Integrand size = 35, antiderivative size = 24 \[ \int \frac {1+x+3 x^2-e^2 x^2}{-x-3 x^2+e^2 x^2} \, dx=-x-\log (2 x)+\log \left (\frac {1}{2} \left (-1+\left (-3+e^2\right ) x\right )\right ) \]
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Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6, 1607, 907} \[ \int \frac {1+x+3 x^2-e^2 x^2}{-x-3 x^2+e^2 x^2} \, dx=-x-\log (x)+\log \left (\left (3-e^2\right ) x+1\right ) \]
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Rule 6
Rule 907
Rule 1607
Rubi steps \begin{align*} \text {integral}& = \int \frac {1+x+\left (3-e^2\right ) x^2}{-x-3 x^2+e^2 x^2} \, dx \\ & = \int \frac {1+x+\left (3-e^2\right ) x^2}{-x+\left (-3+e^2\right ) x^2} \, dx \\ & = \int \frac {1+x+\left (3-e^2\right ) x^2}{x \left (-1+\left (-3+e^2\right ) x\right )} \, dx \\ & = \int \left (-1-\frac {1}{x}+\frac {3-e^2}{1+\left (3-e^2\right ) x}\right ) \, dx \\ & = -x-\log (x)+\log \left (1+\left (3-e^2\right ) x\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1+x+3 x^2-e^2 x^2}{-x-3 x^2+e^2 x^2} \, dx=-x-\log (x)+\log \left (1+3 x-e^2 x\right ) \]
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Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79
method | result | size |
default | \(-x -\ln \left (x \right )+\ln \left ({\mathrm e}^{2} x -3 x -1\right )\) | \(19\) |
norman | \(-x -\ln \left (x \right )+\ln \left ({\mathrm e}^{2} x -3 x -1\right )\) | \(19\) |
risch | \(-x -\ln \left (x \right )+\ln \left (1+x \left (-{\mathrm e}^{2}+3\right )\right )\) | \(20\) |
parallelrisch | \(-x -\ln \left (x \right )+\ln \left (\frac {{\mathrm e}^{2} x -3 x -1}{{\mathrm e}^{2}-3}\right )\) | \(26\) |
meijerg | \(\frac {\left ({\mathrm e}^{2}-3\right ) \left (\ln \left (x \right )+\ln \left (-{\mathrm e}^{2}+3\right )-\ln \left (1+x \left (-{\mathrm e}^{2}+3\right )\right )\right )}{-{\mathrm e}^{2}+3}+\frac {\left ({\mathrm e}^{2}-3\right ) \left (x \left (-{\mathrm e}^{2}+3\right )-\ln \left (1+x \left (-{\mathrm e}^{2}+3\right )\right )\right )}{\left (-{\mathrm e}^{2}+3\right )^{2}}-\frac {\ln \left (1+x \left (-{\mathrm e}^{2}+3\right )\right )}{-{\mathrm e}^{2}+3}\) | \(94\) |
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Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {1+x+3 x^2-e^2 x^2}{-x-3 x^2+e^2 x^2} \, dx=-x + \log \left (x e^{2} - 3 \, x - 1\right ) - \log \left (x\right ) \]
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Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {1+x+3 x^2-e^2 x^2}{-x-3 x^2+e^2 x^2} \, dx=- x - \log {\left (x \right )} + \log {\left (x - \frac {2}{-6 + 2 e^{2}} \right )} \]
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Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {1+x+3 x^2-e^2 x^2}{-x-3 x^2+e^2 x^2} \, dx=-x + \log \left (x {\left (e^{2} - 3\right )} - 1\right ) - \log \left (x\right ) \]
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Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {1+x+3 x^2-e^2 x^2}{-x-3 x^2+e^2 x^2} \, dx=-\frac {x e^{2} - 3 \, x}{e^{2} - 3} + \log \left ({\left | x e^{2} - 3 \, x - 1 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]
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Time = 14.39 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {1+x+3 x^2-e^2 x^2}{-x-3 x^2+e^2 x^2} \, dx=-x-2\,\mathrm {atanh}\left (x\,\left (2\,{\mathrm {e}}^2-6\right )-1\right ) \]
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