Integrand size = 120, antiderivative size = 24 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=-25+5 e^{-5+\frac {1}{\log (x (2+\log (x)))}}+x \left (x+x^2\right ) \]
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Time = 1.84 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2641, 6874, 6820, 45, 6838} \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^3+x^2+5 e^{\frac {1}{\log (x (\log (x)+2))}-5} \]
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Rule 45
Rule 2641
Rule 6820
Rule 6838
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) \left (-15-5 \log (x)+\exp \left (\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{x (2+\log (x)) \log ^2(2 x+x \log (x))} \, dx \\ & = \int \left (\exp \left (5-\frac {1}{\log (x (2+\log (x)))}-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) x (2+3 x)+\frac {5 \exp \left (-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) (-3-\log (x))}{x (2+\log (x)) \log ^2(2 x+x \log (x))}\right ) \, dx \\ & = 5 \int \frac {\exp \left (-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) (-3-\log (x))}{x (2+\log (x)) \log ^2(2 x+x \log (x))} \, dx+\int \exp \left (5-\frac {1}{\log (x (2+\log (x)))}-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) x (2+3 x) \, dx \\ & = 5 \int \frac {e^{-5+\frac {1}{\log (x (2+\log (x)))}} (-3-\log (x))}{x (2+\log (x)) \log ^2(2 x+x \log (x))} \, dx+\int x (2+3 x) \, dx \\ & = 5 e^{-5+\frac {1}{\log (x (2+\log (x)))}}+\int \left (2 x+3 x^2\right ) \, dx \\ & = 5 e^{-5+\frac {1}{\log (x (2+\log (x)))}}+x^2+x^3 \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=\frac {5 e^{\frac {1}{\log (x (2+\log (x)))}}+e^5 x^2 (1+x)}{e^5} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(113\) vs. \(2(27)=54\).
Time = 7.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.75
method | result | size |
parallelrisch | \(\frac {\left (4 \,{\mathrm e}^{\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}} x^{3} \ln \left (\left (\ln \left (x \right )+2\right ) x \right )+4 \ln \left (\left (\ln \left (x \right )+2\right ) x \right ) {\mathrm e}^{\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}} x^{2}+20 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )\right ) {\mathrm e}^{-\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}}}{4 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )}\) | \(114\) |
risch | \(x^{3}+x^{2}+5 \,{\mathrm e}^{-\frac {-5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{3}+5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i x \right )+5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )-5 i \pi \,\operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )+10 \ln \left (x \right )+10 \ln \left (\ln \left (x \right )+2\right )-2}{-i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{3}+i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )-i \pi \,\operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )+2 \ln \left (x \right )+2 \ln \left (\ln \left (x \right )+2\right )}}\) | \(208\) |
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Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (23) = 46\).
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.67 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx={\left ({\left (x^{3} + x^{2}\right )} e^{\left (\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )} + 5\right )} e^{\left (-\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )} \]
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Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^{3} + x^{2} + 5 e^{- \frac {5 \log {\left (x \log {\left (x \right )} + 2 x \right )} - 1}{\log {\left (x \log {\left (x \right )} + 2 x \right )}}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (23) = 46\).
Time = 0.40 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.46 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^{3} + x^{2} + \frac {5 \, e^{\left (\frac {1}{\log \left (x\right ) + \log \left (\log \left (x\right ) + 2\right )}\right )} \log \left (x\right )}{e^{5} \log \left (x\right ) + 3 \, e^{5}} + \frac {15 \, e^{\left (\frac {1}{\log \left (x\right ) + \log \left (\log \left (x\right ) + 2\right )}\right )}}{e^{5} \log \left (x\right ) + 3 \, e^{5}} \]
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\[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=\int { \frac {{\left ({\left (6 \, x^{3} + 4 \, x^{2} + {\left (3 \, x^{3} + 2 \, x^{2}\right )} \log \left (x\right )\right )} e^{\left (\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )} \log \left (x \log \left (x\right ) + 2 \, x\right )^{2} - 5 \, \log \left (x\right ) - 15\right )} e^{\left (-\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )}}{{\left (x \log \left (x\right ) + 2 \, x\right )} \log \left (x \log \left (x\right ) + 2 \, x\right )^{2}} \,d x } \]
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Time = 14.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^2+x^3+5\,{\mathrm {e}}^{\frac {1}{\ln \left (2\,x+x\,\ln \left (x\right )\right )}}\,{\mathrm {e}}^{-5} \]
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