Optimal. Leaf size=36 \[ \frac {1}{8 (1-x)}-\frac {1}{4 (x+1)}-\frac {1}{8 (x+1)^2}+\frac {3}{8} \tanh ^{-1}(x) \]
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Rubi [A] time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {44, 207} \[ \frac {1}{8 (1-x)}-\frac {1}{4 (x+1)}-\frac {1}{8 (x+1)^2}+\frac {3}{8} \tanh ^{-1}(x) \]
Antiderivative was successfully verified.
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Rule 44
Rule 207
Rubi steps
\begin {align*} \int \frac {1}{(-1+x)^2 (1+x)^3} \, dx &=\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}{8} \tanh ^{-1}(x)\\ \end {align*}
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Mathematica [A] time = 0.02, size = 38, normalized size = 1.06 \[ \frac {1}{16} \left (\frac {-6 x^2-6 x+4}{(x-1) (x+1)^2}-3 \log (x-1)+3 \log (x+1)\right ) \]
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.03, size = 31, normalized size = 0.86 \[ \frac {-3 x^2-3 x+2}{8 (x-1) (x+1)^2}+\frac {3}{8} \tanh ^{-1}(x) \]
Antiderivative was successfully verified.
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fricas [B] time = 1.11, size = 59, normalized size = 1.64 \[ -\frac {6 \, x^{2} - 3 \, {\left (x^{3} + x^{2} - x - 1\right )} \log \left (x + 1\right ) + 3 \, {\left (x^{3} + x^{2} - x - 1\right )} \log \left (x - 1\right ) + 6 \, x - 4}{16 \, {\left (x^{3} + x^{2} - x - 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.06, size = 43, normalized size = 1.19 \[ -\frac {1}{8 \, {\left (x - 1\right )}} + \frac {\frac {12}{x - 1} + 5}{32 \, {\left (\frac {2}{x - 1} + 1\right )}^{2}} + \frac {3}{16} \, \log \left ({\left | -\frac {2}{x - 1} - 1 \right |}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.29, size = 35, normalized size = 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 | \(\frac {-\frac {3}{8} x -\frac {3}{8} x^{2}+\frac {1}{4}}{\left (-1+x \right ) \left (1+x \right )^{2}}-\frac {3 \ln \left (-1+x \right )}{16}+\frac {3 \ln \left (1+x \right )}{16}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 38, normalized size = 1.06 \[ -\frac {3 \, x^{2} + 3 \, x - 2}{8 \, {\left (x^{3} + x^{2} - x - 1\right )}} + \frac {3}{16} \, \log \left (x + 1\right ) - \frac {3}{16} \, \log \left (x - 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.04, size = 31, normalized size = 0.86 \[ \frac {3\,\mathrm {atanh}\relax (x)}{8}+\frac {\frac {3\,x^2}{8}+\frac {3\,x}{8}-\frac {1}{4}}{-x^3-x^2+x+1} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 41, normalized size = 1.14 \[ \frac {- 3 x^{2} - 3 x + 2}{8 x^{3} + 8 x^{2} - 8 x - 8} - \frac {3 \log {\left (x - 1 \right )}}{16} + \frac {3 \log {\left (x + 1 \right )}}{16} \]
Verification of antiderivative is not currently implemented for this CAS.
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