Integrand size = 108, antiderivative size = 25 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log \left (\frac {e^{x^2}}{5 x^2}-\log \left (4+e^{2 x}+x\right )\right ) \]
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\[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-5 x^3-10 e^{2 x} x^3-e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{x \left (4+e^{2 x}+x\right ) \left (e^{x^2}-5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx \\ & = \int \frac {-5 x^3-10 e^{2 x} x^3+2 e^{x^2} \left (4+e^{2 x}+x\right ) \left (-1+x^2\right )}{x \left (4+e^{2 x}+x\right ) \left (e^{x^2}-5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx \\ & = \int \left (\frac {2 \left (-1+x^2\right )}{x}-\frac {5 x \left (-x-2 e^{2 x} x-8 \log \left (4+e^{2 x}+x\right )-2 e^{2 x} \log \left (4+e^{2 x}+x\right )-2 x \log \left (4+e^{2 x}+x\right )+8 x^2 \log \left (4+e^{2 x}+x\right )+2 e^{2 x} x^2 \log \left (4+e^{2 x}+x\right )+2 x^3 \log \left (4+e^{2 x}+x\right )\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}\right ) \, dx \\ & = 2 \int \frac {-1+x^2}{x} \, dx-5 \int \frac {x \left (-x-2 e^{2 x} x-8 \log \left (4+e^{2 x}+x\right )-2 e^{2 x} \log \left (4+e^{2 x}+x\right )-2 x \log \left (4+e^{2 x}+x\right )+8 x^2 \log \left (4+e^{2 x}+x\right )+2 e^{2 x} x^2 \log \left (4+e^{2 x}+x\right )+2 x^3 \log \left (4+e^{2 x}+x\right )\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx \\ & = 2 \int \left (-\frac {1}{x}+x\right ) \, dx-5 \int \frac {x \left (\left (1+2 e^{2 x}\right ) x-2 \left (4+e^{2 x}+x\right ) \left (-1+x^2\right ) \log \left (4+e^{2 x}+x\right )\right )}{\left (4+e^{2 x}+x\right ) \left (e^{x^2}-5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx \\ & = x^2-2 \log (x)-5 \int \left (-\frac {x^2}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}-\frac {2 e^{2 x} x^2}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}-\frac {8 x \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}-\frac {2 e^{2 x} x \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}-\frac {2 x^2 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}+\frac {8 x^3 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}+\frac {2 e^{2 x} x^3 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}+\frac {2 x^4 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )}\right ) \, dx \\ & = x^2-2 \log (x)+5 \int \frac {x^2}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx+10 \int \frac {e^{2 x} x^2}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx+10 \int \frac {e^{2 x} x \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx+10 \int \frac {x^2 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx-10 \int \frac {e^{2 x} x^3 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx-10 \int \frac {x^4 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx+40 \int \frac {x \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx-40 \int \frac {x^3 \log \left (4+e^{2 x}+x\right )}{\left (4+e^{2 x}+x\right ) \left (-e^{x^2}+5 x^2 \log \left (4+e^{2 x}+x\right )\right )} \, dx \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=-2 x-2 \log (x)+\log \left (e^{2 x+x^2}-5 e^{2 x} x^2 \log \left (4+e^{2 x}+x\right )\right ) \]
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Time = 3.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80
method | result | size |
risch | \(\ln \left (\ln \left ({\mathrm e}^{2 x}+4+x \right )-\frac {{\mathrm e}^{x^{2}}}{5 x^{2}}\right )\) | \(20\) |
parallelrisch | \(-2 \ln \left (x \right )+\ln \left (\ln \left ({\mathrm e}^{2 x}+4+x \right ) x^{2}-\frac {{\mathrm e}^{x^{2}}}{5}\right )\) | \(26\) |
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Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log \left (\frac {5 \, x^{2} \log \left (x + e^{\left (2 \, x\right )} + 4\right ) - e^{\left (x^{2}\right )}}{x^{2}}\right ) \]
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Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log {\left (\log {\left (x + e^{2 x} + 4 \right )} - \frac {e^{x^{2}}}{5 x^{2}} \right )} \]
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Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log \left (\frac {5 \, x^{2} \log \left (x + e^{\left (2 \, x\right )} + 4\right ) - e^{\left (x^{2}\right )}}{5 \, x^{2}}\right ) \]
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Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log \left (-5 \, x^{2} \log \left (x + e^{\left (2 \, x\right )} + 4\right ) + e^{\left (x^{2}\right )}\right ) - 2 \, \log \left (x\right ) \]
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Time = 8.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\ln \left (5\,x^2\,\ln \left (x+{\mathrm {e}}^{2\,x}+4\right )-{\mathrm {e}}^{x^2}\right )+\ln \left (\frac {1}{x^2}\right ) \]
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