Integrand size = 11, antiderivative size = 6 \[ \int \frac {2}{-1+4 x^2} \, dx=-\text {arctanh}(2 x) \]
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Time = 0.00 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 213} \[ \int \frac {2}{-1+4 x^2} \, dx=-\text {arctanh}(2 x) \]
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Rule 12
Rule 213
Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {1}{-1+4 x^2} \, dx \\ & = -\tanh ^{-1}(2 x) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(23\) vs. \(2(6)=12\).
Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 3.83 \[ \int \frac {2}{-1+4 x^2} \, dx=2 \left (\frac {1}{4} \log (1-2 x)-\frac {1}{4} \log (1+2 x)\right ) \]
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Time = 0.77 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.17
method | result | size |
meijerg | \(-\operatorname {arctanh}\left (2 x \right )\) | \(7\) |
parallelrisch | \(\frac {\ln \left (x -\frac {1}{2}\right )}{2}-\frac {\ln \left (x +\frac {1}{2}\right )}{2}\) | \(14\) |
default | \(-\frac {\ln \left (1+2 x \right )}{2}+\frac {\ln \left (2 x -1\right )}{2}\) | \(18\) |
norman | \(-\frac {\ln \left (1+2 x \right )}{2}+\frac {\ln \left (2 x -1\right )}{2}\) | \(18\) |
risch | \(-\frac {\ln \left (1+2 x \right )}{2}+\frac {\ln \left (2 x -1\right )}{2}\) | \(18\) |
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Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (6) = 12\).
Time = 0.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 2.83 \[ \int \frac {2}{-1+4 x^2} \, dx=-\frac {1}{2} \, \log \left (2 \, x + 1\right ) + \frac {1}{2} \, \log \left (2 \, x - 1\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (5) = 10\).
Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 2.50 \[ \int \frac {2}{-1+4 x^2} \, dx=\frac {\log {\left (x - \frac {1}{2} \right )}}{2} - \frac {\log {\left (x + \frac {1}{2} \right )}}{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (6) = 12\).
Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 2.83 \[ \int \frac {2}{-1+4 x^2} \, dx=-\frac {1}{2} \, \log \left (2 \, x + 1\right ) + \frac {1}{2} \, \log \left (2 \, x - 1\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (6) = 12\).
Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 2.50 \[ \int \frac {2}{-1+4 x^2} \, dx=-\frac {1}{2} \, \log \left ({\left | x + \frac {1}{2} \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x - \frac {1}{2} \right |}\right ) \]
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Time = 9.09 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {2}{-1+4 x^2} \, dx=-\mathrm {atanh}\left (2\,x\right ) \]
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