Integrand size = 89, antiderivative size = 24 \[ \int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx=\frac {x-x^2+\log (x)-\log (\log (x))}{8+2 x+\log (x)} \]
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\[ \int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx=\int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 (4+x)-x (1+2 x) \log ^2(x)+\log (x) \left (7+9 x-15 x^2-2 x^3+(1+2 x) \log (\log (x))\right )}{x \log (x) (8+2 x+\log (x))^2} \, dx \\ & = \int \left (\frac {9}{(8+2 x+\log (x))^2}+\frac {7}{x (8+2 x+\log (x))^2}-\frac {15 x}{(8+2 x+\log (x))^2}-\frac {2 x^2}{(8+2 x+\log (x))^2}-\frac {2 (4+x)}{x \log (x) (8+2 x+\log (x))^2}-\frac {(1+2 x) \log (x)}{(8+2 x+\log (x))^2}+\frac {(1+2 x) \log (\log (x))}{x (8+2 x+\log (x))^2}\right ) \, dx \\ & = -\left (2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx\right )-2 \int \frac {4+x}{x \log (x) (8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx-\int \frac {(1+2 x) \log (x)}{(8+2 x+\log (x))^2} \, dx+\int \frac {(1+2 x) \log (\log (x))}{x (8+2 x+\log (x))^2} \, dx \\ & = -\left (2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx\right )-2 \int \left (\frac {1}{4 x (4+x) \log (x)}-\frac {1}{2 x (8+2 x+\log (x))^2}-\frac {1}{4 x (4+x) (8+2 x+\log (x))}\right ) \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx-\int \left (-\frac {2 \left (4+9 x+2 x^2\right )}{(8+2 x+\log (x))^2}+\frac {1+2 x}{8+2 x+\log (x)}\right ) \, dx+\int \left (\frac {2 \log (\log (x))}{(8+2 x+\log (x))^2}+\frac {\log (\log (x))}{x (8+2 x+\log (x))^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx\right )+\frac {1}{2} \int \frac {1}{x (4+x) (8+2 x+\log (x))} \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \frac {4+9 x+2 x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \frac {1+2 x}{8+2 x+\log (x)} \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx\right )+\frac {1}{2} \int \left (\frac {1}{4 x (8+2 x+\log (x))}-\frac {1}{4 (4+x) (8+2 x+\log (x))}\right ) \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \left (\frac {4}{(8+2 x+\log (x))^2}+\frac {9 x}{(8+2 x+\log (x))^2}+\frac {2 x^2}{(8+2 x+\log (x))^2}\right ) \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \left (\frac {1}{8+2 x+\log (x)}+\frac {2 x}{8+2 x+\log (x)}\right ) \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx \\ & = \frac {1}{8} \int \frac {1}{x (8+2 x+\log (x))} \, dx-\frac {1}{8} \int \frac {1}{(4+x) (8+2 x+\log (x))} \, dx-\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx-2 \int \frac {x}{8+2 x+\log (x)} \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+4 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+8 \int \frac {1}{(8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+18 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \frac {1}{8+2 x+\log (x)} \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx=-\frac {8+x+x^2+\log (\log (x))}{8+2 x+\log (x)} \]
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Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {-8-x^{2}-x -\ln \left (\ln \left (x \right )\right )}{8+2 x +\ln \left (x \right )}\) | \(26\) |
risch | \(-\frac {\ln \left (\ln \left (x \right )\right )}{8+2 x +\ln \left (x \right )}-\frac {x^{2}+x +8}{8+2 x +\ln \left (x \right )}\) | \(33\) |
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx=-\frac {x^{2} + x + \log \left (\log \left (x\right )\right ) + 8}{2 \, x + \log \left (x\right ) + 8} \]
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Exception generated. \[ \int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx=\text {Exception raised: TypeError} \]
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Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx=-\frac {x^{2} + x + \log \left (\log \left (x\right )\right ) + 8}{2 \, x + \log \left (x\right ) + 8} \]
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Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx=-\frac {x^{2} + x + 8}{2 \, x + \log \left (x\right ) + 8} - \frac {\log \left (\log \left (x\right )\right )}{2 \, x + \log \left (x\right ) + 8} \]
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Timed out. \[ \int \frac {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx=\int -\frac {2\,x+{\ln \left (x\right )}^2\,\left (2\,x^2+x\right )-\ln \left (x\right )\,\left (-2\,x^3-15\,x^2+9\,x+7\right )-\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )\,\left (2\,x+1\right )+8}{x\,{\ln \left (x\right )}^3+\left (4\,x^2+16\,x\right )\,{\ln \left (x\right )}^2+\left (4\,x^3+32\,x^2+64\,x\right )\,\ln \left (x\right )} \,d x \]
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