Integrand size = 47, antiderivative size = 18 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=e^{\frac {4 e^{-x^2} (3+\log (5))}{x^3}} \]
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\[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=\int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-12 (3+\log (5))-8 x^2 (3+\log (5))\right )}{x^4} \, dx \\ & = \int \left (-\frac {12 e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} (3+\log (5))}{x^4}-\frac {8 e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} (3+\log (5))}{x^2}\right ) \, dx \\ & = -\left ((8 (3+\log (5))) \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}}}{x^2} \, dx\right )-(12 (3+\log (5))) \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}}}{x^4} \, dx \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=5^{\frac {4 e^{-x^2}}{x^3}} e^{\frac {12 e^{-x^2}}{x^3}} \]
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Time = 4.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94
method | result | size |
risch | \({\mathrm e}^{\frac {4 \left (\ln \left (5\right )+3\right ) {\mathrm e}^{-x^{2}}}{x^{3}}}\) | \(17\) |
norman | \({\mathrm e}^{\frac {2 \left (2 \ln \left (5\right )+6\right ) {\mathrm e}^{-x^{2}}}{x^{3}}}\) | \(20\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=e^{\left (x^{2} - \frac {x^{5} - 4 \, {\left (\log \left (5\right ) + 3\right )} e^{\left (-x^{2}\right )}}{x^{3}}\right )} \]
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Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=e^{\frac {2 \cdot \left (2 \log {\left (5 \right )} + 6\right ) e^{- x^{2}}}{x^{3}}} \]
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Time = 0.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=e^{\left (\frac {4 \, e^{\left (-x^{2}\right )} \log \left (5\right )}{x^{3}} + \frac {12 \, e^{\left (-x^{2}\right )}}{x^{3}}\right )} \]
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\[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=\int { -\frac {4 \, {\left (6 \, x^{2} + {\left (2 \, x^{2} + 3\right )} \log \left (5\right ) + 9\right )} e^{\left (-x^{2} + \frac {4 \, {\left (\log \left (5\right ) + 3\right )} e^{\left (-x^{2}\right )}}{x^{3}}\right )}}{x^{4}} \,d x } \]
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Time = 9.55 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=5^{\frac {4\,{\mathrm {e}}^{-x^2}}{x^3}}\,{\mathrm {e}}^{\frac {12\,{\mathrm {e}}^{-x^2}}{x^3}} \]
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