Integrand size = 80, antiderivative size = 24 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=2+x-\frac {x}{12+e^{-5+2 x} x-x^2} \]
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\[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=\int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{10} \left (132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )\right )}{\left (12 e^5+e^{2 x} x-e^5 x^2\right )^2} \, dx \\ & = e^{10} \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{\left (12 e^5+e^{2 x} x-e^5 x^2\right )^2} \, dx \\ & = e^{10} \int \left (\frac {1}{e^{10}}-\frac {2 x}{e^5 \left (-12 e^5-e^{2 x} x+e^5 x^2\right )}+\frac {-12-24 x-x^2+2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}\right ) \, dx \\ & = x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx+e^{10} \int \frac {-12-24 x-x^2+2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx \\ & = x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx+e^{10} \int \left (-\frac {12}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}-\frac {24 x}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}-\frac {x^2}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}+\frac {2 x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2}\right ) \, dx \\ & = x-\left (2 e^5\right ) \int \frac {x}{-12 e^5-e^{2 x} x+e^5 x^2} \, dx-e^{10} \int \frac {x^2}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx+\left (2 e^{10}\right ) \int \frac {x^3}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx-\left (12 e^{10}\right ) \int \frac {1}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx-\left (24 e^{10}\right ) \int \frac {x}{\left (-12 e^5-e^{2 x} x+e^5 x^2\right )^2} \, dx \\ \end{align*}
Time = 3.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=x+\frac {e^5 x}{-e^{2 x} x+e^5 \left (-12+x^2\right )} \]
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Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
risch | \(x +\frac {x}{x^{2}-x \,{\mathrm e}^{-5+2 x}-12}\) | \(21\) |
norman | \(\frac {x^{3}-11 x -x \,{\mathrm e}^{2 x +\ln \left (x \right )-5}}{x^{2}-{\mathrm e}^{2 x +\ln \left (x \right )-5}-12}\) | \(37\) |
parallelrisch | \(\frac {x^{3}-11 x -x \,{\mathrm e}^{2 x +\ln \left (x \right )-5}}{x^{2}-{\mathrm e}^{2 x +\ln \left (x \right )-5}-12}\) | \(37\) |
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=\frac {x^{3} - x e^{\left (2 \, x + \log \left (x\right ) - 5\right )} - 11 \, x}{x^{2} - e^{\left (2 \, x + \log \left (x\right ) - 5\right )} - 12} \]
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Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=x - \frac {x}{- x^{2} + x e^{2 x - 5} + 12} \]
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Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=\frac {x^{3} e^{5} - x^{2} e^{\left (2 \, x\right )} - 11 \, x e^{5}}{x^{2} e^{5} - x e^{\left (2 \, x\right )} - 12 \, e^{5}} \]
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Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=\frac {x^{3} e^{5} - x^{2} e^{\left (2 \, x\right )} - 11 \, x e^{5}}{x^{2} e^{5} - x e^{\left (2 \, x\right )} - 12 \, e^{5}} \]
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Time = 8.93 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {132-25 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24+2 x-2 x^2\right )}{144-24 x^2+e^{-10+4 x} x^2+x^4+e^{-5+2 x} x \left (24-2 x^2\right )} \, dx=x-\frac {x}{x\,{\mathrm {e}}^{2\,x-5}-x^2+12} \]
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