Integrand size = 98, antiderivative size = 33 \[ \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{3 x^5 \log ^2(x)} \, dx=\frac {4 e^x \left (2-e^{-x+\frac {5}{3 x^2 \log (x)}}-\log (x)\right )}{x^2} \]
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Time = 0.49 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 6874, 2326} \[ \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{3 x^5 \log ^2(x)} \, dx=\frac {4 e^x (2 x-x \log (x))}{x^3}-\frac {4 e^{\frac {5}{3 x^2 \log (x)}} (2 \log (x)+1)}{x^5 \left (\frac {1}{x^3 \log ^2(x)}+\frac {2}{x^3 \log (x)}\right ) \log ^2(x)} \]
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Rule 12
Rule 2326
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{x^5 \log ^2(x)} \, dx \\ & = \frac {1}{3} \int \left (-\frac {12 e^x (5-2 x-2 \log (x)+x \log (x))}{x^3}+\frac {4 e^{\frac {5}{3 x^2 \log (x)}} \left (5+10 \log (x)+6 x^2 \log ^2(x)\right )}{x^5 \log ^2(x)}\right ) \, dx \\ & = \frac {4}{3} \int \frac {e^{\frac {5}{3 x^2 \log (x)}} \left (5+10 \log (x)+6 x^2 \log ^2(x)\right )}{x^5 \log ^2(x)} \, dx-4 \int \frac {e^x (5-2 x-2 \log (x)+x \log (x))}{x^3} \, dx \\ & = -\frac {4 e^{\frac {5}{3 x^2 \log (x)}} (1+2 \log (x))}{x^5 \left (\frac {1}{x^3 \log ^2(x)}+\frac {2}{x^3 \log (x)}\right ) \log ^2(x)}+\frac {4 e^x (2 x-x \log (x))}{x^3} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{3 x^5 \log ^2(x)} \, dx=-\frac {4 \left (-2 e^x+e^{\frac {5}{3 x^2 \log (x)}}+e^x \log (x)\right )}{x^2} \]
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Time = 1.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {4 \,{\mathrm e}^{x} \ln \left (x \right )}{x^{2}}+\frac {8 \,{\mathrm e}^{x}}{x^{2}}-\frac {4 \,{\mathrm e}^{\frac {5}{3 x^{2} \ln \left (x \right )}}}{x^{2}}\) | \(33\) |
parallelrisch | \(\frac {-12 \,{\mathrm e}^{x} \ln \left (x \right )-12 \,{\mathrm e}^{x} {\mathrm e}^{-\frac {3 x^{3} \ln \left (x \right )-5}{3 \ln \left (x \right ) x^{2}}}+24 \,{\mathrm e}^{x}}{3 x^{2}}\) | \(40\) |
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Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{3 x^5 \log ^2(x)} \, dx=-\frac {4 \, {\left (e^{x} \log \left (x\right ) + e^{\left (x - \frac {3 \, x^{3} \log \left (x\right ) - 5}{3 \, x^{2} \log \left (x\right )}\right )} - 2 \, e^{x}\right )}}{x^{2}} \]
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Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{3 x^5 \log ^2(x)} \, dx=\frac {\left (8 - 4 \log {\left (x \right )}\right ) e^{x}}{x^{2}} - \frac {4 e^{x} e^{\frac {- x^{3} \log {\left (x \right )} + \frac {5}{3}}{x^{2} \log {\left (x \right )}}}}{x^{2}} \]
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Exception generated. \[ \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{3 x^5 \log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{3 x^5 \log ^2(x)} \, dx=-\frac {4 \, {\left (e^{x} \log \left (x\right ) - 2 \, e^{x} + e^{\left (\frac {5}{3 \, x^{2} \log \left (x\right )}\right )}\right )}}{x^{2}} \]
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Time = 10.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{3 x^5 \log ^2(x)} \, dx=\frac {8\,{\mathrm {e}}^x}{x^2}-\frac {4\,{\mathrm {e}}^{\frac {5}{3\,x^2\,\ln \left (x\right )}}}{x^2}-\frac {4\,{\mathrm {e}}^x\,\ln \left (x\right )}{x^2} \]
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