Integrand size = 99, antiderivative size = 21 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=\left (-2 x+\log \left (-2+e^x\right )\right ) \log \left (-x+4 x^2\right ) \]
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\[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=\int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-2+e^x\right ) (-1+8 x) \log \left (-2+e^x\right )-x \left (2 \left (-2+e^x\right ) (-1+8 x)+\left (-4+e^x\right ) (-1+4 x) \log (x (-1+4 x))\right )}{\left (2-e^x\right ) (1-4 x) x} \, dx \\ & = \int \left (\frac {2 \log (x (-1+4 x))}{-2+e^x}+\frac {2 x-16 x^2-\log \left (-2+e^x\right )+8 x \log \left (-2+e^x\right )+x \log (x (-1+4 x))-4 x^2 \log (x (-1+4 x))}{x (-1+4 x)}\right ) \, dx \\ & = 2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \frac {2 x-16 x^2-\log \left (-2+e^x\right )+8 x \log \left (-2+e^x\right )+x \log (x (-1+4 x))-4 x^2 \log (x (-1+4 x))}{x (-1+4 x)} \, dx \\ & = 2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \left (-\frac {(-1+8 x) \left (2 x-\log \left (-2+e^x\right )\right )}{x (-1+4 x)}-\log (x (-1+4 x))\right ) \, dx \\ & = 2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx-\int \frac {(-1+8 x) \left (2 x-\log \left (-2+e^x\right )\right )}{x (-1+4 x)} \, dx-\int \log (x (-1+4 x)) \, dx \\ & = -x \log (-((1-4 x) x))+2 \int 1 \, dx+2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \frac {1}{-1+4 x} \, dx-\int \left (\frac {2 (-1+8 x)}{-1+4 x}-\frac {(-1+8 x) \log \left (-2+e^x\right )}{x (-1+4 x)}\right ) \, dx \\ & = 2 x+\frac {1}{4} \log (1-4 x)-x \log (-((1-4 x) x))-2 \int \frac {-1+8 x}{-1+4 x} \, dx+2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \frac {(-1+8 x) \log \left (-2+e^x\right )}{x (-1+4 x)} \, dx \\ & = 2 x+\frac {1}{4} \log (1-4 x)-x \log (-((1-4 x) x))-2 \int \left (2+\frac {1}{-1+4 x}\right ) \, dx+2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+\int \left (\frac {\log \left (-2+e^x\right )}{x}+\frac {4 \log \left (-2+e^x\right )}{-1+4 x}\right ) \, dx \\ & = -2 x-\frac {1}{4} \log (1-4 x)-x \log (-((1-4 x) x))+2 \int \frac {\log (x (-1+4 x))}{-2+e^x} \, dx+4 \int \frac {\log \left (-2+e^x\right )}{-1+4 x} \, dx+\int \frac {\log \left (-2+e^x\right )}{x} \, dx \\ \end{align*}
Time = 0.63 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-2 x \log (x (-1+4 x))+\log \left (-2+e^x\right ) \log (x (-1+4 x)) \]
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Time = 1.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48
method | result | size |
parallelrisch | \(-2 \ln \left (4 x^{2}-x \right ) x +\ln \left ({\mathrm e}^{x}-2\right ) \ln \left (4 x^{2}-x \right )\) | \(31\) |
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-2 \, x \log \left (4 \, x^{2} - x\right ) + \log \left (4 \, x^{2} - x\right ) \log \left (e^{x} - 2\right ) \]
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Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=- 2 x \log {\left (4 x^{2} - x \right )} + \log {\left (4 x^{2} - x \right )} \log {\left (e^{x} - 2 \right )} \]
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Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-2 \, x \log \left (4 \, x - 1\right ) - 2 \, x \log \left (x\right ) + {\left (\log \left (4 \, x - 1\right ) + \log \left (x\right )\right )} \log \left (e^{x} - 2\right ) \]
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Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-2 \, x \log \left (4 \, x^{2} - x\right ) + \log \left (4 \, x^{2} - x\right ) \log \left (e^{x} - 2\right ) \]
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Time = 11.82 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-\ln \left (4\,x^2-x\right )\,\left (2\,x-\ln \left ({\mathrm {e}}^x-2\right )\right ) \]
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