Integrand size = 8, antiderivative size = 81 \[ \int \coth ^{-1}(a+b x)^2 \, dx=\frac {\coth ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}-\frac {2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,-\frac {1+a+b x}{1-a-b x}\right )}{b} \]
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Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6239, 6022, 6132, 6056, 2449, 2352} \[ \int \coth ^{-1}(a+b x)^2 \, dx=-\frac {\operatorname {PolyLog}\left (2,-\frac {a+b x+1}{-a-b x+1}\right )}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}+\frac {\coth ^{-1}(a+b x)^2}{b}-\frac {2 \log \left (\frac {2}{-a-b x+1}\right ) \coth ^{-1}(a+b x)}{b} \]
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Rule 2352
Rule 2449
Rule 6022
Rule 6056
Rule 6132
Rule 6239
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \coth ^{-1}(x)^2 \, dx,x,a+b x\right )}{b} \\ & = \frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}-\frac {2 \text {Subst}\left (\int \frac {x \coth ^{-1}(x)}{1-x^2} \, dx,x,a+b x\right )}{b} \\ & = \frac {\coth ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}-\frac {2 \text {Subst}\left (\int \frac {\coth ^{-1}(x)}{1-x} \, dx,x,a+b x\right )}{b} \\ & = \frac {\coth ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}-\frac {2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b}+\frac {2 \text {Subst}\left (\int \frac {\log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,a+b x\right )}{b} \\ & = \frac {\coth ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}-\frac {2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b}-\frac {2 \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a-b x}\right )}{b} \\ & = \frac {\coth ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \coth ^{-1}(a+b x)^2}{b}-\frac {2 \coth ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1-a-b x}\right )}{b} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \coth ^{-1}(a+b x)^2 \, dx=\frac {\coth ^{-1}(a+b x) \left ((-1+a+b x) \coth ^{-1}(a+b x)-2 \log \left (1-e^{-2 \coth ^{-1}(a+b x)}\right )\right )+\operatorname {PolyLog}\left (2,e^{-2 \coth ^{-1}(a+b x)}\right )}{b} \]
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Time = 0.25 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.64
method | result | size |
derivativedivides | \(\frac {\operatorname {arccoth}\left (b x +a \right )^{2} \left (b x +a -1\right )+2 \operatorname {arccoth}\left (b x +a \right )^{2}-2 \,\operatorname {arccoth}\left (b x +a \right ) \ln \left (1-\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \operatorname {polylog}\left (2, \frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \,\operatorname {arccoth}\left (b x +a \right ) \ln \left (1+\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \operatorname {polylog}\left (2, -\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )}{b}\) | \(133\) |
default | \(\frac {\operatorname {arccoth}\left (b x +a \right )^{2} \left (b x +a -1\right )+2 \operatorname {arccoth}\left (b x +a \right )^{2}-2 \,\operatorname {arccoth}\left (b x +a \right ) \ln \left (1-\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \operatorname {polylog}\left (2, \frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \,\operatorname {arccoth}\left (b x +a \right ) \ln \left (1+\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )-2 \operatorname {polylog}\left (2, -\frac {1}{\sqrt {\frac {b x +a -1}{b x +a +1}}}\right )}{b}\) | \(133\) |
risch | \(\frac {\left (b x +a +1\right ) \ln \left (b x +a +1\right )^{2}}{4 b}+\left (-\frac {x \ln \left (b x +a -1\right )}{2}+\frac {-\ln \left (b x +a -1\right ) a +\ln \left (b x +a -1\right )}{2 b}\right ) \ln \left (b x +a +1\right )+\frac {x \ln \left (b x +a -1\right )^{2}}{4}+\frac {\ln \left (b x +a -1\right )^{2} a}{4 b}-\frac {\ln \left (b x +a -1\right )^{2}}{4 b}-\frac {\ln \left (b x +a -1\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{b}-\frac {\operatorname {dilog}\left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right )}{b}\) | \(142\) |
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\[ \int \coth ^{-1}(a+b x)^2 \, dx=\int { \operatorname {arcoth}\left (b x + a\right )^{2} \,d x } \]
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\[ \int \coth ^{-1}(a+b x)^2 \, dx=\int \operatorname {acoth}^{2}{\left (a + b x \right )}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.72 \[ \int \coth ^{-1}(a+b x)^2 \, dx=-\frac {1}{4} \, b^{2} {\left (\frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a - 1\right )} \log \left (b x + a - 1\right )^{2}}{b^{3}} + \frac {4 \, {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )}}{b^{3}}\right )} + b {\left (\frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac {{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} \operatorname {arcoth}\left (b x + a\right ) + x \operatorname {arcoth}\left (b x + a\right )^{2} \]
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\[ \int \coth ^{-1}(a+b x)^2 \, dx=\int { \operatorname {arcoth}\left (b x + a\right )^{2} \,d x } \]
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Timed out. \[ \int \coth ^{-1}(a+b x)^2 \, dx=\int {\mathrm {acoth}\left (a+b\,x\right )}^2 \,d x \]
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