Integrand size = 14, antiderivative size = 62 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {x}{4 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3+\frac {\text {arctanh}(x)}{4} \]
[Out]
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6104, 6142, 205, 212} \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {\text {arctanh}(x)}{4}+\frac {x}{4 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3 \]
[In]
[Out]
Rule 205
Rule 212
Rule 6104
Rule 6142
Rubi steps \begin{align*} \text {integral}& = \frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3-\int \frac {x \coth ^{-1}(x)}{\left (1-x^2\right )^2} \, dx \\ & = -\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3+\frac {1}{2} \int \frac {1}{\left (1-x^2\right )^2} \, dx \\ & = \frac {x}{4 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3+\frac {1}{4} \int \frac {1}{1-x^2} \, dx \\ & = \frac {x}{4 \left (1-x^2\right )}-\frac {\coth ^{-1}(x)}{2 \left (1-x^2\right )}+\frac {x \coth ^{-1}(x)^2}{2 \left (1-x^2\right )}+\frac {1}{6} \coth ^{-1}(x)^3+\frac {\text {arctanh}(x)}{4} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {-6 x+12 \coth ^{-1}(x)-12 x \coth ^{-1}(x)^2+4 \left (-1+x^2\right ) \coth ^{-1}(x)^3-3 \left (-1+x^2\right ) \log (1-x)+3 \left (-1+x^2\right ) \log (1+x)}{24 \left (-1+x^2\right )} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. \(158\) vs. \(2(52)=104\).
Time = 1.03 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.56
method | result | size |
risch | \(-\frac {\ln \left (x -1\right )^{3}}{48}+\frac {\left (x^{2} \ln \left (1+x \right )-2 x -\ln \left (1+x \right )\right ) \ln \left (x -1\right )^{2}}{16 x^{2}-16}-\frac {\left (x^{2} \ln \left (1+x \right )^{2}-4 \ln \left (1+x \right ) x -\ln \left (1+x \right )^{2}+4\right ) \ln \left (x -1\right )}{16 \left (x -1\right ) \left (1+x \right )}+\frac {x^{2} \ln \left (1+x \right )^{3}+6 x^{2} \ln \left (1+x \right )-6 \ln \left (x -1\right ) x^{2}-6 x \ln \left (1+x \right )^{2}-\ln \left (1+x \right )^{3}+6 \ln \left (1+x \right )+6 \ln \left (x -1\right )-12 x}{48 \left (x -1\right ) \left (1+x \right )}\) | \(159\) |
default | \(-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (1+x \right )}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (1+x \right )}{4}-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (x -1\right )}-\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (x -1\right )}{4}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (\frac {x -1}{1+x}\right )}{4}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{1+x}}}\right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i}{\frac {1+x}{x -1}-1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right ) \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{1+x}}}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right )^{2} \pi }{4}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{2} \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right )^{3} \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{3} \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\frac {1+x}{x -1}-1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{2} \pi }{8}+\frac {\operatorname {arccoth}\left (x \right )^{3}}{6}+\frac {\operatorname {arccoth}\left (x \right ) \left (1+x \right )}{8 x -8}-\frac {1+x}{16 \left (x -1\right )}+\frac {\left (x -1\right ) \operatorname {arccoth}\left (x \right )}{8+8 x}+\frac {x -1}{16+16 x}\) | \(402\) |
parts | \(-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (1+x \right )}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (1+x \right )}{4}-\frac {\operatorname {arccoth}\left (x \right )^{2}}{4 \left (x -1\right )}-\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (x -1\right )}{4}+\frac {\operatorname {arccoth}\left (x \right )^{2} \ln \left (\frac {x -1}{1+x}\right )}{4}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{1+x}}}\right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i}{\frac {1+x}{x -1}-1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right ) \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\sqrt {\frac {x -1}{1+x}}}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right )^{2} \pi }{4}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{2} \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{x -1}\right )^{3} \pi }{8}+\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{3} \pi }{8}-\frac {i \operatorname {arccoth}\left (x \right )^{2} \operatorname {csgn}\left (\frac {i}{\frac {1+x}{x -1}-1}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )}{\left (x -1\right ) \left (\frac {1+x}{x -1}-1\right )}\right )^{2} \pi }{8}+\frac {\operatorname {arccoth}\left (x \right )^{3}}{6}+\frac {\operatorname {arccoth}\left (x \right ) \left (1+x \right )}{8 x -8}-\frac {1+x}{16 \left (x -1\right )}+\frac {\left (x -1\right ) \operatorname {arccoth}\left (x \right )}{8+8 x}+\frac {x -1}{16+16 x}\) | \(402\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {{\left (x^{2} - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{3} - 6 \, x \log \left (\frac {x + 1}{x - 1}\right )^{2} + 6 \, {\left (x^{2} + 1\right )} \log \left (\frac {x + 1}{x - 1}\right ) - 12 \, x}{48 \, {\left (x^{2} - 1\right )}} \]
[In]
[Out]
\[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\int \frac {\operatorname {acoth}^{2}{\left (x \right )}}{\left (x - 1\right )^{2} \left (x + 1\right )^{2}}\, dx \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (46) = 92\).
Time = 0.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.76 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, x}{x^{2} - 1} - \log \left (x + 1\right ) + \log \left (x - 1\right )\right )} \operatorname {arcoth}\left (x\right )^{2} - \frac {{\left ({\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right ) \log \left (x - 1\right ) + {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} - 4\right )} \operatorname {arcoth}\left (x\right )}{8 \, {\left (x^{2} - 1\right )}} + \frac {{\left (x^{2} - 1\right )} \log \left (x + 1\right )^{3} - 3 \, {\left (x^{2} - 1\right )} \log \left (x + 1\right )^{2} \log \left (x - 1\right ) - {\left (x^{2} - 1\right )} \log \left (x - 1\right )^{3} + 3 \, {\left ({\left (x^{2} - 1\right )} \log \left (x - 1\right )^{2} + 2 \, x^{2} - 2\right )} \log \left (x + 1\right ) - 6 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 12 \, x}{48 \, {\left (x^{2} - 1\right )}} \]
[In]
[Out]
none
Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=-\frac {{\left (x - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )^{2}}{16 \, {\left (x + 1\right )}} - \frac {{\left (x - 1\right )} \log \left (\frac {x + 1}{x - 1}\right )}{8 \, {\left (x + 1\right )}} - \frac {x - 1}{8 \, {\left (x + 1\right )}} \]
[In]
[Out]
Time = 5.81 (sec) , antiderivative size = 201, normalized size of antiderivative = 3.24 \[ \int \frac {\coth ^{-1}(x)^2}{\left (1-x^2\right )^2} \, dx=\frac {{\ln \left (\frac {1}{x}+1\right )}^3}{48}-\frac {{\ln \left (1-\frac {1}{x}\right )}^3}{48}-\frac {x}{4\,\left (x^2-1\right )}+\ln \left (1-\frac {1}{x}\right )\,\left (\frac {\frac {3\,x}{32}-\frac {1}{8}}{x^2-1}-\frac {\frac {x}{8}+\frac {1}{8}}{x^2-1}-\frac {{\ln \left (\frac {1}{x}+1\right )}^2}{16}+\frac {x}{32\,\left (x^2-1\right )}+\ln \left (\frac {1}{x}+1\right )\,\left (\frac {\frac {x}{4}+\frac {1}{16}}{x^2-1}-\frac {1}{16\,\left (x^2-1\right )}\right )\right )+{\ln \left (1-\frac {1}{x}\right )}^2\,\left (\frac {\ln \left (\frac {1}{x}+1\right )}{16}-\frac {x}{8\,\left (x^2-1\right )}\right )+\frac {\ln \left (\frac {1}{x}+1\right )}{4\,\left (x^2-1\right )}-\frac {x\,{\ln \left (\frac {1}{x}+1\right )}^2}{8\,\left (x^2-1\right )}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \]
[In]
[Out]