Integrand size = 53, antiderivative size = 16 \[ \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{e^4-2 e^2 x+x^2} \, dx=x \left (4+e^4+\frac {1}{-e^2+x}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {27, 1864} \[ \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{e^4-2 e^2 x+x^2} \, dx=\left (4+e^4\right ) x-\frac {e^2}{e^2-x} \]
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Rule 27
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{\left (-e^2+x\right )^2} \, dx \\ & = \int \left (4+e^4-\frac {e^2}{\left (-e^2+x\right )^2}\right ) \, dx \\ & = -\frac {e^2}{e^2-x}+\left (4+e^4\right ) x \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81 \[ \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{e^4-2 e^2 x+x^2} \, dx=-\frac {e^2}{e^2-x}-\left (4+e^4\right ) \left (e^2-x\right ) \]
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Time = 0.73 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31
method | result | size |
risch | \(x \,{\mathrm e}^{4}+4 x -\frac {{\mathrm e}^{2}}{{\mathrm e}^{2}-x}\) | \(21\) |
norman | \(\frac {\left (-4-{\mathrm e}^{4}\right ) x^{2}+\left ({\mathrm e}^{4}\right )^{2}+4 \,{\mathrm e}^{4}-{\mathrm e}^{2}}{{\mathrm e}^{2}-x}\) | \(38\) |
gosper | \(\frac {\left ({\mathrm e}^{4}\right )^{2}-x^{2} {\mathrm e}^{4}+4 \,{\mathrm e}^{4}-4 x^{2}-{\mathrm e}^{2}}{{\mathrm e}^{2}-x}\) | \(40\) |
parallelrisch | \(\frac {\left ({\mathrm e}^{4}\right )^{2}-x^{2} {\mathrm e}^{4}+4 \,{\mathrm e}^{4}-4 x^{2}-{\mathrm e}^{2}}{{\mathrm e}^{2}-x}\) | \(40\) |
meijerg | \(\frac {4 x}{1-x \,{\mathrm e}^{-2}}+\left (-2 \,{\mathrm e}^{6}-8 \,{\mathrm e}^{2}\right ) \left (\frac {x \,{\mathrm e}^{-2}}{1-x \,{\mathrm e}^{-2}}+\ln \left (1-x \,{\mathrm e}^{-2}\right )\right )-{\mathrm e}^{2} \left (4+{\mathrm e}^{4}\right ) \left (-\frac {x \,{\mathrm e}^{-2} \left (-3 x \,{\mathrm e}^{-2}+6\right )}{3 \left (1-x \,{\mathrm e}^{-2}\right )}-2 \ln \left (1-x \,{\mathrm e}^{-2}\right )\right )+\frac {{\mathrm e}^{4} x}{1-x \,{\mathrm e}^{-2}}-\frac {x \,{\mathrm e}^{-2}}{1-x \,{\mathrm e}^{-2}}\) | \(113\) |
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Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (14) = 28\).
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.19 \[ \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{e^4-2 e^2 x+x^2} \, dx=\frac {x^{2} e^{4} + 4 \, x^{2} - x e^{6} - {\left (4 \, x - 1\right )} e^{2}}{x - e^{2}} \]
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Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{e^4-2 e^2 x+x^2} \, dx=x \left (4 + e^{4}\right ) + \frac {e^{2}}{x - e^{2}} \]
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Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{e^4-2 e^2 x+x^2} \, dx=x {\left (e^{4} + 4\right )} + \frac {e^{2}}{x - e^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{e^4-2 e^2 x+x^2} \, dx=x e^{4} + 4 \, x + \frac {e^{2}}{x - e^{2}} \]
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Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {4 e^4+e^2 (-1-8 x)+4 x^2+e^4 \left (e^4-2 e^2 x+x^2\right )}{e^4-2 e^2 x+x^2} \, dx=x\,\left ({\mathrm {e}}^4+4\right )+\frac {{\mathrm {e}}^2}{x-{\mathrm {e}}^2} \]
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