Integrand size = 59, antiderivative size = 20 \[ \int \frac {16+144 x+36 x^2+e^2 x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx=2+x+\frac {16 x}{1-e+\frac {4+x}{x}} \]
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Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 2009, 27, 697} \[ \int \frac {16+144 x+36 x^2+e^2 x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx=\frac {(18-e) x}{2-e}+\frac {256}{(2-e)^2 ((2-e) x+4)} \]
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Rule 6
Rule 27
Rule 697
Rule 2009
Rubi steps \begin{align*} \text {integral}& = \int \frac {16+144 x+\left (36+e^2\right ) x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx \\ & = \int \frac {16+144 x+\left (36+e^2\right ) x^2+e \left (-8 x-20 x^2\right )}{16+16 x+\left (4+e^2\right ) x^2+e \left (-8 x-4 x^2\right )} \, dx \\ & = \int \frac {16+8 (18-e) x+(2-e) (18-e) x^2}{16+8 (2-e) x+(2-e)^2 x^2} \, dx \\ & = \int \frac {16+8 (18-e) x+(2-e) (18-e) x^2}{(-4-2 x+e x)^2} \, dx \\ & = \int \frac {16+8 (18-e) x+(2-e) (18-e) x^2}{(-4+(-2+e) x)^2} \, dx \\ & = \int \left (\frac {-18+e}{-2+e}+\frac {256}{(-2+e) (4+(2-e) x)^2}\right ) \, dx \\ & = \frac {(18-e) x}{2-e}+\frac {256}{(2-e)^2 (4+(2-e) x)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {16+144 x+36 x^2+e^2 x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx=\frac {(-18+e) x}{-2+e}-\frac {256}{(-2+e)^2 (-4-2 x+e x)} \]
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Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15
method | result | size |
gosper | \(\frac {x \left (x \,{\mathrm e}-18 x -4\right )}{x \,{\mathrm e}-2 x -4}\) | \(23\) |
norman | \(\frac {\left ({\mathrm e}-18\right ) x^{2}-4 x}{x \,{\mathrm e}-2 x -4}\) | \(25\) |
parallelrisch | \(\frac {4 x^{2} {\mathrm e}-72 x^{2}-16 x}{4 x \,{\mathrm e}-8 x -16}\) | \(30\) |
risch | \(\frac {x \,{\mathrm e}}{{\mathrm e}-2}-\frac {18 x}{{\mathrm e}-2}-\frac {256}{\left ({\mathrm e}-2\right )^{2} \left (x \,{\mathrm e}-2 x -4\right )}\) | \(40\) |
meijerg | \(\frac {\left ({\mathrm e}-2\right )^{2} x}{\left ({\mathrm e}^{2}-4 \,{\mathrm e}+4\right ) \left (1+\frac {x \left (2-{\mathrm e}\right )}{4}\right )}+\frac {64 \left (\frac {{\mathrm e}^{2}}{16}-\frac {5 \,{\mathrm e}}{4}+\frac {9}{4}\right ) \left ({\mathrm e}-2\right )^{2} \left (\frac {x \left (2-{\mathrm e}\right ) \left (\frac {3 x \left (2-{\mathrm e}\right )}{4}+6\right )}{12+3 x \left (2-{\mathrm e}\right )}-2 \ln \left (1+\frac {x \left (2-{\mathrm e}\right )}{4}\right )\right )}{\left (2-{\mathrm e}\right )^{3} \left ({\mathrm e}^{2}-4 \,{\mathrm e}+4\right )}+\frac {16 \left (-\frac {{\mathrm e}}{2}+9\right ) \left ({\mathrm e}-2\right )^{2} \left (-\frac {x \left (2-{\mathrm e}\right )}{4 \left (1+\frac {x \left (2-{\mathrm e}\right )}{4}\right )}+\ln \left (1+\frac {x \left (2-{\mathrm e}\right )}{4}\right )\right )}{\left (2-{\mathrm e}\right )^{2} \left ({\mathrm e}^{2}-4 \,{\mathrm e}+4\right )}\) | \(184\) |
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Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (21) = 42\).
Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.60 \[ \int \frac {16+144 x+36 x^2+e^2 x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx=\frac {x^{2} e^{3} - 72 \, x^{2} - 2 \, {\left (11 \, x^{2} + 2 \, x\right )} e^{2} + 4 \, {\left (19 \, x^{2} + 20 \, x\right )} e - 144 \, x - 256}{x e^{3} - 2 \, {\left (3 \, x + 2\right )} e^{2} + 4 \, {\left (3 \, x + 4\right )} e - 8 \, x - 16} \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (15) = 30\).
Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.30 \[ \int \frac {16+144 x+36 x^2+e^2 x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx=x \left (- \frac {18}{-2 + e} + \frac {e}{-2 + e}\right ) - \frac {256}{x \left (- 6 e^{2} - 8 + e^{3} + 12 e\right ) - 4 e^{2} - 16 + 16 e} \]
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none
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05 \[ \int \frac {16+144 x+36 x^2+e^2 x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx=\frac {x {\left (e - 18\right )}}{e - 2} - \frac {256}{x {\left (e^{3} - 6 \, e^{2} + 12 \, e - 8\right )} - 4 \, e^{2} + 16 \, e - 16} \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (21) = 42\).
Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.40 \[ \int \frac {16+144 x+36 x^2+e^2 x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx=\frac {x e^{2} - 20 \, x e + 36 \, x}{e^{2} - 4 \, e + 4} - \frac {256}{{\left (x e - 2 \, x - 4\right )} {\left (e^{2} - 4 \, e + 4\right )}} \]
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Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.15 \[ \int \frac {16+144 x+36 x^2+e^2 x^2+e \left (-8 x-20 x^2\right )}{16+16 x+4 x^2+e^2 x^2+e \left (-8 x-4 x^2\right )} \, dx=\frac {x\,\left ({\mathrm {e}}^2-20\,\mathrm {e}+36\right )}{{\left (\mathrm {e}-2\right )}^2}-\frac {256}{\left (\mathrm {e}-2\right )\,\left (x\,\left ({\mathrm {e}}^2-4\,\mathrm {e}+4\right )-4\,\mathrm {e}+8\right )} \]
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