Integrand size = 66, antiderivative size = 24 \[ \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx=1+\log \left (\frac {\frac {1}{8 \left (-11+e^{x^2}\right )^2}-x}{x}\right ) \]
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\[ \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx=\int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{\left (11-e^{x^2}\right ) x \left (1-8 \left (-11+e^{x^2}\right )^2 x\right )} \, dx \\ & = \int \left (-\frac {44 x}{-11+e^{x^2}}+\frac {1+4 x^2-3872 x^3+352 e^{x^2} x^3}{x \left (-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x\right )}\right ) \, dx \\ & = -\left (44 \int \frac {x}{-11+e^{x^2}} \, dx\right )+\int \frac {1+4 x^2-3872 x^3+352 e^{x^2} x^3}{x \left (-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x\right )} \, dx \\ & = -\left (22 \text {Subst}\left (\int \frac {1}{-11+e^x} \, dx,x,x^2\right )\right )+\int \left (\frac {1}{x \left (-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x\right )}+\frac {4 x}{-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x}-\frac {3872 x^2}{-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x}+\frac {352 e^{x^2} x^2}{-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x}\right ) \, dx \\ & = 4 \int \frac {x}{-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x} \, dx-22 \text {Subst}\left (\int \frac {1}{(-11+x) x} \, dx,x,e^{x^2}\right )+352 \int \frac {e^{x^2} x^2}{-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x} \, dx-3872 \int \frac {x^2}{-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x} \, dx+\int \frac {1}{x \left (-1+968 x-176 e^{x^2} x+8 e^{2 x^2} x\right )} \, dx \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{-11+x} \, dx,x,e^{x^2}\right )\right )+2 \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^{x^2}\right )+4 \int \frac {x}{-1+8 \left (-11+e^{x^2}\right )^2 x} \, dx+352 \int \frac {e^{x^2} x^2}{-1+8 \left (-11+e^{x^2}\right )^2 x} \, dx-3872 \int \frac {x^2}{-1+8 \left (-11+e^{x^2}\right )^2 x} \, dx+\int \frac {1}{x \left (-1+8 \left (-11+e^{x^2}\right )^2 x\right )} \, dx \\ & = 2 x^2-2 \log \left (11-e^{x^2}\right )+4 \int \frac {x}{-1+8 \left (-11+e^{x^2}\right )^2 x} \, dx+352 \int \frac {e^{x^2} x^2}{-1+8 \left (-11+e^{x^2}\right )^2 x} \, dx-3872 \int \frac {x^2}{-1+8 \left (-11+e^{x^2}\right )^2 x} \, dx+\int \frac {1}{x \left (-1+8 \left (-11+e^{x^2}\right )^2 x\right )} \, dx \\ \end{align*}
Time = 0.90 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx=-\log \left (1+\frac {1}{-1+8 \left (-11+e^{x^2}\right )^2 x}\right ) \]
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Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46
method | result | size |
risch | \(\ln \left ({\mathrm e}^{2 x^{2}}-22 \,{\mathrm e}^{x^{2}}+\frac {968 x -1}{8 x}\right )-2 \ln \left ({\mathrm e}^{x^{2}}-11\right )\) | \(35\) |
parallelrisch | \(-\ln \left (x \right )-2 \ln \left ({\mathrm e}^{x^{2}}-11\right )+\ln \left (x \,{\mathrm e}^{2 x^{2}}-22 \,{\mathrm e}^{x^{2}} x +121 x -\frac {1}{8}\right )\) | \(36\) |
norman | \(-\ln \left (x \right )-2 \ln \left ({\mathrm e}^{x^{2}}-11\right )+\ln \left (8 x \,{\mathrm e}^{2 x^{2}}-176 \,{\mathrm e}^{x^{2}} x +968 x -1\right )\) | \(37\) |
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx=\log \left (\frac {8 \, x e^{\left (2 \, x^{2}\right )} - 176 \, x e^{\left (x^{2}\right )} + 968 \, x - 1}{x}\right ) - 2 \, \log \left (e^{\left (x^{2}\right )} - 11\right ) \]
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Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx=- 2 \log {\left (e^{x^{2}} - 11 \right )} + \log {\left (e^{2 x^{2}} - 22 e^{x^{2}} + \frac {968 x - 1}{8 x} \right )} \]
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Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx=\log \left (\frac {8 \, x e^{\left (2 \, x^{2}\right )} - 176 \, x e^{\left (x^{2}\right )} + 968 \, x - 1}{8 \, x}\right ) - 2 \, \log \left (e^{\left (x^{2}\right )} - 11\right ) \]
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Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx=\log \left (8 \, x e^{\left (2 \, x^{2}\right )} - 176 \, x e^{\left (x^{2}\right )} + 968 \, x - 1\right ) - \log \left (x\right ) - 2 \, \log \left (e^{\left (x^{2}\right )} - 11\right ) \]
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Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-11+e^{x^2} \left (1+4 x^2\right )}{11 x-10648 x^2-264 e^{2 x^2} x^2+8 e^{3 x^2} x^2+e^{x^2} \left (-x+2904 x^2\right )} \, dx=\ln \left (968\,x-176\,x\,{\mathrm {e}}^{x^2}+8\,x\,{\mathrm {e}}^{2\,x^2}-1\right )-2\,\ln \left ({\mathrm {e}}^{x^2}-11\right )-\ln \left (x\right ) \]
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