Integrand size = 79, antiderivative size = 24 \[ \int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx=x+\frac {\log (5)}{e^{-x+(3-x) x}-4 x} \]
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\[ \int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx=\int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 x^2} \left (e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))\right )}{\left (e^{2 x}-4 e^{x^2} x\right )^2} \, dx \\ & = \int \left (1+2 e^{-2 x+x^2} (-1+x) \log (5)-\frac {8 e^{2 x^2} (-1+x) x \log (5)}{-e^{4 x}+4 e^{x (2+x)} x}+\frac {4 e^{2 x^2} \left (1-2 x+2 x^2\right ) \log (5)}{\left (e^{2 x}-4 e^{x^2} x\right )^2}\right ) \, dx \\ & = x+(2 \log (5)) \int e^{-2 x+x^2} (-1+x) \, dx+(4 \log (5)) \int \frac {e^{2 x^2} \left (1-2 x+2 x^2\right )}{\left (e^{2 x}-4 e^{x^2} x\right )^2} \, dx-(8 \log (5)) \int \frac {e^{2 x^2} (-1+x) x}{-e^{4 x}+4 e^{x (2+x)} x} \, dx \\ & = x+e^{-2 x+x^2} \log (5)+(4 \log (5)) \int \left (\frac {e^{2 x^2}}{\left (e^{2 x}-4 e^{x^2} x\right )^2}-\frac {2 e^{2 x^2} x}{\left (-e^{2 x}+4 e^{x^2} x\right )^2}+\frac {2 e^{2 x^2} x^2}{\left (-e^{2 x}+4 e^{x^2} x\right )^2}\right ) \, dx-(8 \log (5)) \int \left (-\frac {e^{2 x^2} x}{-e^{4 x}+4 e^{x (2+x)} x}+\frac {e^{2 x^2} x^2}{-e^{4 x}+4 e^{x (2+x)} x}\right ) \, dx \\ & = x+e^{-2 x+x^2} \log (5)+(4 \log (5)) \int \frac {e^{2 x^2}}{\left (e^{2 x}-4 e^{x^2} x\right )^2} \, dx-(8 \log (5)) \int \frac {e^{2 x^2} x}{\left (-e^{2 x}+4 e^{x^2} x\right )^2} \, dx+(8 \log (5)) \int \frac {e^{2 x^2} x^2}{\left (-e^{2 x}+4 e^{x^2} x\right )^2} \, dx+(8 \log (5)) \int \frac {e^{2 x^2} x}{-e^{4 x}+4 e^{x (2+x)} x} \, dx-(8 \log (5)) \int \frac {e^{2 x^2} x^2}{-e^{4 x}+4 e^{x (2+x)} x} \, dx \\ \end{align*}
Time = 1.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx=\frac {e^{2 x} x+e^{x^2} \left (-4 x^2+\log (5)\right )}{e^{2 x}-4 e^{x^2} x} \]
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Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \(x -\frac {\ln \left (5\right )}{4 x -{\mathrm e}^{-\left (-2+x \right ) x}}\) | \(22\) |
parallelrisch | \(-\frac {-4 x^{2}+x \,{\mathrm e}^{-x^{2}+2 x}+\ln \left (5\right )}{4 x -{\mathrm e}^{-x^{2}+2 x}}\) | \(41\) |
norman | \(\frac {4 x^{2}-x \,{\mathrm e}^{-x^{2}+2 x}-\ln \left (5\right )}{4 x -{\mathrm e}^{-x^{2}+2 x}}\) | \(43\) |
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx=\frac {4 \, x^{2} - x e^{\left (-x^{2} + 2 \, x\right )} - \log \left (5\right )}{4 \, x - e^{\left (-x^{2} + 2 \, x\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx=x + \frac {\log {\left (5 \right )}}{- 4 x + e^{- x^{2} + 2 x}} \]
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Time = 0.33 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx=\frac {{\left (4 \, x^{2} - \log \left (5\right )\right )} e^{\left (x^{2}\right )} - x e^{\left (2 \, x\right )}}{4 \, x e^{\left (x^{2}\right )} - e^{\left (2 \, x\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx=\frac {4 \, x^{2} - x e^{\left (-x^{2} + 2 \, x\right )} - \log \left (5\right )}{4 \, x - e^{\left (-x^{2} + 2 \, x\right )}} \]
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Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 x-2 x^2}+16 x^2+4 \log (5)+e^{2 x-x^2} (-8 x+(-2+2 x) \log (5))}{e^{4 x-2 x^2}-8 e^{2 x-x^2} x+16 x^2} \, dx=x-\frac {\ln \left (5\right )}{4\,x-{\mathrm {e}}^{2\,x-x^2}} \]
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