Integrand size = 83, antiderivative size = 22 \[ \int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx=x \left (2+\frac {x}{-2+\frac {3}{\frac {2}{3}+e^x}+x}\right ) \]
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\[ \int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx=\int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {50+60 x+12 x^2+9 e^{2 x} \left (8-12 x+3 x^2\right )+3 e^x \left (-40+6 x+21 x^2\right )}{\left (5+3 e^x (-2+x)+2 x\right )^2} \, dx \\ & = \int \left (\frac {9 x \left (4-2 x+x^2\right )}{(-2+x)^2 \left (5-6 e^x+2 x+3 e^x x\right )}-\frac {9 x^2 \left (-1+x+2 x^2\right )}{(-2+x)^2 \left (5-6 e^x+2 x+3 e^x x\right )^2}+\frac {8-12 x+3 x^2}{(-2+x)^2}\right ) \, dx \\ & = 9 \int \frac {x \left (4-2 x+x^2\right )}{(-2+x)^2 \left (5-6 e^x+2 x+3 e^x x\right )} \, dx-9 \int \frac {x^2 \left (-1+x+2 x^2\right )}{(-2+x)^2 \left (5-6 e^x+2 x+3 e^x x\right )^2} \, dx+\int \frac {8-12 x+3 x^2}{(-2+x)^2} \, dx \\ & = -\left (9 \int \left (\frac {27}{\left (5-6 e^x+2 x+3 e^x x\right )^2}+\frac {36}{(-2+x)^2 \left (5-6 e^x+2 x+3 e^x x\right )^2}+\frac {72}{(-2+x) \left (5-6 e^x+2 x+3 e^x x\right )^2}+\frac {9 x}{\left (5-6 e^x+2 x+3 e^x x\right )^2}+\frac {2 x^2}{\left (5-6 e^x+2 x+3 e^x x\right )^2}\right ) \, dx\right )+9 \int \left (\frac {2}{5-6 e^x+2 x+3 e^x x}+\frac {8}{(-2+x)^2 \left (5-6 e^x+2 x+3 e^x x\right )}+\frac {8}{(-2+x) \left (5-6 e^x+2 x+3 e^x x\right )}+\frac {x}{5-6 e^x+2 x+3 e^x x}\right ) \, dx+\int \left (3-\frac {4}{(-2+x)^2}\right ) \, dx \\ & = -\frac {4}{2-x}+3 x+9 \int \frac {x}{5-6 e^x+2 x+3 e^x x} \, dx-18 \int \frac {x^2}{\left (5-6 e^x+2 x+3 e^x x\right )^2} \, dx+18 \int \frac {1}{5-6 e^x+2 x+3 e^x x} \, dx+72 \int \frac {1}{(-2+x)^2 \left (5-6 e^x+2 x+3 e^x x\right )} \, dx+72 \int \frac {1}{(-2+x) \left (5-6 e^x+2 x+3 e^x x\right )} \, dx-81 \int \frac {x}{\left (5-6 e^x+2 x+3 e^x x\right )^2} \, dx-243 \int \frac {1}{\left (5-6 e^x+2 x+3 e^x x\right )^2} \, dx-324 \int \frac {1}{(-2+x)^2 \left (5-6 e^x+2 x+3 e^x x\right )^2} \, dx-648 \int \frac {1}{(-2+x) \left (5-6 e^x+2 x+3 e^x x\right )^2} \, dx \\ \end{align*}
Time = 2.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx=\frac {4}{-2+x}+3 x-\frac {9 x^2}{(-2+x) \left (5+3 e^x (-2+x)+2 x\right )} \]
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Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73
method | result | size |
risch | \(3 x +\frac {4}{-2+x}-\frac {9 x^{2}}{\left (-2+x \right ) \left (3 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}+2 x +5\right )}\) | \(38\) |
norman | \(\frac {-24 \,{\mathrm e}^{x}+18 x +6 x^{2}+9 \,{\mathrm e}^{x} x^{2}+20}{3 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}+2 x +5}\) | \(39\) |
parallelrisch | \(\frac {27 \,{\mathrm e}^{x} x^{2}+60+18 x^{2}+54 x -72 \,{\mathrm e}^{x}}{9 \,{\mathrm e}^{x} x -18 \,{\mathrm e}^{x}+6 x +15}\) | \(40\) |
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Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx=\frac {6 \, x^{2} + 3 \, {\left (3 \, x^{2} - 6 \, x + 4\right )} e^{x} + 6 \, x - 10}{3 \, {\left (x - 2\right )} e^{x} + 2 \, x + 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx=- \frac {9 x^{2}}{2 x^{2} + x + \left (3 x^{2} - 12 x + 12\right ) e^{x} - 10} + 3 x + \frac {4}{x - 2} \]
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Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx=\frac {6 \, x^{2} + 3 \, {\left (3 \, x^{2} - 6 \, x + 4\right )} e^{x} + 6 \, x - 10}{3 \, {\left (x - 2\right )} e^{x} + 2 \, x + 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (21) = 42\).
Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.95 \[ \int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx=\frac {9 \, x^{2} e^{x} + 6 \, x^{2} - 18 \, x e^{x} + 6 \, x + 12 \, e^{x} - 10}{3 \, x e^{x} + 2 \, x - 6 \, e^{x} + 5} \]
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Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {50+60 x+12 x^2+e^{2 x} \left (72-108 x+27 x^2\right )+e^x \left (-120+18 x+63 x^2\right )}{25+20 x+4 x^2+e^{2 x} \left (36-36 x+9 x^2\right )+e^x \left (-60+6 x+12 x^2\right )} \, dx=\frac {x\,\left (6\,x-12\,{\mathrm {e}}^x+9\,x\,{\mathrm {e}}^x+10\right )}{2\,x-6\,{\mathrm {e}}^x+3\,x\,{\mathrm {e}}^x+5} \]
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