Integrand size = 69, antiderivative size = 27 \[ \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx=4+2 x-x^2-\log (4)-\frac {4}{x (x+4 \log (x))} \]
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\[ \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx=\int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^2 (x+4 \log (x))^2} \, dx \\ & = \int \left (-2 (-1+x)+\frac {4 (4+x)}{x^2 (x+4 \log (x))^2}+\frac {4}{x^2 (x+4 \log (x))}\right ) \, dx \\ & = -(1-x)^2+4 \int \frac {4+x}{x^2 (x+4 \log (x))^2} \, dx+4 \int \frac {1}{x^2 (x+4 \log (x))} \, dx \\ & = -(1-x)^2+4 \int \frac {1}{x^2 (x+4 \log (x))} \, dx+4 \int \left (\frac {4}{x^2 (x+4 \log (x))^2}+\frac {1}{x (x+4 \log (x))^2}\right ) \, dx \\ & = -(1-x)^2+4 \int \frac {1}{x (x+4 \log (x))^2} \, dx+4 \int \frac {1}{x^2 (x+4 \log (x))} \, dx+16 \int \frac {1}{x^2 (x+4 \log (x))^2} \, dx \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx=-2 \left (-x+\frac {x^2}{2}+\frac {2}{x (x+4 \log (x))}\right ) \]
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-x^{2}+2 x -\frac {4}{x \left (4 \ln \left (x \right )+x \right )}\) | \(23\) |
norman | \(\frac {-4+8 x^{2} \ln \left (x \right )+2 x^{3}-x^{4}-4 x^{3} \ln \left (x \right )}{x \left (4 \ln \left (x \right )+x \right )}\) | \(39\) |
default | \(-\frac {2 \left (2-x^{3}+\frac {x^{4}}{2}-4 x^{2} \ln \left (x \right )+2 x^{3} \ln \left (x \right )\right )}{x \left (4 \ln \left (x \right )+x \right )}\) | \(40\) |
parallelrisch | \(\frac {-4 x^{4}-16 x^{3} \ln \left (x \right )+8 x^{3}+32 x^{2} \ln \left (x \right )-16}{4 x \left (4 \ln \left (x \right )+x \right )}\) | \(40\) |
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Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx=-\frac {x^{4} - 2 \, x^{3} + 4 \, {\left (x^{3} - 2 \, x^{2}\right )} \log \left (x\right ) + 4}{x^{2} + 4 \, x \log \left (x\right )} \]
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Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx=- x^{2} + 2 x - \frac {4}{x^{2} + 4 x \log {\left (x \right )}} \]
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Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx=-\frac {x^{4} - 2 \, x^{3} + 4 \, {\left (x^{3} - 2 \, x^{2}\right )} \log \left (x\right ) + 4}{x^{2} + 4 \, x \log \left (x\right )} \]
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx=-x^{2} + 2 \, x - \frac {4}{x^{2} + 4 \, x \log \left (x\right )} \]
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Time = 8.95 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {16+8 x+2 x^4-2 x^5+\left (16+16 x^3-16 x^4\right ) \log (x)+\left (32 x^2-32 x^3\right ) \log ^2(x)}{x^4+8 x^3 \log (x)+16 x^2 \log ^2(x)} \, dx=2\,x-\frac {4}{x\,\left (x+4\,\ln \left (x\right )\right )}-x^2 \]
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