Integrand size = 76, antiderivative size = 23 \[ \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{128 \log (x)} \, dx=\left (-2+\frac {1}{16} e^{-x} x^5 \log \left (\frac {3}{\log (x)}\right )\right )^2 \]
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Time = 0.81 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 6820, 6874, 2326} \[ \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{128 \log (x)} \, dx=\frac {1}{256} e^{-2 x} x^{10} \log ^2\left (\frac {3}{\log (x)}\right )-\frac {1}{4} e^{-x} x^5 \log \left (\frac {3}{\log (x)}\right ) \]
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Rule 12
Rule 2326
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{128} \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{\log (x)} \, dx \\ & = \frac {1}{128} \int \frac {e^{-2 x} x^4 \left (32 e^x-x^5 \log \left (\frac {3}{\log (x)}\right )\right ) \left (1+(-5+x) \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)} \, dx \\ & = \frac {1}{128} \int \left (\frac {32 e^{-x} x^4 \left (1-5 \log (x) \log \left (\frac {3}{\log (x)}\right )+x \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)}-\frac {e^{-2 x} x^9 \log \left (\frac {3}{\log (x)}\right ) \left (1-5 \log (x) \log \left (\frac {3}{\log (x)}\right )+x \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)}\right ) \, dx \\ & = -\left (\frac {1}{128} \int \frac {e^{-2 x} x^9 \log \left (\frac {3}{\log (x)}\right ) \left (1-5 \log (x) \log \left (\frac {3}{\log (x)}\right )+x \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)} \, dx\right )+\frac {1}{4} \int \frac {e^{-x} x^4 \left (1-5 \log (x) \log \left (\frac {3}{\log (x)}\right )+x \log (x) \log \left (\frac {3}{\log (x)}\right )\right )}{\log (x)} \, dx \\ & = -\frac {1}{4} e^{-x} x^5 \log \left (\frac {3}{\log (x)}\right )+\frac {1}{256} e^{-2 x} x^{10} \log ^2\left (\frac {3}{\log (x)}\right ) \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{128 \log (x)} \, dx=\frac {1}{256} e^{-2 x} x^5 \log \left (\frac {3}{\log (x)}\right ) \left (-64 e^x+x^5 \log \left (\frac {3}{\log (x)}\right )\right ) \]
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Time = 4.70 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57
method | result | size |
parallelrisch | \(-\frac {\left (-\ln \left (\frac {3}{\ln \left (x \right )}\right )^{2} x^{10}+64 \ln \left (\frac {3}{\ln \left (x \right )}\right ) {\mathrm e}^{x} x^{5}\right ) {\mathrm e}^{-2 x}}{256}\) | \(36\) |
risch | \(\frac {x^{10} {\mathrm e}^{-2 x} \ln \left (\ln \left (x \right )\right )^{2}}{256}-\frac {x^{5} \left (2 x^{5} \ln \left (3\right )-64 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-2 x} \ln \left (\ln \left (x \right )\right )}{256}+\frac {x^{5} \left (4 x^{5} \ln \left (3\right )^{2}-256 \ln \left (3\right ) {\mathrm e}^{x}\right ) {\mathrm e}^{-2 x}}{1024}\) | \(65\) |
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Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{128 \log (x)} \, dx=\frac {1}{256} \, {\left (x^{10} \log \left (\frac {3}{\log \left (x\right )}\right )^{2} - 64 \, x^{5} e^{x} \log \left (\frac {3}{\log \left (x\right )}\right )\right )} e^{\left (-2 \, x\right )} \]
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Time = 49.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{128 \log (x)} \, dx=\frac {x^{10} e^{- 2 x} \log {\left (\frac {3}{\log {\left (x \right )}} \right )}^{2}}{256} - \frac {x^{5} e^{- x} \log {\left (\frac {3}{\log {\left (x \right )}} \right )}}{4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (21) = 42\).
Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.78 \[ \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{128 \log (x)} \, dx=\frac {1}{256} \, x^{10} e^{\left (-2 \, x\right )} \log \left (3\right )^{2} + \frac {1}{256} \, x^{10} e^{\left (-2 \, x\right )} \log \left (\log \left (x\right )\right )^{2} - \frac {1}{4} \, x^{5} e^{\left (-x\right )} \log \left (3\right ) - \frac {1}{128} \, {\left (x^{10} e^{\left (-2 \, x\right )} \log \left (3\right ) - 32 \, x^{5} e^{\left (-x\right )}\right )} \log \left (\log \left (x\right )\right ) \]
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\[ \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{128 \log (x)} \, dx=\int { \frac {{\left (32 \, x^{4} e^{x} - {\left (x^{10} - 5 \, x^{9}\right )} \log \left (x\right ) \log \left (\frac {3}{\log \left (x\right )}\right )^{2} - {\left (x^{9} - 32 \, {\left (x^{5} - 5 \, x^{4}\right )} e^{x} \log \left (x\right )\right )} \log \left (\frac {3}{\log \left (x\right )}\right )\right )} e^{\left (-2 \, x\right )}}{128 \, \log \left (x\right )} \,d x } \]
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Timed out. \[ \int \frac {e^{-2 x} \left (32 e^x x^4+\left (-x^9+e^x \left (-160 x^4+32 x^5\right ) \log (x)\right ) \log \left (\frac {3}{\log (x)}\right )+\left (5 x^9-x^{10}\right ) \log (x) \log ^2\left (\frac {3}{\log (x)}\right )\right )}{128 \log (x)} \, dx=\int \frac {{\mathrm {e}}^{-2\,x}\,\left (\frac {x^4\,{\mathrm {e}}^x}{4}-\frac {\ln \left (\frac {3}{\ln \left (x\right )}\right )\,\left (x^9+{\mathrm {e}}^x\,\ln \left (x\right )\,\left (160\,x^4-32\,x^5\right )\right )}{128}+\frac {{\ln \left (\frac {3}{\ln \left (x\right )}\right )}^2\,\ln \left (x\right )\,\left (5\,x^9-x^{10}\right )}{128}\right )}{\ln \left (x\right )} \,d x \]
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