Integrand size = 64, antiderivative size = 30 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=-x \left (-6-e^{e^{e^{-x} (4+x)}}-\frac {4 e^{-x}}{5}+x\right ) \]
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\[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=\int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx \\ & = \frac {1}{5} \int \left (4 e^{-x}-10 (-3+x)-4 e^{-x} x+5 e^{e^{e^{-x} (4+x)}-x} \left (e^x-3 e^{4 e^{-x}+e^{-x} x} x-e^{4 e^{-x}+e^{-x} x} x^2\right )\right ) \, dx \\ & = -(3-x)^2+\frac {4}{5} \int e^{-x} \, dx-\frac {4}{5} \int e^{-x} x \, dx+\int e^{e^{e^{-x} (4+x)}-x} \left (e^x-3 e^{4 e^{-x}+e^{-x} x} x-e^{4 e^{-x}+e^{-x} x} x^2\right ) \, dx \\ & = -\frac {4 e^{-x}}{5}-(3-x)^2+\frac {4 e^{-x} x}{5}-\frac {4}{5} \int e^{-x} \, dx+\int e^{e^{e^{-x} (4+x)}-x} \left (e^x-e^{e^{-x} (4+x)} x (3+x)\right ) \, dx \\ & = -(3-x)^2+\frac {4 e^{-x} x}{5}+\int \left (e^{e^{e^{-x} (4+x)}}+e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} (-3-x) x\right ) \, dx \\ & = -(3-x)^2+\frac {4 e^{-x} x}{5}+\int e^{e^{e^{-x} (4+x)}} \, dx+\int e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} (-3-x) x \, dx \\ & = -(3-x)^2+\frac {4 e^{-x} x}{5}+\int e^{e^{e^{-x} (4+x)}} \, dx+\int \left (-3 e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} x-e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} x^2\right ) \, dx \\ & = -(3-x)^2+\frac {4 e^{-x} x}{5}-3 \int e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} x \, dx+\int e^{e^{e^{-x} (4+x)}} \, dx-\int e^{e^{e^{-x} (4+x)}-x+e^{-x} (4+x)} x^2 \, dx \\ \end{align*}
Time = 3.37 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=\frac {1}{5} \left (30 x+5 e^{e^{e^{-x} (4+x)}} x+4 e^{-x} x-5 x^2\right ) \]
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Time = 1.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
method | result | size |
risch | \(-x^{2}+6 x +\frac {4 x \,{\mathrm e}^{-x}}{5}+x \,{\mathrm e}^{{\mathrm e}^{\left (4+x \right ) {\mathrm e}^{-x}}}\) | \(29\) |
norman | \(\left ({\mathrm e}^{x} x \,{\mathrm e}^{{\mathrm e}^{\left (4+x \right ) {\mathrm e}^{-x}}}+\frac {4 x}{5}+6 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}\) | \(36\) |
parallelrisch | \(\frac {{\mathrm e}^{-x} \left (-5 \,{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x} x \,{\mathrm e}^{{\mathrm e}^{\left (4+x \right ) {\mathrm e}^{-x}}}+30 \,{\mathrm e}^{x} x +4 x \right )}{5}\) | \(38\) |
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=\frac {1}{5} \, {\left (5 \, x e^{\left (x + e^{\left ({\left (x + 4\right )} e^{\left (-x\right )}\right )}\right )} - 5 \, {\left (x^{2} - 6 \, x\right )} e^{x} + 4 \, x\right )} e^{\left (-x\right )} \]
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Time = 43.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=- x^{2} + x e^{e^{\left (x + 4\right ) e^{- x}}} + 6 x + \frac {4 x e^{- x}}{5} \]
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Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=-x^{2} + \frac {4}{5} \, {\left (x + 1\right )} e^{\left (-x\right )} + x e^{\left (e^{\left (x e^{\left (-x\right )} + 4 \, e^{\left (-x\right )}\right )}\right )} + 6 \, x - \frac {4}{5} \, e^{\left (-x\right )} \]
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\[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=\int { -\frac {1}{5} \, {\left (10 \, {\left (x - 3\right )} e^{x} + 5 \, {\left ({\left (x^{2} + 3 \, x\right )} e^{\left ({\left (x + 4\right )} e^{\left (-x\right )}\right )} - e^{x}\right )} e^{\left (e^{\left ({\left (x + 4\right )} e^{\left (-x\right )}\right )}\right )} + 4 \, x - 4\right )} e^{\left (-x\right )} \,d x } \]
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Time = 9.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {1}{5} e^{-x} \left (4+e^x (30-10 x)-4 x+e^{e^{e^{-x} (4+x)}} \left (5 e^x+e^{e^{-x} (4+x)} \left (-15 x-5 x^2\right )\right )\right ) \, dx=6\,x+\frac {4\,x\,{\mathrm {e}}^{-x}}{5}+x\,{\mathrm {e}}^{{\mathrm {e}}^{4\,{\mathrm {e}}^{-x}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}}-x^2 \]
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