Integrand size = 51, antiderivative size = 21 \[ \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx=\frac {8}{3} x \left (-\frac {e}{x}+x\right ) \log (4+2 x \log (x)) \]
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\[ \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx=\int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{3 (2+x \log (x))} \, dx \\ & = \frac {1}{3} \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{2+x \log (x)} \, dx \\ & = \frac {1}{3} \int \left (-\frac {8 \left (e-x^2\right ) (1+\log (x))}{2+x \log (x)}+16 x \log (2 (2+x \log (x)))\right ) \, dx \\ & = -\left (\frac {8}{3} \int \frac {\left (e-x^2\right ) (1+\log (x))}{2+x \log (x)} \, dx\right )+\frac {16}{3} \int x \log (2 (2+x \log (x))) \, dx \\ & = \frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {8}{3} \int \left (\frac {e-x^2}{x}-\frac {(-2+x) \left (-e+x^2\right )}{x (2+x \log (x))}\right ) \, dx-\frac {16}{3} \int \frac {x^2 (1+\log (x))}{2 (2+x \log (x))} \, dx \\ & = \frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {8}{3} \int \frac {e-x^2}{x} \, dx+\frac {8}{3} \int \frac {(-2+x) \left (-e+x^2\right )}{x (2+x \log (x))} \, dx-\frac {8}{3} \int \frac {x^2 (1+\log (x))}{2+x \log (x)} \, dx \\ & = \frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {8}{3} \int \left (\frac {e}{x}-x\right ) \, dx-\frac {8}{3} \int \left (x+\frac {(-2+x) x}{2+x \log (x)}\right ) \, dx+\frac {8}{3} \int \left (-\frac {e}{2+x \log (x)}+\frac {2 e}{x (2+x \log (x))}-\frac {2 x}{2+x \log (x)}+\frac {x^2}{2+x \log (x)}\right ) \, dx \\ & = -\frac {8}{3} e \log (x)+\frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {8}{3} \int \frac {(-2+x) x}{2+x \log (x)} \, dx+\frac {8}{3} \int \frac {x^2}{2+x \log (x)} \, dx-\frac {16}{3} \int \frac {x}{2+x \log (x)} \, dx-\frac {1}{3} (8 e) \int \frac {1}{2+x \log (x)} \, dx+\frac {1}{3} (16 e) \int \frac {1}{x (2+x \log (x))} \, dx \\ & = -\frac {8}{3} e \log (x)+\frac {8}{3} x^2 \log (2 (2+x \log (x)))+\frac {8}{3} \int \frac {x^2}{2+x \log (x)} \, dx-\frac {8}{3} \int \left (-\frac {2 x}{2+x \log (x)}+\frac {x^2}{2+x \log (x)}\right ) \, dx-\frac {16}{3} \int \frac {x}{2+x \log (x)} \, dx-\frac {1}{3} (8 e) \int \frac {1}{2+x \log (x)} \, dx+\frac {1}{3} (16 e) \int \frac {1}{x (2+x \log (x))} \, dx \\ & = -\frac {8}{3} e \log (x)+\frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {1}{3} (8 e) \int \frac {1}{2+x \log (x)} \, dx+\frac {1}{3} (16 e) \int \frac {1}{x (2+x \log (x))} \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx=8 \left (-\frac {1}{3} e \log (2+x \log (x))+\frac {1}{3} x^2 \log (4+2 x \log (x))\right ) \]
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Time = 0.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29
method | result | size |
norman | \(-\frac {8 \,{\mathrm e} \ln \left (2 x \ln \left (x \right )+4\right )}{3}+\frac {8 x^{2} \ln \left (2 x \ln \left (x \right )+4\right )}{3}\) | \(27\) |
parallelrisch | \(-\frac {8 \,{\mathrm e} \ln \left (2 x \ln \left (x \right )+4\right )}{3}+\frac {8 x^{2} \ln \left (2 x \ln \left (x \right )+4\right )}{3}\) | \(27\) |
risch | \(\frac {8 x^{2} \ln \left (2 x \ln \left (x \right )+4\right )}{3}-\frac {8 \,{\mathrm e} \ln \left (x \right )}{3}-\frac {8 \,{\mathrm e} \ln \left (\frac {2}{x}+\ln \left (x \right )\right )}{3}\) | \(34\) |
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx=\frac {8}{3} \, {\left (x^{2} - e\right )} \log \left (2 \, x \log \left (x\right ) + 4\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx=\frac {8 x^{2} \log {\left (2 x \log {\left (x \right )} + 4 \right )}}{3} - \frac {8 e \log {\left (x \right )}}{3} - \frac {8 e \log {\left (\log {\left (x \right )} + \frac {2}{x} \right )}}{3} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx=\frac {8}{3} \, x^{2} \log \left (2\right ) + \frac {8}{3} \, x^{2} \log \left (x \log \left (x\right ) + 2\right ) - \frac {8}{3} \, e \log \left (x\right ) - \frac {8}{3} \, e \log \left (\frac {x \log \left (x\right ) + 2}{x}\right ) \]
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Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx=\frac {8}{3} \, x^{2} \log \left (2 \, x \log \left (x\right ) + 4\right ) - \frac {8}{3} \, e \log \left (x \log \left (x\right ) + 2\right ) \]
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Time = 9.54 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx=\frac {8\,x^2\,\ln \left (2\,x\,\ln \left (x\right )+4\right )}{3}-\frac {8\,\mathrm {e}\,\ln \left (x\right )}{3}-\frac {8\,\mathrm {e}\,\ln \left (\frac {x\,\ln \left (x\right )+2}{x}\right )}{3} \]
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