Integrand size = 117, antiderivative size = 25 \[ \int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx=2-x+\frac {x}{-x+\left (x+\frac {e^x x}{2}\right )^2} \]
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\[ \int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx=\int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-8 e^{3 x} x^2-e^{4 x} x^2-16 \left (2-2 x+x^2\right )-16 e^x \left (1-x+2 x^2\right )-4 e^{2 x} \left (1+6 x^2\right )}{\left (4-\left (2+e^x\right )^2 x\right )^2} \, dx \\ & = \int \left (-1-\frac {4 (1+2 x)}{x \left (-4+4 x+4 e^x x+e^{2 x} x\right )}+\frac {16 \left (-1-2 x+2 x^2+e^x x^2\right )}{x \left (-4+4 x+4 e^x x+e^{2 x} x\right )^2}\right ) \, dx \\ & = -x-4 \int \frac {1+2 x}{x \left (-4+4 x+4 e^x x+e^{2 x} x\right )} \, dx+16 \int \frac {-1-2 x+2 x^2+e^x x^2}{x \left (-4+4 x+4 e^x x+e^{2 x} x\right )^2} \, dx \\ & = -x-4 \int \left (\frac {2}{-4+4 x+4 e^x x+e^{2 x} x}+\frac {1}{x \left (-4+4 x+4 e^x x+e^{2 x} x\right )}\right ) \, dx+16 \int \left (-\frac {2}{\left (-4+4 x+4 e^x x+e^{2 x} x\right )^2}-\frac {1}{x \left (-4+4 x+4 e^x x+e^{2 x} x\right )^2}+\frac {2 x}{\left (-4+4 x+4 e^x x+e^{2 x} x\right )^2}+\frac {e^x x}{\left (-4+4 x+4 e^x x+e^{2 x} x\right )^2}\right ) \, dx \\ & = -x-4 \int \frac {1}{x \left (-4+4 x+4 e^x x+e^{2 x} x\right )} \, dx-8 \int \frac {1}{-4+4 x+4 e^x x+e^{2 x} x} \, dx-16 \int \frac {1}{x \left (-4+4 x+4 e^x x+e^{2 x} x\right )^2} \, dx+16 \int \frac {e^x x}{\left (-4+4 x+4 e^x x+e^{2 x} x\right )^2} \, dx-32 \int \frac {1}{\left (-4+4 x+4 e^x x+e^{2 x} x\right )^2} \, dx+32 \int \frac {x}{\left (-4+4 x+4 e^x x+e^{2 x} x\right )^2} \, dx \\ & = -x-4 \int \frac {1}{x \left (-4+\left (2+e^x\right )^2 x\right )} \, dx-8 \int \frac {1}{-4+\left (2+e^x\right )^2 x} \, dx-16 \int \frac {1}{x \left (4-\left (2+e^x\right )^2 x\right )^2} \, dx+16 \int \frac {e^x x}{\left (4-\left (2+e^x\right )^2 x\right )^2} \, dx-32 \int \frac {1}{\left (4-\left (2+e^x\right )^2 x\right )^2} \, dx+32 \int \frac {x}{\left (4-\left (2+e^x\right )^2 x\right )^2} \, dx \\ \end{align*}
Time = 2.89 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx=-\frac {-4-4 x+\left (2+e^x\right )^2 x^2}{-4+\left (2+e^x\right )^2 x} \]
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Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-x +\frac {4}{x \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x} x +4 x -4}\) | \(25\) |
norman | \(\frac {4 x -4 x^{2}-4 \,{\mathrm e}^{x} x^{2}-{\mathrm e}^{2 x} x^{2}+4}{x \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x} x +4 x -4}\) | \(46\) |
parallelrisch | \(-\frac {4 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{2 x} x^{2}+4 x^{2}-4 x -4}{x \,{\mathrm e}^{2 x}+4 \,{\mathrm e}^{x} x +4 x -4}\) | \(46\) |
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Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx=-\frac {x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} e^{x} + 4 \, x^{2} - 4 \, x - 4}{x e^{\left (2 \, x\right )} + 4 \, x e^{x} + 4 \, x - 4} \]
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Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx=- x + \frac {4}{x e^{2 x} + 4 x e^{x} + 4 x - 4} \]
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Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx=-\frac {x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} e^{x} + 4 \, x^{2} - 4 \, x - 4}{x e^{\left (2 \, x\right )} + 4 \, x e^{x} + 4 \, x - 4} \]
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Time = 0.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx=-\frac {x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} e^{x} + 4 \, x^{2} - 4 \, x - 8}{x e^{\left (2 \, x\right )} + 4 \, x e^{x} + 4 \, x - 4} \]
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Timed out. \[ \int \frac {-32+32 x-16 x^2-8 e^{3 x} x^2-e^{4 x} x^2+e^x \left (-16+16 x-32 x^2\right )+e^{2 x} \left (-4-24 x^2\right )}{16-32 x+16 x^2+8 e^{3 x} x^2+e^{4 x} x^2+e^{2 x} \left (-8 x+24 x^2\right )+e^x \left (-32 x+32 x^2\right )} \, dx=\int -\frac {{\mathrm {e}}^{2\,x}\,\left (24\,x^2+4\right )-32\,x+8\,x^2\,{\mathrm {e}}^{3\,x}+x^2\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^x\,\left (32\,x^2-16\,x+16\right )+16\,x^2+32}{8\,x^2\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^{2\,x}\,\left (8\,x-24\,x^2\right )-32\,x+x^2\,{\mathrm {e}}^{4\,x}-{\mathrm {e}}^x\,\left (32\,x-32\,x^2\right )+16\,x^2+16} \,d x \]
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