Integrand size = 57, antiderivative size = 22 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=x+\frac {x}{16 \left (\frac {1}{x^2}+\frac {2}{x}\right ) \log ^4(x)} \]
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\[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{16 (1+2 x)^2 \log ^5(x)} \, dx \\ & = \frac {1}{16} \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{(1+2 x)^2 \log ^5(x)} \, dx \\ & = \frac {1}{16} \int \left (16-\frac {4 x^2}{(1+2 x) \log ^5(x)}+\frac {x^2 (3+4 x)}{(1+2 x)^2 \log ^4(x)}\right ) \, dx \\ & = x+\frac {1}{16} \int \frac {x^2 (3+4 x)}{(1+2 x)^2 \log ^4(x)} \, dx-\frac {1}{4} \int \frac {x^2}{(1+2 x) \log ^5(x)} \, dx \\ \end{align*}
Time = 2.68 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=x+\frac {x^3}{16 (1+2 x) \log ^4(x)} \]
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Time = 6.73 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86
method | result | size |
risch | \(x +\frac {x^{3}}{16 \left (1+2 x \right ) \ln \left (x \right )^{4}}\) | \(19\) |
norman | \(\frac {-\frac {\ln \left (x \right )^{4}}{2}+\frac {x^{3}}{16}+2 x^{2} \ln \left (x \right )^{4}}{\left (1+2 x \right ) \ln \left (x \right )^{4}}\) | \(34\) |
parallelrisch | \(\frac {64 x^{2} \ln \left (x \right )^{4}-16 \ln \left (x \right )^{4}+2 x^{3}}{32 \ln \left (x \right )^{4} \left (1+2 x \right )}\) | \(35\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {16 \, {\left (2 \, x^{2} + x\right )} \log \left (x\right )^{4} + x^{3}}{16 \, {\left (2 \, x + 1\right )} \log \left (x\right )^{4}} \]
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Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {x^{3}}{\left (32 x + 16\right ) \log {\left (x \right )}^{4}} + x \]
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Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {16 \, {\left (2 \, x^{2} + x\right )} \log \left (x\right )^{4} + x^{3}}{16 \, {\left (2 \, x + 1\right )} \log \left (x\right )^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {x^{3}}{16 \, {\left (2 \, x \log \left (x\right )^{4} + \log \left (x\right )^{4}\right )}} + x \]
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Time = 14.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-4 x^2-8 x^3+\left (3 x^2+4 x^3\right ) \log (x)+\left (16+64 x+64 x^2\right ) \log ^5(x)}{\left (16+64 x+64 x^2\right ) \log ^5(x)} \, dx=\frac {x^3}{16\,{\ln \left (x\right )}^4\,\left (2\,x+1\right )}+\frac {x\,\left (32\,x+16\right )}{16\,\left (2\,x+1\right )} \]
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