Integrand size = 85, antiderivative size = 24 \[ \int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx=\log \left (e^3+2 e^{4+\frac {x^2}{e^2}} x \left (3+x^2\right )\right ) \]
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\[ \int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx=\int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {4 e^2+x^2}{e^2}} x \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx \\ & = \int \frac {6 e^2+6 \left (2+e^2\right ) x^2+4 x^4}{e \left (e^{-\frac {x^2}{e^2}}+2 e x \left (3+x^2\right )\right )} \, dx \\ & = \frac {\int \frac {6 e^2+6 \left (2+e^2\right ) x^2+4 x^4}{e^{-\frac {x^2}{e^2}}+2 e x \left (3+x^2\right )} \, dx}{e} \\ & = \frac {\int \left (\frac {-3 e^2-3 \left (2+e^2\right ) x^2-2 x^4}{e x \left (3+x^2\right ) \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )}+\frac {3 e^2+3 \left (2+e^2\right ) x^2+2 x^4}{e x \left (3+x^2\right )}\right ) \, dx}{e} \\ & = \frac {\int \frac {-3 e^2-3 \left (2+e^2\right ) x^2-2 x^4}{x \left (3+x^2\right ) \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )} \, dx}{e^2}+\frac {\int \frac {3 e^2+3 \left (2+e^2\right ) x^2+2 x^4}{x \left (3+x^2\right )} \, dx}{e^2} \\ & = \frac {\text {Subst}\left (\int \frac {3 e^2+3 \left (2+e^2\right ) x+2 x^2}{x (3+x)} \, dx,x,x^2\right )}{2 e^2}+\frac {\int \frac {-3 e^2-3 \left (2+e^2\right ) x^2-2 x^4}{x \left (3+x^2\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx}{e^2} \\ & = \frac {\text {Subst}\left (\int \left (2+\frac {e^2}{x}+\frac {2 e^2}{3+x}\right ) \, dx,x,x^2\right )}{2 e^2}+\frac {\int \left (-\frac {e^2}{x \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )}-\frac {2 x}{1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3}-\frac {2 e^2 x}{\left (3+x^2\right ) \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )}\right ) \, dx}{e^2} \\ & = \frac {x^2}{e^2}+\log (x)+\log \left (3+x^2\right )-2 \int \frac {x}{\left (3+x^2\right ) \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )} \, dx-\frac {2 \int \frac {x}{1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3} \, dx}{e^2}-\int \frac {1}{x \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right )} \, dx \\ & = \frac {x^2}{e^2}+\log (x)+\log \left (3+x^2\right )-2 \int \frac {x}{\left (3+x^2\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx-\frac {2 \int \frac {x}{1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )} \, dx}{e^2}-\int \frac {1}{x \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx \\ & = \frac {x^2}{e^2}+\log (x)+\log \left (3+x^2\right )-2 \int \left (-\frac {1}{2 \left (i \sqrt {3}-x\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )}+\frac {1}{2 \left (i \sqrt {3}+x\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )}\right ) \, dx-\frac {2 \int \frac {x}{1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )} \, dx}{e^2}-\int \frac {1}{x \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx \\ & = \frac {x^2}{e^2}+\log (x)+\log \left (3+x^2\right )-\frac {2 \int \frac {x}{1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )} \, dx}{e^2}+\int \frac {1}{\left (i \sqrt {3}-x\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx-\int \frac {1}{x \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx-\int \frac {1}{\left (i \sqrt {3}+x\right ) \left (1+2 e^{1+\frac {x^2}{e^2}} x \left (3+x^2\right )\right )} \, dx \\ \end{align*}
Time = 7.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx=\log \left (1+6 e^{1+\frac {x^2}{e^2}} x+2 e^{1+\frac {x^2}{e^2}} x^3\right ) \]
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Time = 0.65 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62
method | result | size |
risch | \(-\ln \left (x \right )-4+\ln \left (x^{3}+3 x \right )+\ln \left (x \,{\mathrm e}^{{\mathrm e}^{-2} x^{2}+4}+\frac {{\mathrm e}^{3}}{2 x^{2}+6}\right )\) | \(39\) |
norman | \(\ln \left (2 \,{\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (x \right )+4 \,{\mathrm e}^{2}+x^{2}\right ) {\mathrm e}^{-2}} x^{2}+{\mathrm e}^{3}+6 \,{\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (x \right )+4 \,{\mathrm e}^{2}+x^{2}\right ) {\mathrm e}^{-2}}\right )\) | \(50\) |
parallelrisch | \(\ln \left ({\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (x \right )+4 \,{\mathrm e}^{2}+x^{2}\right ) {\mathrm e}^{-2}} x^{2}+\frac {{\mathrm e}^{3}}{2}+3 \,{\mathrm e}^{\left ({\mathrm e}^{2} \ln \left (x \right )+4 \,{\mathrm e}^{2}+x^{2}\right ) {\mathrm e}^{-2}}\right )\) | \(51\) |
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Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx=\log \left (x^{2} + 3\right ) + \log \left (\frac {2 \, {\left (x^{2} + 3\right )} e^{\left ({\left (x^{2} + e^{2} \log \left (x\right ) + 6 \, e^{2}\right )} e^{\left (-2\right )}\right )} + e^{5}}{x^{2} + 3}\right ) \]
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Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx=\log {\left (x^{2} + 3 \right )} + \log {\left (e^{\frac {x^{2} + e^{2} \log {\left (x \right )} + 4 e^{2}}{e^{2}}} + \frac {e^{3}}{2 x^{2} + 6} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx=\log \left (x^{2} + 3\right ) + \log \left (x\right ) + \log \left (\frac {2 \, {\left (x^{3} e + 3 \, x e\right )} e^{\left (x^{2} e^{\left (-2\right )}\right )} + 1}{2 \, {\left (x^{3} e + 3 \, x e\right )}}\right ) \]
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\[ \int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx=\int { \frac {2 \, {\left (2 \, x^{4} + 6 \, x^{2} + 3 \, {\left (x^{2} + 1\right )} e^{2}\right )} e^{\left ({\left (x^{2} + e^{2} \log \left (x\right ) + 4 \, e^{2}\right )} e^{\left (-2\right )}\right )}}{x e^{5} + 2 \, {\left (x^{3} + 3 \, x\right )} e^{\left ({\left (x^{2} + e^{2} \log \left (x\right ) + 4 \, e^{2}\right )} e^{\left (-2\right )} + 2\right )}} \,d x } \]
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Time = 12.69 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (12 x^2+4 x^4+e^2 \left (6+6 x^2\right )\right )}{e^5 x+e^{2+\frac {4 e^2+x^2+e^2 \log (x)}{e^2}} \left (6 x+2 x^3\right )} \, dx=\ln \left ({\mathrm {e}}^3+6\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,x^2+4}+2\,x^3\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,x^2+4}\right ) \]
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