Integrand size = 82, antiderivative size = 31 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=e+\frac {1}{2} e^{\frac {x}{(2-x) (1+x)}+\frac {4+x}{2}} x \]
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\[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\int \frac {\exp \left (\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}\right ) \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{4 \left (2+x-x^2\right )^2} \, dx \\ & = \frac {1}{4} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{\left (2+x-x^2\right )^2} \, dx \\ & = \frac {1}{4} \int \left (2 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}+\frac {8 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{3 (-2+x)^2}+\frac {4 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{3 (-2+x)}+e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} x-\frac {2 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{3 (1+x)^2}+\frac {2 e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{3 (1+x)}\right ) \, dx \\ & = -\left (\frac {1}{6} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{(1+x)^2} \, dx\right )+\frac {1}{6} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{1+x} \, dx+\frac {1}{4} \int e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} x \, dx+\frac {1}{3} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{-2+x} \, dx+\frac {1}{2} \int e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}} \, dx+\frac {2}{3} \int \frac {e^{\frac {8+8 x-3 x^2-x^3}{4+2 x-2 x^2}}}{(-2+x)^2} \, dx \\ \end{align*}
Time = 1.53 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} e^{2+\frac {x}{2}-\frac {x}{-2-x+x^2}} x \]
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Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35
method | result | size |
parallelrisch | \(x \,{\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}\) | \(42\) |
gosper | \({\mathrm e}^{-\frac {2 x^{2} \ln \left (2\right )-x^{3}-2 x \ln \left (2\right )-3 x^{2}-4 \ln \left (2\right )+8 x +8}{2 \left (x^{2}-x -2\right )}} x\) | \(47\) |
risch | \(x \,{\mathrm e}^{-\frac {2 x^{2} \ln \left (2\right )-x^{3}-2 x \ln \left (2\right )-3 x^{2}-4 \ln \left (2\right )+8 x +8}{2 \left (1+x \right ) \left (-2+x \right )}}\) | \(47\) |
norman | \(\frac {x^{3} {\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}-2 x \,{\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}-x^{2} {\mathrm e}^{\frac {\left (-2 x^{2}+2 x +4\right ) \ln \left (2\right )+x^{3}+3 x^{2}-8 x -8}{2 x^{2}-2 x -4}}}{x^{2}-x -2}\) | \(145\) |
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Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=x e^{\left (\frac {x^{3} + 3 \, x^{2} - 2 \, {\left (x^{2} - x - 2\right )} \log \left (2\right ) - 8 \, x - 8}{2 \, {\left (x^{2} - x - 2\right )}}\right )} \]
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Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=x e^{\frac {x^{3} + 3 x^{2} - 8 x + \left (- 2 x^{2} + 2 x + 4\right ) \log {\left (2 \right )} - 8}{2 x^{2} - 2 x - 4}} \]
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Time = 0.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} \, x e^{\left (\frac {1}{2} \, x - \frac {1}{3 \, {\left (x + 1\right )}} - \frac {2}{3 \, {\left (x - 2\right )}} + 2\right )} \]
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Time = 0.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=\frac {1}{2} \, x e^{\left (\frac {x^{3} - x^{2} - 4 \, x}{2 \, {\left (x^{2} - x - 2\right )}} + 2\right )} \]
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Time = 8.56 (sec) , antiderivative size = 103, normalized size of antiderivative = 3.32 \[ \int \frac {e^{\frac {-8-8 x+3 x^2+x^3+\left (4+2 x-2 x^2\right ) \log (2)}{-4-2 x+2 x^2}} \left (8+16 x-2 x^2-5 x^3+x^5\right )}{8+8 x-6 x^2-4 x^3+2 x^4} \, dx=2^{\frac {x^2-2}{-x^2+x+2}-\frac {2\,x}{-2\,x^2+2\,x+4}}\,x\,{\mathrm {e}}^{\frac {8\,x}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{-\frac {x^3}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{-\frac {3\,x^2}{-2\,x^2+2\,x+4}}\,{\mathrm {e}}^{\frac {8}{-2\,x^2+2\,x+4}} \]
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