Integrand size = 49, antiderivative size = 24 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=-2 e^{-x}+e^{9 e^{-2 x} x^2 \log ^2(x)} \]
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\[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=\int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (2 e^{-x}-18 e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) (-1-\log (x)+x \log (x))\right ) \, dx \\ & = 2 \int e^{-x} \, dx-18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) (-1-\log (x)+x \log (x)) \, dx \\ & = -2 e^{-x}-18 \int \left (-e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x)+e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} (-1+x) x \log ^2(x)\right ) \, dx \\ & = -2 e^{-x}+18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) \, dx-18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} (-1+x) x \log ^2(x) \, dx \\ & = -2 e^{-x}+18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) \, dx-18 \int \left (-e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log ^2(x)+e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x^2 \log ^2(x)\right ) \, dx \\ & = -2 e^{-x}+18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log (x) \, dx+18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x \log ^2(x) \, dx-18 \int e^{-2 x+9 e^{-2 x} x^2 \log ^2(x)} x^2 \log ^2(x) \, dx \\ \end{align*}
Time = 1.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=2 \left (-e^{-x}+\frac {1}{2} e^{9 e^{-2 x} x^2 \log ^2(x)}\right ) \]
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Time = 1.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-2 \,{\mathrm e}^{-x}+{\mathrm e}^{9 x^{2} \ln \left (x \right )^{2} {\mathrm e}^{-2 x}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{-x} \left (-2+{\mathrm e}^{9 x^{2} \ln \left (x \right )^{2} {\mathrm e}^{-2 x}} {\mathrm e}^{x}\right )\) | \(25\) |
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx={\left (e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2} + x\right )} - 2\right )} e^{\left (-x\right )} \]
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=e^{9 x^{2} e^{- 2 x} \log {\left (x \right )}^{2}} - 2 e^{- x} \]
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Time = 0.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2}\right )} - 2 \, e^{\left (-x\right )} \]
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\[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=\int { -2 \, {\left (9 \, {\left ({\left (x^{2} - x\right )} \log \left (x\right )^{2} - x \log \left (x\right )\right )} e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2}\right )} - e^{x}\right )} e^{\left (-2 \, x\right )} \,d x } \]
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Time = 11.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx={\mathrm {e}}^{9\,x^2\,{\mathrm {e}}^{-2\,x}\,{\ln \left (x\right )}^2}-2\,{\mathrm {e}}^{-x} \]
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