Integrand size = 94, antiderivative size = 22 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (\frac {-2 x-\log \left (1+e^{4 x}\right )}{\log \left (x^2\right )}\right ) \]
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\[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (1+e^{4 x}\right ) x \left (2 x+\log \left (1+e^{4 x}\right )\right ) \log \left (x^2\right )} \, dx \\ & = \int \left (-\frac {4}{\left (1+e^{4 x}\right ) \left (2 x+\log \left (1+e^{4 x}\right )\right )}+\frac {2 \left (-2 x-\log \left (1+e^{4 x}\right )+3 x \log \left (x^2\right )\right )}{x \left (2 x+\log \left (1+e^{4 x}\right )\right ) \log \left (x^2\right )}\right ) \, dx \\ & = 2 \int \frac {-2 x-\log \left (1+e^{4 x}\right )+3 x \log \left (x^2\right )}{x \left (2 x+\log \left (1+e^{4 x}\right )\right ) \log \left (x^2\right )} \, dx-4 \int \frac {1}{\left (1+e^{4 x}\right ) \left (2 x+\log \left (1+e^{4 x}\right )\right )} \, dx \\ & = 2 \int \left (\frac {3}{2 x+\log \left (1+e^{4 x}\right )}-\frac {1}{x \log \left (x^2\right )}\right ) \, dx-2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )} \, dx,x,2 x\right ) \\ & = -\left (2 \int \frac {1}{x \log \left (x^2\right )} \, dx\right )-2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )} \, dx,x,2 x\right )+6 \int \frac {1}{2 x+\log \left (1+e^{4 x}\right )} \, dx \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )} \, dx,x,2 x\right )\right )+3 \text {Subst}\left (\int \frac {1}{x+\log \left (1+e^{2 x}\right )} \, dx,x,2 x\right )-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (x^2\right )\right ) \\ & = -\log \left (\log \left (x^2\right )\right )-2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )} \, dx,x,2 x\right )+3 \text {Subst}\left (\int \frac {1}{x+\log \left (1+e^{2 x}\right )} \, dx,x,2 x\right ) \\ \end{align*}
Time = 0.64 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (2 x+\log \left (1+e^{4 x}\right )\right )-\log \left (\log \left (x^2\right )\right ) \]
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Time = 0.80 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(-\ln \left (\ln \left (x^{2}\right )\right )+\ln \left (x +\frac {\ln \left ({\mathrm e}^{4 x}+1\right )}{2}\right )\) | \(21\) |
risch | \(-\ln \left (\ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right )}{4}\right )+\ln \left (\ln \left ({\mathrm e}^{4 x}+1\right )+2 x \right )\) | \(62\) |
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Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x^{2}\right )\right ) \]
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Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log {\left (2 x + \log {\left (e^{4 x} + 1 \right )} \right )} - \log {\left (\log {\left (x^{2} \right )} \right )} \]
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Time = 0.32 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x\right )\right ) \]
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Time = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x^{2}\right )\right ) \]
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Time = 12.72 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\ln \left (2\,x+\ln \left ({\mathrm {e}}^{4\,x}+1\right )\right )-\ln \left (\ln \left (x^2\right )\right ) \]
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