Integrand size = 87, antiderivative size = 28 \[ \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+x^2+e^{10} x^2+e^5 \left (-32 x+2 x^2\right )} \, dx=\frac {e^{4 e^{2 x}+x^2} x}{16-x-e^5 x} \]
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Leaf count is larger than twice the leaf count of optimal. \(95\) vs. \(2(28)=56\).
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.39, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6, 2326} \[ \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+x^2+e^{10} x^2+e^5 \left (-32 x+2 x^2\right )} \, dx=\frac {e^{x^2+4 e^{2 x}} \left (-\left (\left (1+e^5\right ) x^3\right )+16 x^2+4 e^{2 x} \left (-e^5 x^2-x^2+16 x\right )\right )}{\left (x+4 e^{2 x}\right ) \left (\left (1+e^{10}\right ) x^2-2 e^5 \left (16 x-x^2\right )-32 x+256\right )} \]
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Rule 6
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+\left (1+e^{10}\right ) x^2+e^5 \left (-32 x+2 x^2\right )} \, dx \\ & = \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2+\left (-2-2 e^5\right ) x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+\left (1+e^{10}\right ) x^2+e^5 \left (-32 x+2 x^2\right )} \, dx \\ & = \frac {e^{4 e^{2 x}+x^2} \left (16 x^2-\left (1+e^5\right ) x^3+4 e^{2 x} \left (16 x-x^2-e^5 x^2\right )\right )}{\left (4 e^{2 x}+x\right ) \left (256-32 x+\left (1+e^{10}\right ) x^2-2 e^5 \left (16 x-x^2\right )\right )} \\ \end{align*}
Time = 1.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+x^2+e^{10} x^2+e^5 \left (-32 x+2 x^2\right )} \, dx=-\frac {e^{4 e^{2 x}+x^2} x}{-16+x+e^5 x} \]
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Time = 1.55 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86
method | result | size |
norman | \(-\frac {x \,{\mathrm e}^{4 \,{\mathrm e}^{2 x}+x^{2}}}{x \,{\mathrm e}^{5}+x -16}\) | \(24\) |
risch | \(-\frac {x \,{\mathrm e}^{4 \,{\mathrm e}^{2 x}+x^{2}}}{x \,{\mathrm e}^{5}+x -16}\) | \(24\) |
parallelrisch | \(-\frac {x \,{\mathrm e}^{4 \,{\mathrm e}^{2 x}+x^{2}}}{x \,{\mathrm e}^{5}+x -16}\) | \(24\) |
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Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+x^2+e^{10} x^2+e^5 \left (-32 x+2 x^2\right )} \, dx=-\frac {x e^{\left (x^{2} + 4 \, e^{\left (2 \, x\right )}\right )}}{x e^{5} + x - 16} \]
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Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+x^2+e^{10} x^2+e^5 \left (-32 x+2 x^2\right )} \, dx=- \frac {x e^{x^{2} + 4 e^{2 x}}}{x + x e^{5} - 16} \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+x^2+e^{10} x^2+e^5 \left (-32 x+2 x^2\right )} \, dx=-\frac {x e^{\left (x^{2} + 4 \, e^{\left (2 \, x\right )}\right )}}{x {\left (e^{5} + 1\right )} - 16} \]
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\[ \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+x^2+e^{10} x^2+e^5 \left (-32 x+2 x^2\right )} \, dx=\int { -\frac {2 \, {\left (x^{3} e^{5} + x^{3} - 16 \, x^{2} + 4 \, {\left (x^{2} e^{5} + x^{2} - 16 \, x\right )} e^{\left (2 \, x\right )} - 8\right )} e^{\left (x^{2} + 4 \, e^{\left (2 \, x\right )}\right )}}{x^{2} e^{10} + x^{2} + 2 \, {\left (x^{2} - 16 \, x\right )} e^{5} - 32 \, x + 256} \,d x } \]
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Time = 15.58 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{4 e^{2 x}+x^2} \left (16+32 x^2-2 x^3-2 e^5 x^3+e^{2 x} \left (128 x-8 x^2-8 e^5 x^2\right )\right )}{256-32 x+x^2+e^{10} x^2+e^5 \left (-32 x+2 x^2\right )} \, dx=-\frac {x\,{\mathrm {e}}^{4\,{\mathrm {e}}^{2\,x}+x^2}}{\left (x-\frac {16}{{\mathrm {e}}^5+1}\right )\,\left ({\mathrm {e}}^5+1\right )} \]
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