Integrand size = 136, antiderivative size = 28 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=x \left (\log (x)+\frac {x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}\right ) \]
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\[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (1+\log (x)+\frac {2 e^{2 x} x^5}{\left (x-\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}+\frac {5 x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}\right ) \, dx \\ & = x+2 \int \frac {e^{2 x} x^5}{\left (x-\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx+5 \int \frac {x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx+\int \log (x) \, dx \\ & = x \log (x)+2 \int \frac {e^{2 x} x^5}{\left (x-\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx+5 \int \frac {x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=x \log (x)+\frac {x^5}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \]
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Time = 26.72 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04
method | result | size |
risch | \(x \ln \left (x \right )+\frac {x^{5}}{\ln \left (2 \ln \left (2\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right )}\) | \(29\) |
parallelrisch | \(\frac {120 x^{5}+120 \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right ) x \ln \left (x \right )}{120 \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right )}\) | \(46\) |
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Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \]
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Timed out. \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\text {Timed out} \]
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Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \]
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \]
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Time = 9.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx=\frac {x^5}{\ln \left ({\mathrm {e}}^{2\,x}+\ln \left (4\right )\right )}+x\,\ln \left (x\right ) \]
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