Integrand size = 13, antiderivative size = 70 \[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]
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Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2194, 2191, 2188, 29} \[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]
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Rule 29
Rule 2188
Rule 2191
Rule 2194
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\int \frac {1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{-b x+\coth ^{-1}(\tanh (a+b x))} \\ & = -\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac {\int \frac {1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )} \\ & = -\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )} \\ & = -\frac {1}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {-b x+\coth ^{-1}(\tanh (a+b x)) \left (1+\log (b x)-\log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )}{\coth ^{-1}(\tanh (a+b x)) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2} \]
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Timed out.
\[\int \frac {1}{x \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}d x\]
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Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {4 \, {\left (-i \, \pi + {\left (i \, \pi + 2 \, b x + 2 \, a\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right ) + {\left (-i \, \pi - 2 \, b x - 2 \, a\right )} \log \left (x\right ) - 2 \, a\right )}}{8 \, a^{2} b x - i \, \pi ^{3} - 2 \, \pi ^{2} {\left (b x + 3 \, a\right )} + 8 \, a^{3} + 4 i \, \pi {\left (2 \, a b x + 3 \, a^{2}\right )}} \]
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\[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx=\int \frac {1}{x \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
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Result contains complex when optimal does not.
Time = 0.52 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {4 \, \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} - \frac {4 \, \log \left (x\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} - \frac {4}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2} - 2 \, {\left (-i \, \pi b + 2 \, a b\right )} x} \]
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Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {4 \, \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}} - \frac {4 \, \log \left (x\right )}{\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}} + \frac {4}{2 i \, \pi b x + 4 \, a b x - \pi ^{2} + 4 i \, \pi a + 4 \, a^{2}} \]
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Time = 7.72 (sec) , antiderivative size = 421, normalized size of antiderivative = 6.01 \[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {4\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-4\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+8\,b\,x+\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}-\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,\left (\ln \left (2\right )+\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,8{}\mathrm {i}}{\left (\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,{\left (\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2} \]
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