Integrand size = 11, antiderivative size = 4 \[ \int \frac {1}{1-(1+x)^2} \, dx=\text {arctanh}(1+x) \]
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Time = 0.00 (sec) , antiderivative size = 4, 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 = {253, 212} \[ \int \frac {1}{1-(1+x)^2} \, dx=\text {arctanh}(x+1) \]
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Rule 212
Rule 253
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,1+x\right ) \\ & = \tanh ^{-1}(1+x) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(15\) vs. \(2(4)=8\).
Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 3.75 \[ \int \frac {1}{1-(1+x)^2} \, dx=-\frac {\log (x)}{2}+\frac {1}{2} \log (2+x) \]
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Leaf count of result is larger than twice the leaf count of optimal. \(11\) vs. \(2(4)=8\).
Time = 0.71 (sec) , antiderivative size = 12, normalized size of antiderivative = 3.00
method | result | size |
default | \(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x +2\right )}{2}\) | \(12\) |
norman | \(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x +2\right )}{2}\) | \(12\) |
risch | \(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x +2\right )}{2}\) | \(12\) |
parallelrisch | \(-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x +2\right )}{2}\) | \(12\) |
meijerg | \(\frac {\ln \left (1+\frac {x}{2}\right )}{2}-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (2\right )}{2}\) | \(18\) |
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Leaf count of result is larger than twice the leaf count of optimal. 11 vs. \(2 (4) = 8\).
Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 2.75 \[ \int \frac {1}{1-(1+x)^2} \, dx=\frac {1}{2} \, \log \left (x + 2\right ) - \frac {1}{2} \, \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 10 vs. \(2 (3) = 6\).
Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 2.50 \[ \int \frac {1}{1-(1+x)^2} \, dx=- \frac {\log {\left (x \right )}}{2} + \frac {\log {\left (x + 2 \right )}}{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 11 vs. \(2 (4) = 8\).
Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 2.75 \[ \int \frac {1}{1-(1+x)^2} \, dx=\frac {1}{2} \, \log \left (x + 2\right ) - \frac {1}{2} \, \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 13 vs. \(2 (4) = 8\).
Time = 0.30 (sec) , antiderivative size = 13, normalized size of antiderivative = 3.25 \[ \int \frac {1}{1-(1+x)^2} \, dx=\frac {1}{2} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.11 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1-(1+x)^2} \, dx=\mathrm {atanh}\left (x+1\right ) \]
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