Integrand size = 78, antiderivative size = 27 \[ \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx=e^{10}+3 x+\frac {(4-x) x}{-3-e^x+x+x^2} \]
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\[ \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx=\int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4-7 e^x \left (-2+x^2\right )}{\left (3+e^x-x-x^2\right )^2} \, dx \\ & = \int \left (3-\frac {4-6 x+x^2}{3+e^x-x-x^2}-\frac {x \left (16-5 x^2+x^3\right )}{\left (-3-e^x+x+x^2\right )^2}\right ) \, dx \\ & = 3 x-\int \frac {4-6 x+x^2}{3+e^x-x-x^2} \, dx-\int \frac {x \left (16-5 x^2+x^3\right )}{\left (-3-e^x+x+x^2\right )^2} \, dx \\ & = 3 x-\int \left (\frac {16 x}{\left (-3-e^x+x+x^2\right )^2}-\frac {5 x^3}{\left (-3-e^x+x+x^2\right )^2}+\frac {x^4}{\left (-3-e^x+x+x^2\right )^2}\right ) \, dx-\int \left (\frac {4}{3+e^x-x-x^2}+\frac {6 x}{-3-e^x+x+x^2}-\frac {x^2}{-3-e^x+x+x^2}\right ) \, dx \\ & = 3 x-4 \int \frac {1}{3+e^x-x-x^2} \, dx+5 \int \frac {x^3}{\left (-3-e^x+x+x^2\right )^2} \, dx-6 \int \frac {x}{-3-e^x+x+x^2} \, dx-16 \int \frac {x}{\left (-3-e^x+x+x^2\right )^2} \, dx-\int \frac {x^4}{\left (-3-e^x+x+x^2\right )^2} \, dx+\int \frac {x^2}{-3-e^x+x+x^2} \, dx \\ \end{align*}
Time = 2.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx=x \left (3-\frac {-4+x}{-3-e^x+x+x^2}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
risch | \(3 x -\frac {\left (x -4\right ) x}{x^{2}-3-{\mathrm e}^{x}+x}\) | \(23\) |
norman | \(\frac {-7 x +2 \,{\mathrm e}^{x}+3 x^{3}-3 \,{\mathrm e}^{x} x +6}{x^{2}-3-{\mathrm e}^{x}+x}\) | \(33\) |
parallelrisch | \(\frac {3 x^{3}+2 x^{2}-3 \,{\mathrm e}^{x} x -5 x}{x^{2}-3-{\mathrm e}^{x}+x}\) | \(33\) |
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Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx=\frac {3 \, x^{3} + 2 \, x^{2} - 3 \, x e^{x} - 5 \, x}{x^{2} + x - e^{x} - 3} \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx=3 x + \frac {x^{2} - 4 x}{- x^{2} - x + e^{x} + 3} \]
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Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx=\frac {3 \, x^{3} + 2 \, x^{2} - 3 \, x e^{x} - 5 \, x}{x^{2} + x - e^{x} - 3} \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx=\frac {3 \, x^{3} + 2 \, x^{2} - 3 \, x e^{x} - 5 \, x}{x^{2} + x - e^{x} - 3} \]
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Time = 8.59 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {15+3 e^{2 x}-12 x-20 x^2+6 x^3+3 x^4+e^x \left (14-7 x^2\right )}{9+e^{2 x}-6 x-5 x^2+2 x^3+x^4+e^x \left (6-2 x-2 x^2\right )} \, dx=3\,x-\frac {{\mathrm {e}}^x-5\,x+3}{x-{\mathrm {e}}^x+x^2-3} \]
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