Integrand size = 46, antiderivative size = 23 \[ \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx=\frac {2 e^{-e^x-x} (5+x)}{-4+2 x} \]
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\[ \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx=\int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{(-2+x)^2} \, dx \\ & = \int \left (-\frac {e^{-e^x} (5+x)}{-2+x}+\frac {e^{-e^x-x} \left (3-3 x-x^2\right )}{(-2+x)^2}\right ) \, dx \\ & = -\int \frac {e^{-e^x} (5+x)}{-2+x} \, dx+\int \frac {e^{-e^x-x} \left (3-3 x-x^2\right )}{(-2+x)^2} \, dx \\ & = -\int \left (e^{-e^x}+\frac {7 e^{-e^x}}{-2+x}\right ) \, dx+\int \left (-e^{-e^x-x}-\frac {7 e^{-e^x-x}}{(-2+x)^2}-\frac {7 e^{-e^x-x}}{-2+x}\right ) \, dx \\ & = -\left (7 \int \frac {e^{-e^x-x}}{(-2+x)^2} \, dx\right )-7 \int \frac {e^{-e^x}}{-2+x} \, dx-7 \int \frac {e^{-e^x-x}}{-2+x} \, dx-\int e^{-e^x} \, dx-\int e^{-e^x-x} \, dx \\ & = -\left (7 \int \frac {e^{-e^x-x}}{(-2+x)^2} \, dx\right )-7 \int \frac {e^{-e^x}}{-2+x} \, dx-7 \int \frac {e^{-e^x-x}}{-2+x} \, dx-\text {Subst}\left (\int \frac {e^{-x}}{x^2} \, dx,x,e^x\right )-\text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,e^x\right ) \\ & = e^{-e^x-x}-\operatorname {ExpIntegralEi}\left (-e^x\right )-7 \int \frac {e^{-e^x-x}}{(-2+x)^2} \, dx-7 \int \frac {e^{-e^x}}{-2+x} \, dx-7 \int \frac {e^{-e^x-x}}{-2+x} \, dx+\text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,e^x\right ) \\ & = e^{-e^x-x}-7 \int \frac {e^{-e^x-x}}{(-2+x)^2} \, dx-7 \int \frac {e^{-e^x}}{-2+x} \, dx-7 \int \frac {e^{-e^x-x}}{-2+x} \, dx \\ \end{align*}
Time = 1.57 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx=e^{-e^x-x} \left (1+\frac {7}{-2+x}\right ) \]
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Time = 1.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74
method | result | size |
norman | \(\frac {\left (5+x \right ) {\mathrm e}^{-{\mathrm e}^{x}-x}}{-2+x}\) | \(17\) |
parallelrisch | \(\frac {\left (5+x \right ) {\mathrm e}^{-{\mathrm e}^{x}-x}}{-2+x}\) | \(17\) |
risch | \(\frac {\left (5+x \right ) {\mathrm e}^{-{\mathrm e}^{x}-x}}{-2+x}\) | \(19\) |
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx=\frac {{\left (x + 5\right )} e^{\left (-x - e^{x}\right )}}{x - 2} \]
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Time = 0.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx=\frac {\left (x + 5\right ) e^{- x - e^{x}}}{x - 2} \]
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Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx=\frac {{\left (x + 5\right )} e^{\left (-x - e^{x}\right )}}{x - 2} \]
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Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx=\frac {x e^{\left (-x - e^{x}\right )} + 5 \, e^{\left (-x - e^{x}\right )}}{x - 2} \]
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Time = 14.93 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-e^x-x} \left (3-3 x-x^2+e^x \left (10-3 x-x^2\right )\right )}{4-4 x+x^2} \, dx=\frac {{\mathrm {e}}^{-x-{\mathrm {e}}^x}\,\left (x+5\right )}{x-2} \]
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