Integrand size = 13, antiderivative size = 12 \[ \int \frac {e^x}{-4+e^{2 x}} \, dx=-\frac {1}{2} \text {arctanh}\left (\frac {e^x}{2}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2281, 213} \[ \int \frac {e^x}{-4+e^{2 x}} \, dx=-\frac {1}{2} \text {arctanh}\left (\frac {e^x}{2}\right ) \]
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Rule 213
Rule 2281
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{-4+x^2} \, dx,x,e^x\right ) \\ & = -\frac {1}{2} \tanh ^{-1}\left (\frac {e^x}{2}\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {e^x}{-4+e^{2 x}} \, dx=-\frac {1}{2} \text {arctanh}\left (\frac {e^x}{2}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(15\) vs. \(2(7)=14\).
Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33
method | result | size |
default | \(\frac {\ln \left (-2+{\mathrm e}^{x}\right )}{4}-\frac {\ln \left (2+{\mathrm e}^{x}\right )}{4}\) | \(16\) |
norman | \(\frac {\ln \left (-2+{\mathrm e}^{x}\right )}{4}-\frac {\ln \left (2+{\mathrm e}^{x}\right )}{4}\) | \(16\) |
risch | \(\frac {\ln \left (-2+{\mathrm e}^{x}\right )}{4}-\frac {\ln \left (2+{\mathrm e}^{x}\right )}{4}\) | \(16\) |
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Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (7) = 14\).
Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {e^x}{-4+e^{2 x}} \, dx=-\frac {1}{4} \, \log \left (e^{x} + 2\right ) + \frac {1}{4} \, \log \left (e^{x} - 2\right ) \]
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Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {e^x}{-4+e^{2 x}} \, dx=\frac {\log {\left (e^{x} - 2 \right )}}{4} - \frac {\log {\left (e^{x} + 2 \right )}}{4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (7) = 14\).
Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {e^x}{-4+e^{2 x}} \, dx=-\frac {1}{4} \, \log \left (e^{x} + 2\right ) + \frac {1}{4} \, \log \left (e^{x} - 2\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 16 vs. \(2 (7) = 14\).
Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \frac {e^x}{-4+e^{2 x}} \, dx=-\frac {1}{4} \, \log \left (e^{x} + 2\right ) + \frac {1}{4} \, \log \left ({\left | e^{x} - 2 \right |}\right ) \]
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Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25 \[ \int \frac {e^x}{-4+e^{2 x}} \, dx=\frac {\ln \left ({\mathrm {e}}^x-2\right )}{4}-\frac {\ln \left ({\mathrm {e}}^x+2\right )}{4} \]
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