Integrand size = 11, antiderivative size = 45 \[ \int \frac {1}{\left (1-(1+x)^2\right )^3} \, dx=\frac {1+x}{4 \left (1-(1+x)^2\right )^2}+\frac {3 (1+x)}{8 \left (1-(1+x)^2\right )}+\frac {3}{8} \text {arctanh}(1+x) \]
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Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {253, 205, 212} \[ \int \frac {1}{\left (1-(1+x)^2\right )^3} \, dx=\frac {3}{8} \text {arctanh}(x+1)+\frac {3 (x+1)}{8 \left (1-(x+1)^2\right )}+\frac {x+1}{4 \left (1-(x+1)^2\right )^2} \]
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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 )^3} \, dx,x,1+x\right ) \\ & = \frac {1+x}{4 \left (1-(1+x)^2\right )^2}+\frac {3}{4} \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2} \, dx,x,1+x\right ) \\ & = \frac {1+x}{4 \left (1-(1+x)^2\right )^2}+\frac {3 (1+x)}{8 \left (1-(1+x)^2\right )}+\frac {3}{8} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,1+x\right ) \\ & = \frac {1+x}{4 \left (1-(1+x)^2\right )^2}+\frac {3 (1+x)}{8 \left (1-(1+x)^2\right )}+\frac {3}{8} \tanh ^{-1}(1+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (1-(1+x)^2\right )^3} \, dx=\frac {1}{16} \left (\frac {1}{x^2}-\frac {3}{x}-\frac {1}{(2+x)^2}-\frac {3}{2+x}-3 \log (x)+3 \log (2+x)\right ) \]
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Time = 0.73 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {1}{16 x^{2}}-\frac {3}{16 x}-\frac {3 \ln \left (x \right )}{16}-\frac {1}{16 \left (x +2\right )^{2}}-\frac {3}{16 \left (x +2\right )}+\frac {3 \ln \left (x +2\right )}{16}\) | \(36\) |
norman | \(\frac {\frac {1}{4}-\frac {9}{8} x^{2}-\frac {3}{8} x^{3}-\frac {1}{2} x}{x^{2} \left (x +2\right )^{2}}-\frac {3 \ln \left (x \right )}{16}+\frac {3 \ln \left (x +2\right )}{16}\) | \(36\) |
risch | \(\frac {\frac {1}{4}-\frac {9}{8} x^{2}-\frac {3}{8} x^{3}-\frac {1}{2} x}{x^{2} \left (x +2\right )^{2}}-\frac {3 \ln \left (x \right )}{16}+\frac {3 \ln \left (x +2\right )}{16}\) | \(36\) |
meijerg | \(\frac {x \left (\frac {7 x}{2}+8\right )}{128 \left (1+\frac {x}{2}\right )^{2}}+\frac {3 \ln \left (1+\frac {x}{2}\right )}{16}-\frac {7}{64}-\frac {3 \ln \left (x \right )}{16}+\frac {3 \ln \left (2\right )}{16}+\frac {1}{16 x^{2}}-\frac {3}{16 x}\) | \(44\) |
parallelrisch | \(-\frac {3 \ln \left (x \right ) x^{4}-3 \ln \left (x +2\right ) x^{4}-4+12 \ln \left (x \right ) x^{3}-12 \ln \left (x +2\right ) x^{3}+12 \ln \left (x \right ) x^{2}-12 \ln \left (x +2\right ) x^{2}+6 x^{3}+18 x^{2}+8 x}{16 x^{2} \left (x +2\right )^{2}}\) | \(74\) |
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (35) = 70\).
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\left (1-(1+x)^2\right )^3} \, dx=-\frac {6 \, x^{3} + 18 \, x^{2} - 3 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (x + 2\right ) + 3 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (x\right ) + 8 \, x - 4}{16 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (1-(1+x)^2\right )^3} \, dx=- \frac {3 \log {\left (x \right )}}{16} + \frac {3 \log {\left (x + 2 \right )}}{16} - \frac {3 x^{3} + 9 x^{2} + 4 x - 2}{8 x^{4} + 32 x^{3} + 32 x^{2}} \]
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none
Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (1-(1+x)^2\right )^3} \, dx=-\frac {3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2}{8 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} + \frac {3}{16} \, \log \left (x + 2\right ) - \frac {3}{16} \, \log \left (x\right ) \]
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none
Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (1-(1+x)^2\right )^3} \, dx=-\frac {3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2}{8 \, {\left (x^{2} + 2 \, x\right )}^{2}} + \frac {3}{16} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {3}{16} \, \log \left ({\left | x \right |}\right ) \]
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Time = 10.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (1-(1+x)^2\right )^3} \, dx=\frac {3\,\mathrm {atanh}\left (x+1\right )}{8}+\frac {\frac {5\,x}{8}-\frac {3\,{\left (x+1\right )}^3}{8}+\frac {5}{8}}{{\left (x+1\right )}^4-2\,{\left (x+1\right )}^2+1} \]
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