Integrand size = 81, antiderivative size = 32 \[ \int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx=x-\frac {1}{2} \left (-e^{-2-e^x-x}+x\right )^2-\log (-4+3 x) \]
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\[ \int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx=\int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (e^{-2 \left (2+e^x+x\right )}-e^{-4-2 e^x-x} \left (-1-e^{2+e^x}+e^{2+e^x} x\right )-\frac {e^{-2-e^x} \left (7 e^{2+e^x}-4 x-7 e^{2+e^x} x+3 x^2+3 e^{2+e^x} x^2\right )}{-4+3 x}\right ) \, dx \\ & = \int e^{-2 \left (2+e^x+x\right )} \, dx-\int e^{-4-2 e^x-x} \left (-1-e^{2+e^x}+e^{2+e^x} x\right ) \, dx-\int \frac {e^{-2-e^x} \left (7 e^{2+e^x}-4 x-7 e^{2+e^x} x+3 x^2+3 e^{2+e^x} x^2\right )}{-4+3 x} \, dx \\ & = \int e^{-2 \left (2+e^x+x\right )} \, dx-\int e^{-4-2 e^x-x} \left (-1+e^{2+e^x} (-1+x)\right ) \, dx-\int \left (e^{-2-e^x} x+\frac {7-7 x+3 x^2}{-4+3 x}\right ) \, dx \\ & = \int e^{-2 \left (2+e^x+x\right )} \, dx-\int \left (-e^{-4-2 e^x-x}+e^{-2-e^x-x} (-1+x)\right ) \, dx-\int e^{-2-e^x} x \, dx-\int \frac {7-7 x+3 x^2}{-4+3 x} \, dx \\ & = \int e^{-4-2 e^x-x} \, dx+\int e^{-2 \left (2+e^x+x\right )} \, dx-\int e^{-2-e^x-x} (-1+x) \, dx-\int e^{-2-e^x} x \, dx-\int \left (-1+x+\frac {3}{-4+3 x}\right ) \, dx \\ & = x-\frac {x^2}{2}-\log (4-3 x)+\int e^{-2 \left (2+e^x+x\right )} \, dx-\int e^{-2-e^x} x \, dx-\int \left (-e^{-2-e^x-x}+e^{-2-e^x-x} x\right ) \, dx+\text {Subst}\left (\int \frac {e^{-4-2 x}}{x^2} \, dx,x,e^x\right ) \\ & = -e^{-4-2 e^x-x}+x-\frac {x^2}{2}-\log (4-3 x)-2 \text {Subst}\left (\int \frac {e^{-4-2 x}}{x} \, dx,x,e^x\right )+\int e^{-2-e^x-x} \, dx+\int e^{-2 \left (2+e^x+x\right )} \, dx-\int e^{-2-e^x} x \, dx-\int e^{-2-e^x-x} x \, dx \\ & = -e^{-4-2 e^x-x}+x-\frac {x^2}{2}-\frac {2 \operatorname {ExpIntegralEi}\left (-2 e^x\right )}{e^4}-\log (4-3 x)+\int e^{-2 \left (2+e^x+x\right )} \, dx-\int e^{-2-e^x} x \, dx-\int e^{-2-e^x-x} x \, dx+\text {Subst}\left (\int \frac {e^{-2-x}}{x^2} \, dx,x,e^x\right ) \\ & = -e^{-4-2 e^x-x}-e^{-2-e^x-x}+x-\frac {x^2}{2}-\frac {2 \operatorname {ExpIntegralEi}\left (-2 e^x\right )}{e^4}-\log (4-3 x)+\int e^{-2 \left (2+e^x+x\right )} \, dx-\int e^{-2-e^x} x \, dx-\int e^{-2-e^x-x} x \, dx-\text {Subst}\left (\int \frac {e^{-2-x}}{x} \, dx,x,e^x\right ) \\ & = -e^{-4-2 e^x-x}-e^{-2-e^x-x}+x-\frac {x^2}{2}-\frac {2 \operatorname {ExpIntegralEi}\left (-2 e^x\right )}{e^4}-\frac {\operatorname {ExpIntegralEi}\left (-e^x\right )}{e^2}-\log (4-3 x)+\int e^{-2 \left (2+e^x+x\right )} \, dx-\int e^{-2-e^x} x \, dx-\int e^{-2-e^x-x} x \, dx \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx=-\frac {1}{2} e^{-2 \left (2+e^x+x\right )}+x+e^{-2-e^x-x} x-\frac {x^2}{2}-\log (4-3 x) \]
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Time = 0.17 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25
method | result | size |
risch | \(x +{\mathrm e}^{-{\mathrm e}^{x}-x -2} x -\frac {x^{2}}{2}-\frac {{\mathrm e}^{-2 \,{\mathrm e}^{x}-2 x -4}}{2}-\ln \left (-4+3 x \right )\) | \(40\) |
parallelrisch | \(x -\frac {x^{2}}{2}-\frac {{\mathrm e}^{-2 \,{\mathrm e}^{x}-2 x -4}}{2}+{\mathrm e}^{-{\mathrm e}^{x}-x -2} x -\ln \left (x -\frac {4}{3}\right )\) | \(40\) |
norman | \(x +{\mathrm e}^{-{\mathrm e}^{x}-x -2} x -\frac {x^{2}}{2}-\frac {{\mathrm e}^{-2 \,{\mathrm e}^{x}-2 x -4}}{2}-\ln \left (-4+3 x \right )\) | \(42\) |
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Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx=-\frac {1}{2} \, x^{2} + x e^{\left (-x - e^{x} - 2\right )} + x - \frac {1}{2} \, e^{\left (-2 \, x - 2 \, e^{x} - 4\right )} - \log \left (3 \, x - 4\right ) \]
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Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx=- \frac {x^{2}}{2} + x e^{- x - e^{x} - 2} + x - \frac {e^{- 2 x - 2 e^{x} - 4}}{2} - \log {\left (3 x - 4 \right )} \]
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx=-\frac {1}{2} \, x^{2} + \frac {1}{2} \, {\left (2 \, x e^{\left (x - e^{x} + 2\right )} - e^{\left (-2 \, e^{x}\right )}\right )} e^{\left (-2 \, x - 4\right )} + x - \log \left (3 \, x - 4\right ) \]
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\[ \int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx=\int { -\frac {3 \, x^{2} + {\left (3 \, x^{2} + {\left (3 \, x^{2} - 4 \, x\right )} e^{x} - 7 \, x + 4\right )} e^{\left (-x - e^{x} - 2\right )} - {\left ({\left (3 \, x - 4\right )} e^{x} + 3 \, x - 4\right )} e^{\left (-2 \, x - 2 \, e^{x} - 4\right )} - 7 \, x + 7}{3 \, x - 4} \,d x } \]
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Time = 7.99 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {-7+7 x-3 x^2+e^{-4-2 e^x-2 x} \left (-4+3 x+e^x (-4+3 x)\right )+e^{-2-e^x-x} \left (-4+7 x-3 x^2+e^x \left (4 x-3 x^2\right )\right )}{-4+3 x} \, dx=x-\ln \left (x-\frac {4}{3}\right )-\frac {{\mathrm {e}}^{-2\,x-2\,{\mathrm {e}}^x-4}}{2}+x\,{\mathrm {e}}^{-x-{\mathrm {e}}^x-2}-\frac {x^2}{2} \]
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