Integrand size = 11, antiderivative size = 27 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {1+x}{2 \left (1-(1+x)^2\right )}+\frac {1}{2} \text {arctanh}(1+x) \]
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Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {253, 205, 212} \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {1}{2} \text {arctanh}(x+1)+\frac {x+1}{2 \left (1-(x+1)^2\right )} \]
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Rule 205
Rule 212
Rule 253
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2} \, dx,x,1+x\right ) \\ & = \frac {1+x}{2 \left (1-(1+x)^2\right )}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,1+x\right ) \\ & = \frac {1+x}{2 \left (1-(1+x)^2\right )}+\frac {1}{2} \tanh ^{-1}(1+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {1}{4} \left (-\frac {2 (1+x)}{x (2+x)}-\log (x)+\log (2+x)\right ) \]
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Time = 0.76 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {1}{4 x}-\frac {\ln \left (x \right )}{4}-\frac {1}{4 \left (x +2\right )}+\frac {\ln \left (x +2\right )}{4}\) | \(24\) |
norman | \(\frac {-\frac {1}{2}-\frac {x}{2}}{x \left (x +2\right )}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x +2\right )}{4}\) | \(26\) |
risch | \(\frac {-\frac {1}{2}-\frac {x}{2}}{x \left (x +2\right )}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x +2\right )}{4}\) | \(26\) |
meijerg | \(\frac {3 x}{16 \left (\frac {3 x}{2}+3\right )}+\frac {\ln \left (1+\frac {x}{2}\right )}{4}-\frac {1}{8}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (2\right )}{4}-\frac {1}{4 x}\) | \(34\) |
parallelrisch | \(-\frac {\ln \left (x \right ) x^{2}-\ln \left (x +2\right ) x^{2}+2+2 \ln \left (x \right ) x -2 \ln \left (x +2\right ) x +2 x}{4 x \left (x +2\right )}\) | \(43\) |
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {{\left (x^{2} + 2 \, x\right )} \log \left (x + 2\right ) - {\left (x^{2} + 2 \, x\right )} \log \left (x\right ) - 2 \, x - 2}{4 \, {\left (x^{2} + 2 \, x\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {- x - 1}{2 x^{2} + 4 x} - \frac {\log {\left (x \right )}}{4} + \frac {\log {\left (x + 2 \right )}}{4} \]
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Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=-\frac {x + 1}{2 \, {\left (x^{2} + 2 \, x\right )}} + \frac {1}{4} \, \log \left (x + 2\right ) - \frac {1}{4} \, \log \left (x\right ) \]
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Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=-\frac {x + 1}{2 \, {\left (x^{2} + 2 \, x\right )}} + \frac {1}{4} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {1}{4} \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {\mathrm {atanh}\left (x+1\right )}{2}-\frac {x+1}{2\,\left ({\left (x+1\right )}^2-1\right )} \]
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