Integrand size = 56, antiderivative size = 30 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=4+e^{1+\frac {1}{5} e^{15+2 x^2}-x}+\log (4 (2-x)) \]
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\[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=\int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{5} e^{\frac {1}{5} \left (80+e^{15+2 x^2}-5 x+10 x^2\right )} x+\frac {e^{-x} \left (2 e^{1+\frac {1}{5} e^{15+2 x^2}}+e^x-e^{1+\frac {1}{5} e^{15+2 x^2}} x\right )}{-2+x}\right ) \, dx \\ & = \frac {4}{5} \int e^{\frac {1}{5} \left (80+e^{15+2 x^2}-5 x+10 x^2\right )} x \, dx+\int \frac {e^{-x} \left (2 e^{1+\frac {1}{5} e^{15+2 x^2}}+e^x-e^{1+\frac {1}{5} e^{15+2 x^2}} x\right )}{-2+x} \, dx \\ & = \frac {4}{5} \int e^{\frac {1}{5} \left (80+e^{15+2 x^2}-5 x+10 x^2\right )} x \, dx+\int \left (-e^{\frac {1}{5} \left (5+e^{15+2 x^2}-5 x\right )}+\frac {1}{-2+x}\right ) \, dx \\ & = \log (2-x)+\frac {4}{5} \int e^{\frac {1}{5} \left (80+e^{15+2 x^2}-5 x+10 x^2\right )} x \, dx-\int e^{\frac {1}{5} \left (5+e^{15+2 x^2}-5 x\right )} \, dx \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=e^{1+\frac {1}{5} e^{15+2 x^2}-x}+\log (-2+x) \]
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Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73
method | result | size |
risch | \({\mathrm e}^{\frac {{\mathrm e}^{2 x^{2}+15}}{5}-x +1}+\ln \left (-2+x \right )\) | \(22\) |
norman | \({\mathrm e}^{{\mathrm e}^{-\ln \left (5\right )+2 x^{2}+15}-x +1}+\ln \left (-2+x \right )\) | \(24\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{-\ln \left (5\right )+2 x^{2}+15}-x +1}+\ln \left (-2+x \right )\) | \(24\) |
parts | \({\mathrm e}^{{\mathrm e}^{-\ln \left (5\right )+2 x^{2}+15}-x +1}+\ln \left (-2+x \right )\) | \(24\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=e^{\left (-x + e^{\left (2 \, x^{2} - \log \left (5\right ) + 15\right )} + 1\right )} + \log \left (x - 2\right ) \]
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Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=e^{- x + \frac {e^{2 x^{2} + 15}}{5} + 1} + \log {\left (x - 2 \right )} \]
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Time = 0.35 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=e^{\left (-x + \frac {1}{5} \, e^{\left (2 \, x^{2} + 15\right )} + 1\right )} + \log \left (x - 2\right ) \]
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\[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=\int { \frac {{\left (4 \, {\left (x^{2} - 2 \, x\right )} e^{\left (2 \, x^{2} - \log \left (5\right ) + 15\right )} - x + 2\right )} e^{\left (-x + e^{\left (2 \, x^{2} - \log \left (5\right ) + 15\right )} + 1\right )} + 1}{x - 2} \,d x } \]
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Time = 12.53 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {1+e^{1+\frac {1}{5} e^{15+2 x^2}-x} \left (2-x+\frac {1}{5} e^{15+2 x^2} \left (-8 x+4 x^2\right )\right )}{-2+x} \, dx=\ln \left (x-2\right )+{\mathrm {e}}^{\frac {{\mathrm {e}}^{15}\,{\mathrm {e}}^{2\,x^2}}{5}-x+1} \]
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