Integrand size = 93, antiderivative size = 31 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\frac {4 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)} \]
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\[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{3 \left (e^x-2 x\right )^2 x^2 \log ^2(x)} \, dx \\ & = \frac {1}{3} \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (e^x-2 x\right )^2 x^2 \log ^2(x)} \, dx \\ & = \frac {1}{3} \int \left (-\frac {24 e^{-\frac {x^2}{3}} (-1+x) \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)}-\frac {4 e^{-\frac {x^2}{3}} \left (-3 \log (x)+3 \log (2 x)+3 \log (x) \log (2 x)+3 x \log (x) \log (2 x)+2 x^2 \log (x) \log (2 x)\right )}{\left (e^x-2 x\right ) x^2 \log ^2(x)}\right ) \, dx \\ & = -\left (\frac {4}{3} \int \frac {e^{-\frac {x^2}{3}} \left (-3 \log (x)+3 \log (2 x)+3 \log (x) \log (2 x)+3 x \log (x) \log (2 x)+2 x^2 \log (x) \log (2 x)\right )}{\left (e^x-2 x\right ) x^2 \log ^2(x)} \, dx\right )-8 \int \frac {e^{-\frac {x^2}{3}} (-1+x) \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)} \, dx \\ & = -\left (\frac {4}{3} \int \frac {e^{-\frac {x^2}{3}} \left (3 \log (2 x)+\log (x) \left (-3+\left (3+3 x+2 x^2\right ) \log (2 x)\right )\right )}{\left (e^x-2 x\right ) x^2 \log ^2(x)} \, dx\right )-8 \int \left (\frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 \log (x)}-\frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)}\right ) \, dx \\ & = -\left (\frac {4}{3} \int \left (-\frac {3 e^{-\frac {x^2}{3}}}{\left (e^x-2 x\right ) x^2 \log (x)}+\frac {3 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x^2 \log ^2(x)}+\frac {2 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) \log (x)}+\frac {3 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x^2 \log (x)}+\frac {3 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)}\right ) \, dx\right )-8 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 \log (x)} \, dx+8 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)} \, dx \\ & = -\left (\frac {8}{3} \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) \log (x)} \, dx\right )+4 \int \frac {e^{-\frac {x^2}{3}}}{\left (e^x-2 x\right ) x^2 \log (x)} \, dx-4 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x^2 \log ^2(x)} \, dx-4 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x^2 \log (x)} \, dx-4 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)} \, dx-8 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 \log (x)} \, dx+8 \int \frac {e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right )^2 x \log (x)} \, dx \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\frac {4 e^{-\frac {x^2}{3}} \log (2 x)}{\left (e^x-2 x\right ) x \log (x)} \]
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Time = 20.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(-\frac {4 \ln \left (2 x \right ) {\mathrm e}^{-\frac {x^{2}}{3}}}{x \ln \left (x \right ) \left (2 x -{\mathrm e}^{x}\right )}\) | \(32\) |
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Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=-\frac {4 \, {\left (e^{\left (-\frac {1}{3} \, x^{2}\right )} \log \left (2\right ) + e^{\left (-\frac {1}{3} \, x^{2}\right )} \log \left (x\right )\right )}}{{\left (2 \, x^{2} - x e^{x}\right )} \log \left (x\right )} \]
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Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\frac {\left (- 4 \log {\left (x \right )} - 4 \log {\left (2 \right )}\right ) e^{- \frac {x^{2}}{3}}}{2 x^{2} \log {\left (x \right )} - x e^{x} \log {\left (x \right )}} \]
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Time = 0.35 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=-\frac {4 \, {\left (\log \left (2\right ) + \log \left (x\right )\right )} e^{\left (-\frac {1}{3} \, x^{2}\right )}}{2 \, x^{2} \log \left (x\right ) - x e^{x} \log \left (x\right )} \]
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Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=-\frac {4 \, {\left (e^{\left (-\frac {1}{3} \, x^{2} + x\right )} \log \left (2\right ) + e^{\left (-\frac {1}{3} \, x^{2} + x\right )} \log \left (x\right )\right )}}{2 \, x^{2} e^{x} \log \left (x\right ) - x e^{\left (2 \, x\right )} \log \left (x\right )} \]
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Timed out. \[ \int \frac {e^{-\frac {x^2}{3}} \left (\left (12 e^x-24 x\right ) \log (x)+\left (-12 e^x+24 x+\left (48 x+16 x^3+e^x \left (-12-12 x-8 x^2\right )\right ) \log (x)\right ) \log (2 x)\right )}{\left (3 e^{2 x} x^2-12 e^x x^3+12 x^4\right ) \log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^{-\frac {x^2}{3}}\,\left (\ln \left (2\,x\right )\,\left (24\,x-12\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (48\,x-{\mathrm {e}}^x\,\left (8\,x^2+12\,x+12\right )+16\,x^3\right )\right )-\ln \left (x\right )\,\left (24\,x-12\,{\mathrm {e}}^x\right )\right )}{{\ln \left (x\right )}^2\,\left (3\,x^2\,{\mathrm {e}}^{2\,x}-12\,x^3\,{\mathrm {e}}^x+12\,x^4\right )} \,d x \]
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