Integrand size = 11, antiderivative size = 36 \[ \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx=\frac {1}{8 (1-x)}-\frac {1}{8 (1+x)^2}-\frac {1}{4 (1+x)}+\frac {3 \text {arctanh}(x)}{8} \]
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Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {46, 213} \[ \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx=\frac {3 \text {arctanh}(x)}{8}+\frac {1}{8 (1-x)}-\frac {1}{4 (x+1)}-\frac {1}{8 (x+1)^2} \]
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Rule 46
Rule 213
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{8 (-1+x)^2}+\frac {1}{4 (1+x)^3}+\frac {1}{4 (1+x)^2}-\frac {3}{8 \left (-1+x^2\right )}\right ) \, dx \\ & = \frac {1}{8 (1-x)}-\frac {1}{8 (1+x)^2}-\frac {1}{4 (1+x)}-\frac {3}{8} \int \frac {1}{-1+x^2} \, dx \\ & = \frac {1}{8 (1-x)}-\frac {1}{8 (1+x)^2}-\frac {1}{4 (1+x)}+\frac {3 \text {arctanh}(x)}{8} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx=\frac {1}{16} \left (\frac {4-6 x-6 x^2}{(-1+x) (1+x)^2}-3 \log (-1+x)+3 \log (1+x)\right ) \]
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Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97
method | result | size |
default | \(-\frac {1}{8 \left (-1+x \right )}-\frac {3 \ln \left (-1+x \right )}{16}-\frac {1}{8 \left (1+x \right )^{2}}-\frac {1}{4 \left (1+x \right )}+\frac {3 \ln \left (1+x \right )}{16}\) | \(35\) |
norman | \(\frac {-\frac {3}{8} x -\frac {3}{8} x^{2}+\frac {1}{4}}{\left (-1+x \right ) \left (1+x \right )^{2}}-\frac {3 \ln \left (-1+x \right )}{16}+\frac {3 \ln \left (1+x \right )}{16}\) | \(35\) |
risch | \(\frac {-\frac {3}{8} x -\frac {3}{8} x^{2}+\frac {1}{4}}{\left (-1+x \right ) \left (1+x \right )^{2}}-\frac {3 \ln \left (-1+x \right )}{16}+\frac {3 \ln \left (1+x \right )}{16}\) | \(35\) |
parallelrisch | \(-\frac {3 \ln \left (-1+x \right ) x^{3}-3 \ln \left (1+x \right ) x^{3}-4+3 \ln \left (-1+x \right ) x^{2}-3 \ln \left (1+x \right ) x^{2}-3 \ln \left (-1+x \right ) x +3 \ln \left (1+x \right ) x +6 x^{2}-3 \ln \left (-1+x \right )+3 \ln \left (1+x \right )+6 x}{16 \left (-1+x \right ) \left (1+x \right )^{2}}\) | \(85\) |
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Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (26) = 52\).
Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.64 \[ \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx=-\frac {6 \, x^{2} - 3 \, {\left (x^{3} + x^{2} - x - 1\right )} \log \left (x + 1\right ) + 3 \, {\left (x^{3} + x^{2} - x - 1\right )} \log \left (x - 1\right ) + 6 \, x - 4}{16 \, {\left (x^{3} + x^{2} - x - 1\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx=\frac {- 3 x^{2} - 3 x + 2}{8 x^{3} + 8 x^{2} - 8 x - 8} - \frac {3 \log {\left (x - 1 \right )}}{16} + \frac {3 \log {\left (x + 1 \right )}}{16} \]
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none
Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx=-\frac {3 \, x^{2} + 3 \, x - 2}{8 \, {\left (x^{3} + x^{2} - x - 1\right )}} + \frac {3}{16} \, \log \left (x + 1\right ) - \frac {3}{16} \, \log \left (x - 1\right ) \]
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none
Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx=-\frac {1}{8 \, {\left (x - 1\right )}} + \frac {\frac {12}{x - 1} + 5}{32 \, {\left (\frac {2}{x - 1} + 1\right )}^{2}} + \frac {3}{16} \, \log \left ({\left | -\frac {2}{x - 1} - 1 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx=\frac {3\,\mathrm {atanh}\left (x\right )}{8}+\frac {\frac {3\,x^2}{8}+\frac {3\,x}{8}-\frac {1}{4}}{-x^3-x^2+x+1} \]
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