln \left (x^{2}\right )+16 x \ln \left (2\right )+16 x^{2}+16 x +4\right )\) | \(32\) |
parallelrisch | \(-\frac {\ln \left (x^{2}\right )}{2}+\ln \left (x^{2} \ln \left (x^{2}\right )+16 x \ln \left (2\right )+16 x^{2}+16 x +4\right )\) | \(32\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx=\frac {1}{2} \, \log \left (x^{2}\right ) + \log \left (\frac {x^{2} \log \left (x^{2}\right ) + 16 \, x^{2} + 16 \, x \log \left (2\right ) + 16 \, x + 4}{x^{2}}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (x^{2} \right )} + \frac {16 x^{2} + 16 x \log {\left (2 \right )} + 16 x + 4}{x^{2}} \right )} \]
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Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx=\log \left (x\right ) + \log \left (\frac {x^{2} \log \left (x\right ) + 8 \, x^{2} + 8 \, x {\left (\log \left (2\right ) + 1\right )} + 2}{x^{2}}\right ) \]
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx=\log \left (x^{2} \log \left (x^{2}\right ) + 16 \, x^{2} + 16 \, x \log \left (2\right ) + 16 \, x + 4\right ) - \log \left (x\right ) \]
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Time = 8.84 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-4+18 x^2+x^2 \log \left (x^2\right )}{4 x+16 x^2+16 x^3+8 x^2 \log (4)+x^3 \log \left (x^2\right )} \, dx=\ln \left (16\,x+16\,x\,\ln \left (2\right )+x^2\,\ln \left (x^2\right )+16\,x^2+4\right )-\frac {\ln \left (x^2\right )}{2} \]
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