Integrand size = 13, antiderivative size = 65 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx=\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]
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Time = 0.03 (sec) , antiderivative size = 65, 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} \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx=\frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \]
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Rule 29
Rule 2188
Rule 2191
Rule 2194
Rubi steps \begin{align*} \text {integral}& = \frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}+\frac {b \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{b x-\coth ^{-1}(\tanh (a+b x))} \\ & = \frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b \int \frac {1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {b^2 \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \\ & = \frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \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 )^2} \\ & = \frac {1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}-\frac {b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2}+\frac {b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx=\frac {-\coth ^{-1}(\tanh (a+b x))+b x \left (1-\log (x)+\log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )}{x \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2} \]
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Timed out.
\[\int \frac {1}{x^{2} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )}d x\]
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Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx=\frac {2 \, {\left (i \, \pi - 2 \, b x \log \left (i \, \pi + 2 \, b x + 2 \, a\right ) + 2 \, b x \log \left (x\right ) + 2 \, a\right )}}{\pi ^{2} x - 4 i \, \pi a x - 4 \, a^{2} x} \]
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\[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx=\int \frac {1}{x^{2} \operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
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Result contains complex when optimal does not.
Time = 0.36 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx=-\frac {4 \, b \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} + \frac {4 \, b \log \left (x\right )}{\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}} + \frac {2}{{\left (i \, \pi - 2 \, a\right )} x} \]
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Result contains complex when optimal does not.
Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx=-\frac {4 \, b \log \left (\pi - 2 i \, b x - 2 i \, a\right )}{\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}} + \frac {4 \, b \log \left (x\right )}{\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}} + \frac {2}{-i \, \pi x - 2 \, a x} \]
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Time = 6.12 (sec) , antiderivative size = 220, normalized size of antiderivative = 3.38 \[ \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))} \, dx=\frac {2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+4\,b\,x+b\,x\,\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,8{}\mathrm {i}}{x\,{\left (\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2} \]
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