Integrand size = 59, antiderivative size = 31 \[ \int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx=4-\frac {-x+\frac {e^{-x} x}{5}}{\frac {e^{-2+x}}{5}+x} \]
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\[ \int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx=\int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^2 \left (-1-5 e^x (-1+x)+2 x+5 e^{2-x} x^2\right )}{\left (e^x+5 e^2 x\right )^2} \, dx \\ & = e^2 \int \frac {-1-5 e^x (-1+x)+2 x+5 e^{2-x} x^2}{\left (e^x+5 e^2 x\right )^2} \, dx \\ & = e^2 \int \left (\frac {e^{-2-x}}{5}+\frac {(-1+x) \left (1+25 e^2 x\right )}{\left (e^x+5 e^2 x\right )^2}-\frac {1-25 e^2+25 e^2 x}{5 e^2 \left (e^x+5 e^2 x\right )}\right ) \, dx \\ & = -\left (\frac {1}{5} \int \frac {1-25 e^2+25 e^2 x}{e^x+5 e^2 x} \, dx\right )+\frac {1}{5} e^2 \int e^{-2-x} \, dx+e^2 \int \frac {(-1+x) \left (1+25 e^2 x\right )}{\left (e^x+5 e^2 x\right )^2} \, dx \\ & = -\frac {e^{-x}}{5}-\frac {1}{5} \int \left (\frac {1-25 e^2}{e^x+5 e^2 x}+\frac {25 e^2 x}{e^x+5 e^2 x}\right ) \, dx+e^2 \int \left (-\frac {1}{\left (e^x+5 e^2 x\right )^2}-\frac {\left (-1+25 e^2\right ) x}{\left (e^x+5 e^2 x\right )^2}+\frac {25 e^2 x^2}{\left (e^x+5 e^2 x\right )^2}\right ) \, dx \\ & = -\frac {e^{-x}}{5}-e^2 \int \frac {1}{\left (e^x+5 e^2 x\right )^2} \, dx-\left (5 e^2\right ) \int \frac {x}{e^x+5 e^2 x} \, dx+\left (25 e^4\right ) \int \frac {x^2}{\left (e^x+5 e^2 x\right )^2} \, dx-\frac {1}{5} \left (1-25 e^2\right ) \int \frac {1}{e^x+5 e^2 x} \, dx+\left (e^2 \left (1-25 e^2\right )\right ) \int \frac {x}{\left (e^x+5 e^2 x\right )^2} \, dx \\ \end{align*}
Time = 1.96 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx=-\frac {e^2 \left (e^{-2+x}+e^{-x} x\right )}{e^x+5 e^2 x} \]
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Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-\frac {5 x \,{\mathrm e}^{-x}-25 x}{5 \left ({\mathrm e}^{-2+x}+5 x \right )}\) | \(24\) |
parallelrisch | \(\frac {\left (5 \,{\mathrm e}^{x} x -x \right ) {\mathrm e}^{-x}}{{\mathrm e}^{-2+x}+5 x}\) | \(25\) |
norman | \(\frac {\left (5 x \,{\mathrm e}^{2} {\mathrm e}^{x}-{\mathrm e}^{2} x \right ) {\mathrm e}^{-x}}{5 \,{\mathrm e}^{2} x +{\mathrm e}^{x}}\) | \(29\) |
parts | \(-\frac {{\mathrm e}^{-2+x}}{{\mathrm e}^{-2+x}+5 x}-\frac {{\mathrm e}^{2} x \,{\mathrm e}^{-x}}{5 \,{\mathrm e}^{2} x +{\mathrm e}^{x}}\) | \(37\) |
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Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx=\frac {5 \, x e^{x} - x}{5 \, x e^{x} + e^{\left (2 \, x - 2\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx=\frac {25 x e^{2} + 1}{25 x e^{2} + 5 e^{x}} - \frac {e^{- x}}{5} \]
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx=-\frac {x e^{2} - 5 \, x e^{\left (x + 2\right )}}{5 \, x e^{\left (x + 2\right )} + e^{\left (2 \, x\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx=\frac {5 \, {\left (x - 2\right )} e^{x} - x + 10 \, e^{x}}{5 \, {\left (x - 2\right )} e^{x} + e^{\left (2 \, x - 2\right )} + 10 \, e^{x}} \]
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Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-x} \left (e^{-2+2 x} (5-5 x)+5 x^2+e^{-2+x} (-1+2 x)\right )}{e^{-4+2 x}+10 e^{-2+x} x+25 x^2} \, dx=\frac {x\,{\mathrm {e}}^{2-x}\,\left (5\,{\mathrm {e}}^x-1\right )}{{\mathrm {e}}^x+5\,x\,{\mathrm {e}}^2} \]
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