Integrand size = 55, antiderivative size = 25 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=5+e^{\frac {x^2}{2}+\frac {15 \left (e^3-x\right )}{\log (x)}} \]
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\[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=\int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x-\frac {15 e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (e^3-x\right )}{x \log ^2(x)}-\frac {15 e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)}\right ) \, dx \\ & = -\left (15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (e^3-x\right )}{x \log ^2(x)} \, dx\right )-15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)} \, dx+\int e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x \, dx \\ & = -\left (15 \int \left (-\frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log ^2(x)}+\frac {e^{3+\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{x \log ^2(x)}\right ) \, dx\right )-15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)} \, dx+\int e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x \, dx \\ & = 15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log ^2(x)} \, dx-15 \int \frac {e^{3+\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{x \log ^2(x)} \, dx-15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)} \, dx+\int e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x \, dx \\ & = 15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log ^2(x)} \, dx-15 \int \frac {e^{\frac {30 e^3-30 x+6 \log (x)+x^2 \log (x)}{2 \log (x)}}}{x \log ^2(x)} \, dx-15 \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}}}{\log (x)} \, dx+\int e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} x \, dx \\ \end{align*}
Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\frac {x^2}{2}+\frac {15 \left (e^3-x\right )}{\log (x)}} \]
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Time = 0.90 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\) | \(22\) |
default | \({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\) | \(22\) |
norman | \({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\) | \(22\) |
risch | \({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{\frac {x^{2} \ln \left (x \right )+30 \,{\mathrm e}^{3}-30 x}{2 \ln \left (x \right )}}\) | \(22\) |
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\left (\frac {x^{2} \log \left (x\right ) - 30 \, x + 30 \, e^{3}}{2 \, \log \left (x\right )}\right )} \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\frac {\frac {x^{2} \log {\left (x \right )}}{2} - 15 x + 15 e^{3}}{\log {\left (x \right )}}} \]
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\left (\frac {1}{2} \, x^{2} - \frac {15 \, x}{\log \left (x\right )} + \frac {15 \, e^{3}}{\log \left (x\right )}\right )} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx=e^{\left (\frac {x^{2} \log \left (x\right ) - 30 \, x + 30 \, e^{3}}{2 \, \log \left (x\right )}\right )} \]
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Time = 7.77 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {30 e^3-30 x+x^2 \log (x)}{2 \log (x)}} \left (-15 e^3+15 x-15 x \log (x)+x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx={\mathrm {e}}^{\frac {15\,{\mathrm {e}}^3}{\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {15\,x}{\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {x^2}{2}} \]
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