Integrand size = 151, antiderivative size = 22 \[ \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\left (-x+\frac {x}{\log \left (e^{x^2}+2 x+\log (8)\right )}\right )^2 \]
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\[ \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x \left (1-\log \left (e^{x^2}+2 x+\log (8)\right )\right ) \left (-2 x \left (1+e^{x^2} x\right )+\left (e^{x^2}+2 x+\log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )-\left (e^{x^2}+2 x+\log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx \\ & = 2 \int \frac {x \left (1-\log \left (e^{x^2}+2 x+\log (8)\right )\right ) \left (-2 x \left (1+e^{x^2} x\right )+\left (e^{x^2}+2 x+\log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )-\left (e^{x^2}+2 x+\log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx \\ & = 2 \int \left (-\frac {2 x^2 \left (-1+2 x^2+x \log (8)\right ) \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}+\frac {x \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right ) \left (2 x^2-\log \left (e^{x^2}+2 x+\log (8)\right )+\log ^2\left (e^{x^2}+2 x+\log (8)\right )\right )}{\log ^3\left (e^{x^2}+2 x+\log (8)\right )}\right ) \, dx \\ & = 2 \int \frac {x \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right ) \left (2 x^2-\log \left (e^{x^2}+2 x+\log (8)\right )+\log ^2\left (e^{x^2}+2 x+\log (8)\right )\right )}{\log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx-4 \int \frac {x^2 \left (-1+2 x^2+x \log (8)\right ) \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx \\ & = 2 \int \left (x-\frac {2 x^3}{\log ^3\left (e^{x^2}+2 x+\log (8)\right )}+\frac {x \left (1+2 x^2\right )}{\log ^2\left (e^{x^2}+2 x+\log (8)\right )}-\frac {2 x}{\log \left (e^{x^2}+2 x+\log (8)\right )}\right ) \, dx-4 \int \left (-\frac {x^2 \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}+\frac {2 x^4 \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}+\frac {x^3 \log (8) \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}\right ) \, dx \\ & = x^2+2 \int \frac {x \left (1+2 x^2\right )}{\log ^2\left (e^{x^2}+2 x+\log (8)\right )} \, dx-4 \int \frac {x^3}{\log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx+4 \int \frac {x^2 \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx-4 \int \frac {x}{\log \left (e^{x^2}+2 x+\log (8)\right )} \, dx-8 \int \frac {x^4 \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx-(4 \log (8)) \int \frac {x^3 \left (-1+\log \left (e^{x^2}+2 x+\log (8)\right )\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx \\ & = x^2+2 \int \left (\frac {x}{\log ^2\left (e^{x^2}+2 x+\log (8)\right )}+\frac {2 x^3}{\log ^2\left (e^{x^2}+2 x+\log (8)\right )}\right ) \, dx+4 \int \left (-\frac {x^2}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}+\frac {x^2}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )}\right ) \, dx-4 \int \frac {x^3}{\log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx-4 \int \frac {x}{\log \left (e^{x^2}+2 x+\log (8)\right )} \, dx-8 \int \left (-\frac {x^4}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}+\frac {x^4}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )}\right ) \, dx-(4 \log (8)) \int \left (-\frac {x^3}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}+\frac {x^3}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )}\right ) \, dx \\ & = x^2+2 \int \frac {x}{\log ^2\left (e^{x^2}+2 x+\log (8)\right )} \, dx-4 \int \frac {x^3}{\log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx-4 \int \frac {x^2}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx+4 \int \frac {x^3}{\log ^2\left (e^{x^2}+2 x+\log (8)\right )} \, dx+4 \int \frac {x^2}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )} \, dx-4 \int \frac {x}{\log \left (e^{x^2}+2 x+\log (8)\right )} \, dx+8 \int \frac {x^4}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx-8 \int \frac {x^4}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )} \, dx+(4 \log (8)) \int \frac {x^3}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx-(4 \log (8)) \int \frac {x^3}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )} \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(50\) vs. \(2(22)=44\).
Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.27 \[ \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=2 \left (\frac {x^2}{2}+\frac {x^2}{2 \log ^2\left (e^{x^2}+2 x+\log (8)\right )}-\frac {x^2}{\log \left (e^{x^2}+2 x+\log (8)\right )}\right ) \]
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Time = 0.95 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91
method | result | size |
risch | \(x^{2}-\frac {\left (2 \ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right )-1\right ) x^{2}}{\ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right )^{2}}\) | \(42\) |
parallelrisch | \(\frac {\ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right )^{2} x^{2}-2 \ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right ) x^{2}+x^{2}}{\ln \left ({\mathrm e}^{x^{2}}+3 \ln \left (2\right )+2 x \right )^{2}}\) | \(58\) |
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (22) = 44\).
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \[ \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=x^{2} + \frac {- 2 x^{2} \log {\left (2 x + e^{x^{2}} + 3 \log {\left (2 \right )} \right )} + x^{2}}{\log {\left (2 x + e^{x^{2}} + 3 \log {\left (2 \right )} \right )}^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (22) = 44\).
Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \[ \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (22) = 44\).
Time = 0.71 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59 \[ \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=\frac {x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2} - 2 \, x^{2} \log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right ) + x^{2}}{\log \left (2 \, x + e^{\left (x^{2}\right )} + 3 \, \log \left (2\right )\right )^{2}} \]
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Timed out. \[ \int \frac {-4 x^2-4 e^{x^2} x^3+\left (8 x^2+e^{x^2} \left (2 x+4 x^3\right )+2 x \log (8)\right ) \log \left (e^{x^2}+2 x+\log (8)\right )+\left (-4 e^{x^2} x-8 x^2-4 x \log (8)\right ) \log ^2\left (e^{x^2}+2 x+\log (8)\right )+\left (2 e^{x^2} x+4 x^2+2 x \log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )}{\left (e^{x^2}+2 x+\log (8)\right ) \log ^3\left (e^{x^2}+2 x+\log (8)\right )} \, dx=-\int \frac {{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )}^2\,\left (4\,x\,{\mathrm {e}}^{x^2}+12\,x\,\ln \left (2\right )+8\,x^2\right )-{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )}^3\,\left (2\,x\,{\mathrm {e}}^{x^2}+6\,x\,\ln \left (2\right )+4\,x^2\right )+4\,x^3\,{\mathrm {e}}^{x^2}+4\,x^2-\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^{x^2}\,\left (4\,x^3+2\,x\right )+6\,x\,\ln \left (2\right )+8\,x^2\right )}{{\ln \left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )}^3\,\left (2\,x+{\mathrm {e}}^{x^2}+3\,\ln \left (2\right )\right )} \,d x \]
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