Integrand size = 72, antiderivative size = 36 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=x-\frac {e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (x+\frac {e^{-x} x}{3}\right )}{x} \]
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\[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=\int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} \int e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx \\ & = \frac {1}{6} \int \left (6+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \left (2-5 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}-15 e^{\frac {e^{x^2}}{2}+\frac {3 x}{2}}-10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}+x^2} x-30 e^{\frac {e^{x^2}}{2}+\frac {3 x}{2}+x^2} x\right )\right ) \, dx \\ & = x+\frac {1}{6} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \left (2-5 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}-15 e^{\frac {e^{x^2}}{2}+\frac {3 x}{2}}-10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}+x^2} x-30 e^{\frac {e^{x^2}}{2}+\frac {3 x}{2}+x^2} x\right ) \, dx \\ & = x+\frac {1}{6} \int \left (2 e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x}-5 e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+x\right )}-15 \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+3 x\right )\right )-10 \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+x+2 x^2\right )\right ) x-30 \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+3 x+2 x^2\right )\right ) x\right ) \, dx \\ & = x+\frac {1}{3} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \, dx-\frac {5}{6} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+x\right )} \, dx-\frac {5}{3} \int \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+x+2 x^2\right )\right ) x \, dx-\frac {5}{2} \int \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+3 x\right )\right ) \, dx-5 \int \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+3 x+2 x^2\right )\right ) x \, dx \\ & = x+\frac {1}{3} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \, dx-\frac {5}{6} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}-x\right )} \, dx-\frac {5}{3} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}-x+2 x^2\right )} x \, dx-\frac {5}{2} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}+x\right )} \, dx-5 \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}+x+2 x^2\right )} x \, dx \\ & = x+\frac {1}{3} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \, dx-\frac {5}{3} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}-x+2 x^2\right )} x \, dx-\frac {5}{3} \text {Subst}\left (\int e^{\frac {e^{4 x^2}}{2}+5 e^{\frac {e^{4 x^2}}{2}+x}-x} \, dx,x,\frac {x}{2}\right )-5 \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}+x+2 x^2\right )} x \, dx-5 \text {Subst}\left (\int e^{\frac {e^{4 x^2}}{2}+5 e^{\frac {e^{4 x^2}}{2}+x}+x} \, dx,x,\frac {x}{2}\right ) \\ & = x+\frac {1}{3} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \, dx-\frac {5}{3} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}-x+2 x^2\right )} x \, dx-\frac {5}{3} \text {Subst}\left (\int e^{\frac {1}{2} \left (e^{4 x^2}+10 e^{\frac {e^{4 x^2}}{2}+x}-2 x\right )} \, dx,x,\frac {x}{2}\right )-5 \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}+x+2 x^2\right )} x \, dx-5 \text {Subst}\left (\int e^{\frac {1}{2} \left (e^{4 x^2}+10 e^{\frac {e^{4 x^2}}{2}+x}+2 x\right )} \, dx,x,\frac {x}{2}\right ) \\ \end{align*}
Time = 3.92 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=\frac {1}{6} e^{5 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}} \left (-6-2 e^{-x}\right )+x \]
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Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81
method | result | size |
risch | \(x -\frac {\left (1+3 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x +5 \,{\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}+\frac {x}{2}}}}{3}\) | \(29\) |
parallelrisch | \(-\frac {\left (-6 \ln \left ({\mathrm e}^{x}\right ) {\mathrm e}^{x}+6 \,{\mathrm e}^{5 \,{\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}+\frac {x}{2}}} {\mathrm e}^{x}+2 \,{\mathrm e}^{5 \,{\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}+\frac {x}{2}}}\right ) {\mathrm e}^{-x}}{6}\) | \(49\) |
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Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=\frac {1}{3} \, {\left (3 \, x e^{x} - {\left (3 \, e^{x} + 1\right )} e^{\left (5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )}\right )} e^{\left (-x\right )} \]
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Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=x + \frac {\left (- 3 e^{x} - 1\right ) e^{- x} e^{5 e^{\frac {x}{2} + \frac {e^{x^{2}}}{2}}}}{3} \]
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=-\frac {1}{3} \, {\left (3 \, e^{x} + 1\right )} e^{\left (-x + 5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )} + x \]
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\[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=\int { -\frac {1}{6} \, {\left ({\left (5 \, {\left (2 \, {\left (3 \, x e^{x} + x\right )} e^{\left (x^{2}\right )} + 3 \, e^{x} + 1\right )} e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )} - 2\right )} e^{\left (5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )} - 6 \, e^{x}\right )} e^{\left (-x\right )} \,d x } \]
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Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.72 \[ \int \frac {1}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx=x-{\mathrm {e}}^{5\,\sqrt {{\mathrm {e}}^{{\mathrm {e}}^{x^2}}}\,\sqrt {{\mathrm {e}}^x}-x}\,\left ({\mathrm {e}}^x+\frac {1}{3}\right ) \]
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