Integrand size = 64, antiderivative size = 32 \[ \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx=4-x+\frac {3 x (-x+4 (1+x))}{e^{2+2 x}-x^2} \]
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\[ \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx=\int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{\left (e^{2+2 x}-x^2\right )^2} \, dx \\ & = \int \left (-1-\frac {3 (1+3 x)}{2 \left (e^{1+x}-x\right )}+\frac {3 (1+3 x)}{2 \left (e^{1+x}+x\right )}-\frac {3 \left (-4+x+3 x^2\right )}{2 \left (e^{1+x}-x\right )^2}-\frac {3 \left (-4+x+3 x^2\right )}{2 \left (e^{1+x}+x\right )^2}\right ) \, dx \\ & = -x-\frac {3}{2} \int \frac {1+3 x}{e^{1+x}-x} \, dx+\frac {3}{2} \int \frac {1+3 x}{e^{1+x}+x} \, dx-\frac {3}{2} \int \frac {-4+x+3 x^2}{\left (e^{1+x}-x\right )^2} \, dx-\frac {3}{2} \int \frac {-4+x+3 x^2}{\left (e^{1+x}+x\right )^2} \, dx \\ & = -x-\frac {3}{2} \int \left (\frac {1}{e^{1+x}-x}+\frac {3 x}{e^{1+x}-x}\right ) \, dx-\frac {3}{2} \int \left (-\frac {4}{\left (e^{1+x}-x\right )^2}+\frac {x}{\left (e^{1+x}-x\right )^2}+\frac {3 x^2}{\left (e^{1+x}-x\right )^2}\right ) \, dx-\frac {3}{2} \int \left (-\frac {4}{\left (e^{1+x}+x\right )^2}+\frac {x}{\left (e^{1+x}+x\right )^2}+\frac {3 x^2}{\left (e^{1+x}+x\right )^2}\right ) \, dx+\frac {3}{2} \int \left (\frac {1}{e^{1+x}+x}+\frac {3 x}{e^{1+x}+x}\right ) \, dx \\ & = -x-\frac {3}{2} \int \frac {1}{e^{1+x}-x} \, dx-\frac {3}{2} \int \frac {x}{\left (e^{1+x}-x\right )^2} \, dx-\frac {3}{2} \int \frac {x}{\left (e^{1+x}+x\right )^2} \, dx+\frac {3}{2} \int \frac {1}{e^{1+x}+x} \, dx-\frac {9}{2} \int \frac {x}{e^{1+x}-x} \, dx-\frac {9}{2} \int \frac {x^2}{\left (e^{1+x}-x\right )^2} \, dx-\frac {9}{2} \int \frac {x^2}{\left (e^{1+x}+x\right )^2} \, dx+\frac {9}{2} \int \frac {x}{e^{1+x}+x} \, dx+6 \int \frac {1}{\left (e^{1+x}-x\right )^2} \, dx+6 \int \frac {1}{\left (e^{1+x}+x\right )^2} \, dx \\ \end{align*}
Time = 5.60 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx=-x-\frac {3 x (4+3 x)}{-e^{2+2 x}+x^2} \]
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Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84
method | result | size |
risch | \(-x -\frac {3 \left (4+3 x \right ) x}{x^{2}-{\mathrm e}^{2+2 x}}\) | \(27\) |
parallelrisch | \(-\frac {x^{3}-x \,{\mathrm e}^{2+2 x}+9 x^{2}+12 x}{x^{2}-{\mathrm e}^{2+2 x}}\) | \(38\) |
norman | \(\frac {x \,{\mathrm e}^{2+2 x}-9 \,{\mathrm e}^{2+2 x}-12 x -x^{3}}{x^{2}-{\mathrm e}^{2+2 x}}\) | \(41\) |
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Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx=-\frac {x^{3} + 9 \, x^{2} - x e^{\left (2 \, x + 2\right )} + 12 \, x}{x^{2} - e^{\left (2 \, x + 2\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59 \[ \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx=- x + \frac {9 x^{2} + 12 x}{- x^{2} + e^{2 x + 2}} \]
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Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx=-\frac {x^{3} + 9 \, x^{2} - x e^{\left (2 \, x + 2\right )} + 12 \, x}{x^{2} - e^{\left (2 \, x + 2\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx=-\frac {x^{3} + 9 \, x^{2} - x e^{\left (2 \, x + 2\right )} + 12 \, x}{x^{2} - e^{\left (2 \, x + 2\right )}} \]
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Time = 13.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-e^{4+4 x}+12 x^2-x^4+e^{2+2 x} \left (12-6 x-16 x^2\right )}{e^{4+4 x}-2 e^{2+2 x} x^2+x^4} \, dx=\frac {9\,x^2+12\,x}{{\mathrm {e}}^{2\,x+2}-x^2}-x \]
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