Integrand size = 60, antiderivative size = 19 \[ \int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx=\frac {4}{2+\frac {3 e^{-x}}{16 x}+x} \]
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\[ \int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx=\int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {64 e^x \left (3+3 x-16 e^x x^2\right )}{\left (3+16 e^x x (2+x)\right )^2} \, dx \\ & = 64 \int \frac {e^x \left (3+3 x-16 e^x x^2\right )}{\left (3+16 e^x x (2+x)\right )^2} \, dx \\ & = 64 \int \left (\frac {3 e^x \left (2+4 x+x^2\right )}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )^2}-\frac {e^x x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )}\right ) \, dx \\ & = -\left (64 \int \frac {e^x x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )} \, dx\right )+192 \int \frac {e^x \left (2+4 x+x^2\right )}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )^2} \, dx \\ & = -\left (64 \int \frac {e^x x}{(2+x) \left (3+16 e^x x (2+x)\right )} \, dx\right )+192 \int \frac {e^x \left (2+4 x+x^2\right )}{(2+x) \left (3+16 e^x x (2+x)\right )^2} \, dx \\ & = -\left (64 \int \left (\frac {e^x}{3+32 e^x x+16 e^x x^2}-\frac {2 e^x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )}\right ) \, dx\right )+192 \int \left (\frac {2 e^x}{\left (3+32 e^x x+16 e^x x^2\right )^2}+\frac {e^x x}{\left (3+32 e^x x+16 e^x x^2\right )^2}-\frac {2 e^x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )^2}\right ) \, dx \\ & = -\left (64 \int \frac {e^x}{3+32 e^x x+16 e^x x^2} \, dx\right )+128 \int \frac {e^x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )} \, dx+192 \int \frac {e^x x}{\left (3+32 e^x x+16 e^x x^2\right )^2} \, dx+384 \int \frac {e^x}{\left (3+32 e^x x+16 e^x x^2\right )^2} \, dx-384 \int \frac {e^x}{(2+x) \left (3+32 e^x x+16 e^x x^2\right )^2} \, dx \\ & = -\left (64 \int \frac {e^x}{3+16 e^x x (2+x)} \, dx\right )+128 \int \frac {e^x}{(2+x) \left (3+16 e^x x (2+x)\right )} \, dx+192 \int \frac {e^x x}{\left (3+16 e^x x (2+x)\right )^2} \, dx+384 \int \frac {e^x}{\left (3+16 e^x x (2+x)\right )^2} \, dx-384 \int \frac {e^x}{(2+x) \left (3+16 e^x x (2+x)\right )^2} \, dx \\ \end{align*}
Time = 1.64 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx=\frac {64 e^x x}{3+16 e^x x (2+x)} \]
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Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16
method | result | size |
norman | \(\frac {64 \,{\mathrm e}^{x} x}{16 \,{\mathrm e}^{x} x^{2}+32 \,{\mathrm e}^{x} x +3}\) | \(22\) |
parallelrisch | \(\frac {64 \,{\mathrm e}^{x} x}{16 \,{\mathrm e}^{x} x^{2}+32 \,{\mathrm e}^{x} x +3}\) | \(22\) |
risch | \(\frac {4}{2+x}-\frac {12}{\left (2+x \right ) \left (16 \,{\mathrm e}^{x} x^{2}+32 \,{\mathrm e}^{x} x +3\right )}\) | \(32\) |
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Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx=\frac {64 \, x e^{x}}{16 \, {\left (x^{2} + 2 \, x\right )} e^{x} + 3} \]
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Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\).
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx=- \frac {12}{3 x + \left (16 x^{3} + 64 x^{2} + 64 x\right ) e^{x} + 6} + \frac {4}{x + 2} \]
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Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx=\frac {64 \, x e^{x}}{16 \, {\left (x^{2} + 2 \, x\right )} e^{x} + 3} \]
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Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx=\frac {64 \, x e^{x}}{16 \, x^{2} e^{x} + 32 \, x e^{x} + 3} \]
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Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {-1024 e^{2 x} x^2+e^x (192+192 x)}{9+e^x \left (192 x+96 x^2\right )+e^{2 x} \left (1024 x^2+1024 x^3+256 x^4\right )} \, dx=\frac {4}{x+2}-\frac {12}{\left ({\mathrm {e}}^x\,\left (16\,x^2+32\,x\right )+3\right )\,\left (x+2\right )} \]
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