Integrand size = 139, antiderivative size = 32 \[ \int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx=\frac {3 e^{-x} x}{1+\frac {e^x}{-4+3 \left (x-e^{-x} x\right )}} \]
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\[ \int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx=\int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = 3 \int \frac {e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx \\ & = 3 \int \frac {e^{-x} \left (-9 (-1+x) x^2-2 e^{3 x} \left (2-7 x+3 x^2\right )+6 e^x x \left (4-7 x+3 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )\right )}{\left (e^{2 x}-3 x+e^x (-4+3 x)\right )^2} \, dx \\ & = 3 \int \left (-\frac {e^{-x} \left (4 e^x-22 x-14 e^x x+24 x^2+6 e^x x^2-9 x^3\right )}{-4 e^x+e^{2 x}-3 x+3 e^x x}-\frac {e^{-x} x \left (-100 e^x-75 x+162 e^x x+81 x^2-108 e^x x^2-27 x^3+27 e^x x^3\right )}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2}\right ) \, dx \\ & = -\left (3 \int \frac {e^{-x} \left (4 e^x-22 x-14 e^x x+24 x^2+6 e^x x^2-9 x^3\right )}{-4 e^x+e^{2 x}-3 x+3 e^x x} \, dx\right )-3 \int \frac {e^{-x} x \left (-100 e^x-75 x+162 e^x x+81 x^2-108 e^x x^2-27 x^3+27 e^x x^3\right )}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2} \, dx \\ & = -\left (3 \int \left (-\frac {100 x}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2}+\frac {162 x^2}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2}-\frac {75 e^{-x} x^2}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2}-\frac {108 x^3}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2}+\frac {81 e^{-x} x^3}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2}+\frac {27 x^4}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2}-\frac {27 e^{-x} x^4}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2}\right ) \, dx\right )-3 \int \left (\frac {4}{-4 e^x+e^{2 x}-3 x+3 e^x x}-\frac {14 x}{-4 e^x+e^{2 x}-3 x+3 e^x x}-\frac {22 e^{-x} x}{-4 e^x+e^{2 x}-3 x+3 e^x x}+\frac {6 x^2}{-4 e^x+e^{2 x}-3 x+3 e^x x}+\frac {24 e^{-x} x^2}{-4 e^x+e^{2 x}-3 x+3 e^x x}-\frac {9 e^{-x} x^3}{-4 e^x+e^{2 x}-3 x+3 e^x x}\right ) \, dx \\ & = -\left (12 \int \frac {1}{-4 e^x+e^{2 x}-3 x+3 e^x x} \, dx\right )-18 \int \frac {x^2}{-4 e^x+e^{2 x}-3 x+3 e^x x} \, dx+27 \int \frac {e^{-x} x^3}{-4 e^x+e^{2 x}-3 x+3 e^x x} \, dx+42 \int \frac {x}{-4 e^x+e^{2 x}-3 x+3 e^x x} \, dx+66 \int \frac {e^{-x} x}{-4 e^x+e^{2 x}-3 x+3 e^x x} \, dx-72 \int \frac {e^{-x} x^2}{-4 e^x+e^{2 x}-3 x+3 e^x x} \, dx-81 \int \frac {x^4}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2} \, dx+81 \int \frac {e^{-x} x^4}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2} \, dx+225 \int \frac {e^{-x} x^2}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2} \, dx-243 \int \frac {e^{-x} x^3}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2} \, dx+300 \int \frac {x}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2} \, dx+324 \int \frac {x^3}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2} \, dx-486 \int \frac {x^2}{\left (-4 e^x+e^{2 x}-3 x+3 e^x x\right )^2} \, dx \\ \end{align*}
Time = 3.60 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx=\frac {3 e^{-x} x \left (-3 x+e^x (-4+3 x)\right )}{e^{2 x}-3 x+e^x (-4+3 x)} \]
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Time = 3.74 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
risch | \(\frac {3 x \left (3 x -4-3 x \,{\mathrm e}^{-x}\right )}{{\mathrm e}^{2 x}+3 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-3 x}\) | \(35\) |
norman | \(\frac {\left (-12 \,{\mathrm e}^{x} x -9 x^{2}+9 \,{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}}{{\mathrm e}^{2 x}+3 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-3 x}\) | \(43\) |
parallelrisch | \(\frac {16 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (3\right )-x} x \,{\mathrm e}^{2 x}-12 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (3\right )-x} {\mathrm e}^{2 x} x^{2}-4 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (3\right )-x} {\mathrm e}^{3 x} x +9 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (3\right )-x} x^{3} {\mathrm e}^{x}-9 x^{3} {\mathrm e}^{\ln \left (x \right )+\ln \left (3\right )-x}}{3 x^{2} \left ({\mathrm e}^{2 x}+3 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x}-3 x \right )}\) | \(106\) |
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Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (32) = 64\).
Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.25 \[ \int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx=\frac {{\left (3 \, x - 4\right )} e^{\left (-2 \, x + 2 \, \log \left (3\right ) + 2 \, \log \left (x\right )\right )} - e^{\left (-3 \, x + 3 \, \log \left (3\right ) + 3 \, \log \left (x\right )\right )}}{{\left (3 \, x - 4\right )} e^{\left (-x + \log \left (3\right ) + \log \left (x\right )\right )} + 3 \, x - e^{\left (-2 \, x + 2 \, \log \left (3\right ) + 2 \, \log \left (x\right )\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx=3 x e^{- x} - \frac {3 x e^{x}}{- 3 x + \left (3 x - 4\right ) e^{x} + e^{2 x}} \]
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Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx=\frac {3 \, {\left (3 \, x^{2} - 4 \, x\right )}}{{\left (3 \, x - 4\right )} e^{x} - 3 \, x + e^{\left (2 \, x\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx=\frac {3 \, {\left (3 \, x^{2} e^{x} - 3 \, x^{2} - x e^{\left (2 \, x\right )} - 4 \, x e^{x}\right )}}{3 \, x e^{\left (2 \, x\right )} - 3 \, x e^{x} + e^{\left (3 \, x\right )} - 4 \, e^{\left (2 \, x\right )}} \]
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Time = 9.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {3 e^{-x} x \left (9 x^2-9 x^3+e^{3 x} \left (-4+14 x-6 x^2\right )+e^{2 x} \left (16-46 x+42 x^2-9 x^3\right )+e^x \left (24 x-42 x^2+18 x^3\right )\right )}{e^{4 x} x+9 x^3+e^{3 x} \left (-8 x+6 x^2\right )+e^x \left (24 x^2-18 x^3\right )+e^{2 x} \left (16 x-30 x^2+9 x^3\right )} \, dx=\frac {12\,x\,{\mathrm {e}}^x-9\,x^2\,{\mathrm {e}}^x+9\,x^2}{4\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{3\,x}-3\,x\,{\mathrm {e}}^{2\,x}+3\,x\,{\mathrm {e}}^x} \]
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