Integrand size = 71, antiderivative size = 29 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=e^{e^{\left (\frac {e^{2 x}}{x}-x\right ) \left (x+x^2\right )}}+5 e^x \]
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\[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=\int \left (5 e^x+\exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx \]
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Rubi steps \begin{align*} \text {integral}& = 5 \int e^x \, dx+\int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right ) \, dx \\ & = 5 e^x+\int \left (-2 \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) x-3 \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) x^2+\exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}+2 x-x^2-x^3+e^{2 x} (1+x)\right ) (3+2 x)\right ) \, dx \\ & = 5 e^x-2 \int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) x \, dx-3 \int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) x^2 \, dx+\int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}+2 x-x^2-x^3+e^{2 x} (1+x)\right ) (3+2 x) \, dx \\ & = 5 e^x-2 \int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) x \, dx-3 \int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) x^2 \, dx+\int \left (3 \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}+2 x-x^2-x^3+e^{2 x} (1+x)\right )+2 \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}+2 x-x^2-x^3+e^{2 x} (1+x)\right ) x\right ) \, dx \\ & = 5 e^x-2 \int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) x \, dx+2 \int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}+2 x-x^2-x^3+e^{2 x} (1+x)\right ) x \, dx+3 \int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}+2 x-x^2-x^3+e^{2 x} (1+x)\right ) \, dx-3 \int \exp \left (e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)\right ) x^2 \, dx \\ \end{align*}
Time = 3.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=e^{e^{-x^2-x^3+e^{2 x} (1+x)}}+5 e^x \]
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Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{-\left (1+x \right ) \left (x^{2}-{\mathrm e}^{2 x}\right )}}+5 \,{\mathrm e}^{x}\) | \(23\) |
default | \({\mathrm e}^{{\mathrm e}^{\left (1+x \right ) {\mathrm e}^{2 x}-x^{3}-x^{2}}}+5 \,{\mathrm e}^{x}\) | \(27\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\left (1+x \right ) {\mathrm e}^{2 x}-x^{3}-x^{2}}}+5 \,{\mathrm e}^{x}\) | \(27\) |
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Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.83 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx={\left (5 \, e^{\left (-x^{3} - x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )} + x\right )} + e^{\left (-x^{3} - x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )} + e^{\left (-x^{3} - x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )}\right )}\right )}\right )} e^{\left (x^{3} + x^{2} - {\left (x + 1\right )} e^{\left (2 \, x\right )}\right )} \]
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Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=5 e^{x} + e^{e^{- x^{3} - x^{2} + \left (x + 1\right ) e^{2 x}}} \]
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none
Time = 0.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=5 \, e^{x} + e^{\left (e^{\left (-x^{3} - x^{2} + x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )}\right )}\right )} \]
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\[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx=\int { -{\left (3 \, x^{2} - {\left (2 \, x + 3\right )} e^{\left (2 \, x\right )} + 2 \, x\right )} e^{\left (-x^{3} - x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )} + e^{\left (-x^{3} - x^{2} + {\left (x + 1\right )} e^{\left (2 \, x\right )}\right )}\right )} + 5 \, e^{x} \,d x } \]
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Time = 12.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \left (5 e^x+e^{e^{-x^2-x^3+e^{2 x} (1+x)}-x^2-x^3+e^{2 x} (1+x)} \left (-2 x-3 x^2+e^{2 x} (3+2 x)\right )\right ) \, dx={\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}}+5\,{\mathrm {e}}^x \]
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