Integrand size = 137, antiderivative size = 24 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\frac {x \log (-8+x-x (4 x-\log (x)))}{e^2+x} \]
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\[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x \left (e^2+x\right ) (-1+4 x)+e^2 \left (8-x+4 x^2\right ) \log \left (-8+x-4 x^2+x \log (x)\right )-x \log (x) \left (e^2+x+e^2 \log \left (-8+x-4 x^2+x \log (x)\right )\right )}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )} \, dx \\ & = \int \left (\frac {2 x (-1+4 x)}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )}-\frac {e^2 x \log (x)}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )}-\frac {x^2 \log (x)}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )}+\frac {e^2 \log \left (-8+x-4 x^2+x \log (x)\right )}{\left (e^2+x\right )^2}\right ) \, dx \\ & = 2 \int \frac {x (-1+4 x)}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )} \, dx-e^2 \int \frac {x \log (x)}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )} \, dx+e^2 \int \frac {\log \left (-8+x-4 x^2+x \log (x)\right )}{\left (e^2+x\right )^2} \, dx-\int \frac {x^2 \log (x)}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )} \, dx \\ & = 2 \int \left (-\frac {4 \left (1+\frac {1}{4 e^2}\right ) e^2}{8-x+4 x^2-x \log (x)}+\frac {4 x}{8-x+4 x^2-x \log (x)}+\frac {e^2+4 e^4}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )}\right ) \, dx-e^2 \int \left (-\frac {1}{\left (e^2+x\right )^2}+\frac {8-x+4 x^2}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )}\right ) \, dx+e^2 \int \frac {\log \left (-8+x-4 x^2+x \log (x)\right )}{\left (e^2+x\right )^2} \, dx-\int \left (-\frac {x}{\left (e^2+x\right )^2}+\frac {x \left (8-x+4 x^2\right )}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )}\right ) \, dx \\ & = -\frac {e^2}{e^2+x}+8 \int \frac {x}{8-x+4 x^2-x \log (x)} \, dx-e^2 \int \frac {8-x+4 x^2}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )} \, dx+e^2 \int \frac {\log \left (-8+x-4 x^2+x \log (x)\right )}{\left (e^2+x\right )^2} \, dx-\left (2 \left (1+4 e^2\right )\right ) \int \frac {1}{8-x+4 x^2-x \log (x)} \, dx+\left (2 e^2 \left (1+4 e^2\right )\right ) \int \frac {1}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )} \, dx+\int \frac {x}{\left (e^2+x\right )^2} \, dx-\int \frac {x \left (8-x+4 x^2\right )}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )} \, dx \\ & = -\frac {e^2}{e^2+x}+8 \int \frac {x}{8-x+4 x^2-x \log (x)} \, dx-e^2 \int \left (\frac {4}{8-x+4 x^2-x \log (x)}+\frac {8+e^2+4 e^4}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )}+\frac {-1-8 e^2}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )}\right ) \, dx+e^2 \int \frac {\log \left (-8+x-4 x^2+x \log (x)\right )}{\left (e^2+x\right )^2} \, dx-\left (2 \left (1+4 e^2\right )\right ) \int \frac {1}{8-x+4 x^2-x \log (x)} \, dx+\left (2 e^2 \left (1+4 e^2\right )\right ) \int \frac {1}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )} \, dx+\int \left (-\frac {e^2}{\left (e^2+x\right )^2}+\frac {1}{e^2+x}\right ) \, dx-\int \left (-\frac {8 \left (1+\frac {1}{8 e^2}\right ) e^2}{8-x+4 x^2-x \log (x)}+\frac {4 x}{8-x+4 x^2-x \log (x)}-\frac {e^2 \left (8+e^2+4 e^4\right )}{\left (e^2+x\right )^2 \left (8-x+4 x^2-x \log (x)\right )}+\frac {2 \left (4+e^2+6 e^4\right )}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )}\right ) \, dx \\ & = \log \left (e^2+x\right )-4 \int \frac {x}{8-x+4 x^2-x \log (x)} \, dx+8 \int \frac {x}{8-x+4 x^2-x \log (x)} \, dx+e^2 \int \frac {\log \left (-8+x-4 x^2+x \log (x)\right )}{\left (e^2+x\right )^2} \, dx-\left (4 e^2\right ) \int \frac {1}{8-x+4 x^2-x \log (x)} \, dx-\left (2 \left (1+4 e^2\right )\right ) \int \frac {1}{8-x+4 x^2-x \log (x)} \, dx+\left (2 e^2 \left (1+4 e^2\right )\right ) \int \frac {1}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )} \, dx+\left (1+8 e^2\right ) \int \frac {1}{8-x+4 x^2-x \log (x)} \, dx+\left (e^2 \left (1+8 e^2\right )\right ) \int \frac {1}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )} \, dx-\left (2 \left (4+e^2+6 e^4\right )\right ) \int \frac {1}{\left (e^2+x\right ) \left (8-x+4 x^2-x \log (x)\right )} \, dx \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\log \left (8-x+4 x^2-x \log (x)\right )-\frac {e^2 \log \left (-8+x-4 x^2+x \log (x)\right )}{e^2+x} \]
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Time = 23.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {\ln \left (x \ln \left (x \right )-4 x^{2}+x -8\right ) x}{x +{\mathrm e}^{2}}\) | \(22\) |
risch | \(-\frac {{\mathrm e}^{2} \ln \left (x \ln \left (x \right )-4 x^{2}+x -8\right )}{x +{\mathrm e}^{2}}+\ln \left (x \right )+\ln \left (\ln \left (x \right )-\frac {4 x^{2}-x +8}{x}\right )\) | \(46\) |
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Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\frac {x \log \left (-4 \, x^{2} + x \log \left (x\right ) + x - 8\right )}{x + e^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (x \right )} + \frac {- 4 x^{2} + x - 8}{x} \right )} - \frac {e^{2} \log {\left (- 4 x^{2} + x \log {\left (x \right )} + x - 8 \right )}}{x + e^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=-\frac {e^{2} \log \left (-4 \, x^{2} + x \log \left (x\right ) + x - 8\right )}{x + e^{2}} + \log \left (x\right ) + \log \left (-\frac {4 \, x^{2} - x \log \left (x\right ) - x + 8}{x}\right ) \]
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Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\frac {x \log \left (-4 \, x^{2} + x \log \left (x\right ) + x - 8\right )}{x + e^{2}} \]
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Time = 15.77 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {2 x^2-8 x^3+e^2 \left (2 x-8 x^2\right )+\left (e^2 x+x^2\right ) \log (x)+\left (e^2 \left (-8+x-4 x^2\right )+e^2 x \log (x)\right ) \log \left (-8+x-4 x^2+x \log (x)\right )}{-8 x^2+x^3-4 x^4+e^4 \left (-8+x-4 x^2\right )+e^2 \left (-16 x+2 x^2-8 x^3\right )+\left (e^4 x+2 e^2 x^2+x^3\right ) \log (x)} \, dx=\ln \left (\frac {x+x\,\ln \left (x\right )-4\,x^2-8}{x}\right )+\ln \left (x\right )-\frac {\ln \left (x+x\,\ln \left (x\right )-4\,x^2-8\right )\,{\mathrm {e}}^2}{x+{\mathrm {e}}^2} \]
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