Integrand size = 97, antiderivative size = 29 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=-2+x+\left (256+2 \left (3+e^{\frac {1}{-e^x+x}}\right )+\frac {x}{4}\right ) (3+x) \]
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\[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 \left (e^x-x\right )^2} \, dx \\ & = \frac {1}{4} \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{\left (e^x-x\right )^2} \, dx \\ & = \frac {1}{4} \int \left (\frac {8 e^{-\frac {1}{e^x-x}} (3+x)}{e^x-x}+e^{-\frac {1}{e^x-x}} \left (8+1055 e^{\frac {1}{e^x-x}}+2 e^{\frac {1}{e^x-x}} x\right )+\frac {8 e^{-\frac {1}{e^x-x}} \left (-3+2 x+x^2\right )}{\left (e^x-x\right )^2}\right ) \, dx \\ & = \frac {1}{4} \int e^{-\frac {1}{e^x-x}} \left (8+1055 e^{\frac {1}{e^x-x}}+2 e^{\frac {1}{e^x-x}} x\right ) \, dx+2 \int \frac {e^{-\frac {1}{e^x-x}} (3+x)}{e^x-x} \, dx+2 \int \frac {e^{-\frac {1}{e^x-x}} \left (-3+2 x+x^2\right )}{\left (e^x-x\right )^2} \, dx \\ & = \frac {1}{4} \int \left (1055+8 e^{-\frac {1}{e^x-x}}+2 x\right ) \, dx+2 \int \left (\frac {3 e^{-\frac {1}{e^x-x}}}{e^x-x}+\frac {e^{-\frac {1}{e^x-x}} x}{e^x-x}\right ) \, dx+2 \int \left (-\frac {3 e^{-\frac {1}{e^x-x}}}{\left (e^x-x\right )^2}+\frac {2 e^{-\frac {1}{e^x-x}} x}{\left (e^x-x\right )^2}+\frac {e^{-\frac {1}{e^x-x}} x^2}{\left (e^x-x\right )^2}\right ) \, dx \\ & = \frac {1055 x}{4}+\frac {x^2}{4}+2 \int e^{-\frac {1}{e^x-x}} \, dx+2 \int \frac {e^{-\frac {1}{e^x-x}} x}{e^x-x} \, dx+2 \int \frac {e^{-\frac {1}{e^x-x}} x^2}{\left (e^x-x\right )^2} \, dx+4 \int \frac {e^{-\frac {1}{e^x-x}} x}{\left (e^x-x\right )^2} \, dx-6 \int \frac {e^{-\frac {1}{e^x-x}}}{\left (e^x-x\right )^2} \, dx+6 \int \frac {e^{-\frac {1}{e^x-x}}}{e^x-x} \, dx \\ \end{align*}
Time = 5.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {1}{4} \left (1055 x+x^2+e^{-\frac {1}{e^x-x}} (24+8 x)\right ) \]
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Time = 0.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {x^{2}}{4}+\frac {1055 x}{4}+\left (2 x +6\right ) {\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}}\) | \(27\) |
parallelrisch | \(\frac {x^{2}}{4}+\frac {1055 x}{4}+2 \,{\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}} x +6 \,{\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}}\) | \(37\) |
norman | \(\frac {\frac {1055 x^{2}}{4}+\frac {x^{3}}{4}-\frac {1055 \,{\mathrm e}^{x} x}{4}-\frac {{\mathrm e}^{x} x^{2}}{4}-6 \,{\mathrm e}^{x} {\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}}+6 \,{\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}} x +2 \,{\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}} x^{2}-2 \,{\mathrm e}^{x} {\mathrm e}^{-\frac {1}{{\mathrm e}^{x}-x}} x}{x -{\mathrm e}^{x}}\) | \(94\) |
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Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {1}{4} \, x^{2} + 2 \, {\left (x + 3\right )} e^{\left (\frac {1}{x - e^{x}}\right )} + \frac {1055}{4} \, x \]
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Time = 0.77 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {x^{2}}{4} + \frac {1055 x}{4} + \left (2 x + 6\right ) e^{- \frac {1}{- x + e^{x}}} \]
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\[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\int { \frac {2 \, x^{3} + 1055 \, x^{2} + {\left (2 \, x + 1055\right )} e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x^{2} + 1055 \, x\right )} e^{x} + 8 \, {\left (x^{2} - {\left (x - 3\right )} e^{x} - x + e^{\left (2 \, x\right )} - 3\right )} e^{\left (\frac {1}{x - e^{x}}\right )}}{4 \, {\left (x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {1}{4} \, {\left (x^{2} e^{x} + 1055 \, x e^{x} + 8 \, x e^{\left (\frac {x^{2} - x e^{x} + 1}{x - e^{x}}\right )} + 24 \, e^{\left (\frac {x^{2} - x e^{x} + 1}{x - e^{x}}\right )}\right )} e^{\left (-x\right )} \]
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Time = 8.86 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {1055 x^2+2 x^3+e^{2 x} (1055+2 x)+e^x \left (-2110 x-4 x^2\right )+e^{-\frac {1}{e^x-x}} \left (-24+8 e^{2 x}+e^x (24-8 x)-8 x+8 x^2\right )}{4 e^{2 x}-8 e^x x+4 x^2} \, dx=\frac {1055\,x}{4}+{\mathrm {e}}^{\frac {1}{x-{\mathrm {e}}^x}}\,\left (2\,x+6\right )+\frac {x^2}{4} \]
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