Integrand size = 151, antiderivative size = 36 \[ \int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=-x+\frac {5 x}{-x+\frac {2+e^{6-x-x^2} x}{\log \left (x^2\right )}} \]
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\[ \int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 x (1+x)} \left (16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )\right )}{\left (2 e^{x+x^2}+e^6 x-e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx \\ & = \int \left (-1+\frac {5 e^{-6-x-x^2+2 x (1+x)} \left (2+x \log \left (x^2\right )+2 x^2 \log \left (x^2\right )\right )}{x}+\frac {5 e^{2 x (1+x)} \log \left (x^2\right ) \left (2-4 x^2+x^2 \log \left (x^2\right )+2 x^3 \log \left (x^2\right )\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2}+\frac {5 e^{-6+2 x (1+x)} \left (-4-4 x^2 \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )+2 x^3 \log ^2\left (x^2\right )\right )}{x \left (2 e^{x+x^2}+e^6 x-e^{x+x^2} x \log \left (x^2\right )\right )}\right ) \, dx \\ & = -x+5 \int \frac {e^{-6-x-x^2+2 x (1+x)} \left (2+x \log \left (x^2\right )+2 x^2 \log \left (x^2\right )\right )}{x} \, dx+5 \int \frac {e^{2 x (1+x)} \log \left (x^2\right ) \left (2-4 x^2+x^2 \log \left (x^2\right )+2 x^3 \log \left (x^2\right )\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx+5 \int \frac {e^{-6+2 x (1+x)} \left (-4-4 x^2 \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )+2 x^3 \log ^2\left (x^2\right )\right )}{x \left (2 e^{x+x^2}+e^6 x-e^{x+x^2} x \log \left (x^2\right )\right )} \, dx \\ & = -x-\frac {5 e^{-6-x-x^2+2 x (1+x)} \left (x \log \left (x^2\right )+2 x^2 \log \left (x^2\right )\right )}{x (1-2 (1+x))}+5 \int \frac {e^{-6+2 x+2 x^2} \left (-4-4 x^2 \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )+2 x^3 \log ^2\left (x^2\right )\right )}{x \left (2 e^{x+x^2}+e^6 x-e^{x+x^2} x \log \left (x^2\right )\right )} \, dx+5 \int \left (\frac {2 e^{2 x (1+x)} \log \left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2}-\frac {4 e^{2 x (1+x)} x^2 \log \left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2}+\frac {e^{2 x (1+x)} x^2 \log ^2\left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2}+\frac {2 e^{2 x (1+x)} x^3 \log ^2\left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2}\right ) \, dx \\ & = -x-\frac {5 e^{-6-x-x^2+2 x (1+x)} \left (x \log \left (x^2\right )+2 x^2 \log \left (x^2\right )\right )}{x (1-2 (1+x))}+5 \int \frac {e^{2 x (1+x)} x^2 \log ^2\left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx+5 \int \left (\frac {4 e^{-6+2 x+2 x^2}}{x \left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )}+\frac {4 e^{-6+2 x+2 x^2} x \log \left (x^2\right )}{-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )}-\frac {e^{-6+2 x+2 x^2} x \log ^2\left (x^2\right )}{-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )}-\frac {2 e^{-6+2 x+2 x^2} x^2 \log ^2\left (x^2\right )}{-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )}\right ) \, dx+10 \int \frac {e^{2 x (1+x)} \log \left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx+10 \int \frac {e^{2 x (1+x)} x^3 \log ^2\left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx-20 \int \frac {e^{2 x (1+x)} x^2 \log \left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx \\ & = -x-\frac {5 e^{-6-x-x^2+2 x (1+x)} \left (x \log \left (x^2\right )+2 x^2 \log \left (x^2\right )\right )}{x (1-2 (1+x))}+5 \int \frac {e^{2 x (1+x)} x^2 \log ^2\left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx-5 \int \frac {e^{-6+2 x+2 x^2} x \log ^2\left (x^2\right )}{-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )} \, dx+10 \int \frac {e^{2 x (1+x)} \log \left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx+10 \int \frac {e^{2 x (1+x)} x^3 \log ^2\left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx-10 \int \frac {e^{-6+2 x+2 x^2} x^2 \log ^2\left (x^2\right )}{-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )} \, dx-20 \int \frac {e^{2 x (1+x)} x^2 \log \left (x^2\right )}{\left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )^2} \, dx+20 \int \frac {e^{-6+2 x+2 x^2}}{x \left (-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )\right )} \, dx+20 \int \frac {e^{-6+2 x+2 x^2} x \log \left (x^2\right )}{-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )} \, dx \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.44 \[ \int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=-x-\frac {5 \left (2 e^{x+x^2}+e^6 x\right )}{-2 e^{x+x^2}-e^6 x+e^{x+x^2} x \log \left (x^2\right )} \]
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Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.69
method | result | size |
norman | \(\frac {x^{2} \ln \left (x^{2}\right )+5 x \ln \left (x^{2}\right )-2 x -x^{2} {\mathrm e}^{-x^{2}-x +6}}{x \,{\mathrm e}^{-x^{2}-x +6}-x \ln \left (x^{2}\right )+2}\) | \(61\) |
parallelrisch | \(\frac {-2 x^{2} \ln \left (x^{2}\right )+2 x^{2} {\mathrm e}^{-x^{2}-x +6}-10 x \ln \left (x^{2}\right )+4 x}{2 x \ln \left (x^{2}\right )-2 x \,{\mathrm e}^{-x^{2}-x +6}-4}\) | \(63\) |
default | \(\frac {x^{2} \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )+\left (5 \ln \left (x^{2}\right )-10 \ln \left (x \right )-2\right ) x +10 x \ln \left (x \right )-x^{2} {\mathrm e}^{-x^{2}-x +6}+2 x^{2} \ln \left (x \right )}{x \,{\mathrm e}^{-x^{2}-x +6}-2 x \ln \left (x \right )-x \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )+2}\) | \(92\) |
risch | \(-x +\frac {20+10 x \,{\mathrm e}^{-\left (3+x \right ) \left (-2+x \right )}}{i x \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-2 i x \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 x \,{\mathrm e}^{-\left (3+x \right ) \left (-2+x \right )}-4 x \ln \left (x \right )+4}\) | \(93\) |
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Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.61 \[ \int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\frac {x^{2} \log \left (x^{2}\right ) - {\left (x^{2} - 5 \, x\right )} e^{\left (-x^{2} - x + 6\right )} - 2 \, x + 10}{x e^{\left (-x^{2} - x + 6\right )} - x \log \left (x^{2}\right ) + 2} \]
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Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=- x + \frac {5 x \log {\left (x^{2} \right )}}{x e^{- x^{2} - x + 6} - x \log {\left (x^{2} \right )} + 2} \]
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Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.50 \[ \int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=-\frac {x^{2} e^{6} - 5 \, x e^{6} - 2 \, {\left (x^{2} \log \left (x\right ) - x + 5\right )} e^{\left (x^{2} + x\right )}}{x e^{6} - 2 \, {\left (x \log \left (x\right ) - 1\right )} e^{\left (x^{2} + x\right )}} \]
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Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.69 \[ \int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=-\frac {x^{2} e^{\left (-x^{2} - x + 6\right )} - x^{2} \log \left (x^{2}\right ) - 5 \, x \log \left (x^{2}\right ) + 2 \, x}{x e^{\left (-x^{2} - x + 6\right )} - x \log \left (x^{2}\right ) + 2} \]
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Time = 7.70 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.89 \[ \int \frac {16+6 e^{6-x-x^2} x-e^{12-2 x-2 x^2} x^2+\left (10+4 x+e^{6-x-x^2} \left (7 x^2+10 x^3\right )\right ) \log \left (x^2\right )-x^2 \log ^2\left (x^2\right )}{4+4 e^{6-x-x^2} x+e^{12-2 x-2 x^2} x^2+\left (-4 x-2 e^{6-x-x^2} x^2\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=-\frac {x\,\left (x+2\,{\mathrm {e}}^{x^2+x-6}-5\,\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+x-6}-x\,\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+x-6}\right )}{x+2\,{\mathrm {e}}^{x^2+x-6}-x\,\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+x-6}} \]
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