Integrand size = 47, antiderivative size = 23 \[ \int \frac {-34-8 x-x^2+e^5 \left (16+8 x+x^2\right )}{-16-8 x-x^2+e^5 \left (16+8 x+x^2\right )} \, dx=x-\frac {2 (5-x)}{\left (1-e^5\right ) (4+x)} \]
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Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {2009, 27, 12, 697} \[ \int \frac {-34-8 x-x^2+e^5 \left (16+8 x+x^2\right )}{-16-8 x-x^2+e^5 \left (16+8 x+x^2\right )} \, dx=x-\frac {18}{\left (1-e^5\right ) (x+4)} \]
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Rule 12
Rule 27
Rule 697
Rule 2009
Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 \left (17-8 e^5\right )-8 \left (1-e^5\right ) x-\left (1-e^5\right ) x^2}{-16 \left (1-e^5\right )-8 \left (1-e^5\right ) x+\left (-1+e^5\right ) x^2} \, dx \\ & = \int \frac {-2 \left (17-8 e^5\right )-8 \left (1-e^5\right ) x-\left (1-e^5\right ) x^2}{\left (-1+e^5\right ) (4+x)^2} \, dx \\ & = \frac {\int \frac {-2 \left (17-8 e^5\right )-8 \left (1-e^5\right ) x-\left (1-e^5\right ) x^2}{(4+x)^2} \, dx}{-1+e^5} \\ & = \frac {\int \left (-1+e^5-\frac {18}{(4+x)^2}\right ) \, dx}{-1+e^5} \\ & = x-\frac {18}{\left (1-e^5\right ) (4+x)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-34-8 x-x^2+e^5 \left (16+8 x+x^2\right )}{-16-8 x-x^2+e^5 \left (16+8 x+x^2\right )} \, dx=-\frac {-\frac {18}{4+x}+\left (1-e^5\right ) (4+x)}{-1+e^5} \]
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Time = 0.57 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
risch | \(x +\frac {18}{x \,{\mathrm e}^{5}+4 \,{\mathrm e}^{5}-x -4}\) | \(20\) |
default | \(\frac {x \,{\mathrm e}^{5}-x +\frac {18}{4+x}}{{\mathrm e}^{5}-1}\) | \(23\) |
norman | \(\frac {x^{2}-\frac {2 \left (8 \,{\mathrm e}^{5}-17\right )}{{\mathrm e}^{5}-1}}{4+x}\) | \(25\) |
parallelrisch | \(\frac {x^{2} {\mathrm e}^{5}-x^{2}-16 \,{\mathrm e}^{5}+34}{\left ({\mathrm e}^{5}-1\right ) \left (4+x \right )}\) | \(30\) |
gosper | \(\frac {x^{2} {\mathrm e}^{5}-x^{2}-16 \,{\mathrm e}^{5}+34}{x \,{\mathrm e}^{5}+4 \,{\mathrm e}^{5}-x -4}\) | \(34\) |
meijerg | \(-\frac {17 x}{8 \left ({\mathrm e}^{5}-1\right ) \left (1+\frac {x}{4}\right )}+\frac {x \left (\frac {3 x}{4}+6\right )}{3+\frac {3 x}{4}}-8 \ln \left (1+\frac {x}{4}\right )+\frac {16 \left (\frac {{\mathrm e}^{5}}{2}-\frac {1}{2}\right ) \left (-\frac {x}{4 \left (1+\frac {x}{4}\right )}+\ln \left (1+\frac {x}{4}\right )\right )}{{\mathrm e}^{5}-1}+\frac {{\mathrm e}^{5} x}{\left ({\mathrm e}^{5}-1\right ) \left (1+\frac {x}{4}\right )}\) | \(89\) |
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-34-8 x-x^2+e^5 \left (16+8 x+x^2\right )}{-16-8 x-x^2+e^5 \left (16+8 x+x^2\right )} \, dx=-\frac {x^{2} - {\left (x^{2} + 4 \, x\right )} e^{5} + 4 \, x - 18}{{\left (x + 4\right )} e^{5} - x - 4} \]
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Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {-34-8 x-x^2+e^5 \left (16+8 x+x^2\right )}{-16-8 x-x^2+e^5 \left (16+8 x+x^2\right )} \, dx=x + \frac {18}{x \left (-1 + e^{5}\right ) - 4 + 4 e^{5}} \]
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Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-34-8 x-x^2+e^5 \left (16+8 x+x^2\right )}{-16-8 x-x^2+e^5 \left (16+8 x+x^2\right )} \, dx=x + \frac {18}{x {\left (e^{5} - 1\right )} + 4 \, e^{5} - 4} \]
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-34-8 x-x^2+e^5 \left (16+8 x+x^2\right )}{-16-8 x-x^2+e^5 \left (16+8 x+x^2\right )} \, dx=\frac {x e^{5} - x}{e^{5} - 1} + \frac {18}{{\left (x + 4\right )} {\left (e^{5} - 1\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {-34-8 x-x^2+e^5 \left (16+8 x+x^2\right )}{-16-8 x-x^2+e^5 \left (16+8 x+x^2\right )} \, dx=x+\frac {18}{\left ({\mathrm {e}}^5-1\right )\,\left (x+4\right )} \]
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