Integrand size = 88, antiderivative size = 33 \[ \int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx=x \left (\frac {x}{2}+e^{-e^{2 x}-x^2} \left (e^5+x+\log (4 x)\right )\right ) \]
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\[ \int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx=\int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (e^{-e^{2 x}-x^2}+x+2 e^{-e^{2 x}-x^2} x-2 e^{-e^{2 x}-x^2} x^3-2 e^{-e^{2 x}+2 x-x^2} x \left (e^5+x\right )-e^{5-e^{2 x}-x^2} \left (-1+2 x^2\right )-e^{-e^{2 x}-x^2} \left (-1+2 e^{2 x} x+2 x^2\right ) \log (4 x)\right ) \, dx \\ & = \frac {x^2}{2}+2 \int e^{-e^{2 x}-x^2} x \, dx-2 \int e^{-e^{2 x}-x^2} x^3 \, dx-2 \int e^{-e^{2 x}+2 x-x^2} x \left (e^5+x\right ) \, dx+\int e^{-e^{2 x}-x^2} \, dx-\int e^{5-e^{2 x}-x^2} \left (-1+2 x^2\right ) \, dx-\int e^{-e^{2 x}-x^2} \left (-1+2 e^{2 x} x+2 x^2\right ) \log (4 x) \, dx \\ & = \frac {x^2}{2}+\frac {e^{-e^{2 x}-x^2} \left (e^{2 x} x+x^2\right ) \log (4 x)}{e^{2 x}+x}+2 \int e^{-e^{2 x}-x^2} x \, dx-2 \int e^{-e^{2 x}-x^2} x^3 \, dx-2 \int \left (e^{5-e^{2 x}+2 x-x^2} x+e^{-e^{2 x}+2 x-x^2} x^2\right ) \, dx-\int \left (-e^{5-e^{2 x}-x^2}+2 e^{5-e^{2 x}-x^2} x^2\right ) \, dx \\ & = \frac {x^2}{2}+\frac {e^{-e^{2 x}-x^2} \left (e^{2 x} x+x^2\right ) \log (4 x)}{e^{2 x}+x}+2 \int e^{-e^{2 x}-x^2} x \, dx-2 \int e^{5-e^{2 x}+2 x-x^2} x \, dx-2 \int e^{5-e^{2 x}-x^2} x^2 \, dx-2 \int e^{-e^{2 x}+2 x-x^2} x^2 \, dx-2 \int e^{-e^{2 x}-x^2} x^3 \, dx+\int e^{5-e^{2 x}-x^2} \, dx \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx=\frac {x^2}{2}+e^{-e^{2 x}} \left (e^{-x^2} \left (e^5 x+x^2\right )+e^{-x^2} x \log (4 x)\right ) \]
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Time = 0.46 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06
method | result | size |
risch | \(\frac {x^{2}}{2}+\left (x \,{\mathrm e}^{5}+x^{2}+x \ln \left (4 x \right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}-x^{2}}\) | \(35\) |
parallelrisch | \(\frac {\left ({\mathrm e}^{{\mathrm e}^{2 x}+x^{2}} x^{2}+2 x \,{\mathrm e}^{5}+2 x^{2}+2 x \ln \left (4 x \right )\right ) {\mathrm e}^{-{\mathrm e}^{2 x}-x^{2}}}{2}\) | \(45\) |
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Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx=\frac {1}{2} \, {\left (x^{2} e^{\left (x^{2} + e^{\left (2 \, x\right )}\right )} + 2 \, x^{2} + 2 \, x e^{5} + 2 \, x \log \left (4 \, x\right )\right )} e^{\left (-x^{2} - e^{\left (2 \, x\right )}\right )} \]
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Time = 4.98 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx=\frac {x^{2}}{2} + \left (x^{2} + x \log {\left (4 x \right )} + x e^{5}\right ) e^{- x^{2} - e^{2 x}} \]
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Time = 0.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx=\frac {1}{2} \, x^{2} + {\left (x^{2} + x {\left (e^{5} + 2 \, \log \left (2\right )\right )} + x \log \left (x\right )\right )} e^{\left (-x^{2} - e^{\left (2 \, x\right )}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (28) = 56\).
Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.76 \[ \int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx=x^{2} e^{\left (-x^{2} - e^{\left (2 \, x\right )}\right )} + x e^{\left (-x^{2} - e^{\left (2 \, x\right )}\right )} \log \left (4 \, x\right ) + \frac {1}{2} \, x^{2} + x e^{\left (-x^{2} - e^{\left (2 \, x\right )} + 5\right )} \]
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Timed out. \[ \int e^{-e^{2 x}-x^2} \left (1+2 x+e^{e^{2 x}+x^2} x-2 x^3+e^5 \left (1-2 x^2\right )+e^{2 x} \left (-2 e^5 x-2 x^2\right )+\left (1-2 e^{2 x} x-2 x^2\right ) \log (4 x)\right ) \, dx=\int -{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}-x^2}\,\left (\ln \left (4\,x\right )\,\left (2\,x\,{\mathrm {e}}^{2\,x}+2\,x^2-1\right )-2\,x+{\mathrm {e}}^5\,\left (2\,x^2-1\right )+{\mathrm {e}}^{2\,x}\,\left (2\,x^2+2\,{\mathrm {e}}^5\,x\right )-x\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}+x^2}+2\,x^3-1\right ) \,d x \]
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