Optimal. Leaf size=102 \[ -\frac{2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac{2 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac{2 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]
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Rubi [A] time = 0.0626622, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2171, 2163, 2160, 2157, 29} \[ -\frac{2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac{2 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac{2 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3} \]
Antiderivative was successfully verified.
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Rule 2171
Rule 2163
Rule 2160
Rule 2157
Rule 29
Rubi steps
\begin{align*} \int \frac{1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx &=\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac{(2 b) \int \frac{1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx}{-b x+\coth ^{-1}(\tanh (a+b x))}\\ &=-\frac{2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac{(2 b) \int \frac{1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{\left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac{2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac{(2 b) \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 )^2}-\frac{\left (2 b^2\right ) \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 )^2}\\ &=-\frac{2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac{2 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac{(2 b) \operatorname{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 )^2}\\ &=-\frac{2 b}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac{1}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac{2 b \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}-\frac{2 b \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3}\\ \end{align*}
Mathematica [A] time = 0.0619438, size = 70, normalized size = 0.69 \[ \frac{\coth ^{-1}(\tanh (a+b x))^2+2 b x \coth ^{-1}(\tanh (a+b x)) \left (\log (x)-\log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )-b^2 x^2}{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))} \]
Antiderivative was successfully verified.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2} \left ({\rm arccoth} \left (\tanh \left ( bx+a \right ) \right ) \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] time = 2.45069, size = 182, normalized size = 1.78 \begin{align*} -\frac{16 \, b \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} + \frac{16 \, b \log \left (x\right )}{-i \, \pi ^{3} + 6 \, \pi ^{2} a + 12 i \, \pi a^{2} - 8 \, a^{3}} - \frac{4 \,{\left (i \, \pi - 4 \, b x - 2 \, a\right )}}{{\left (2 \, \pi ^{2} b + 8 i \, \pi a b - 8 \, a^{2} b\right )} x^{2} -{\left (i \, \pi ^{3} - 6 \, \pi ^{2} a - 12 i \, \pi a^{2} + 8 \, a^{3}\right )} x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.81465, size = 1065, normalized size = 10.44 \begin{align*} \frac{4 \,{\left (\pi ^{6} + 4 \, \pi ^{4} a^{2} - 16 \, \pi ^{2} a^{4} - 64 \, a^{6} + 8 \,{\left (\pi ^{4} b^{2} - 16 \, a^{4} b^{2}\right )} x^{2} + 4 \,{\left (5 \, \pi ^{4} a b + 8 \, \pi ^{2} a^{3} b - 48 \, a^{5} b\right )} x - 8 \,{\left (4 \,{\left (\pi ^{3} b^{3} - 12 \, \pi a^{2} b^{3}\right )} x^{3} + 8 \,{\left (\pi ^{3} a b^{2} - 12 \, \pi a^{3} b^{2}\right )} x^{2} +{\left (\pi ^{5} b - 8 \, \pi ^{3} a^{2} b - 48 \, \pi a^{4} b\right )} x\right )} \arctan \left (-\frac{2 \, b x + 2 \, a - \sqrt{4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 4 \,{\left (4 \,{\left (3 \, \pi ^{2} a b^{3} - 4 \, a^{3} b^{3}\right )} x^{3} + 8 \,{\left (3 \, \pi ^{2} a^{2} b^{2} - 4 \, a^{4} b^{2}\right )} x^{2} +{\left (3 \, \pi ^{4} a b + 8 \, \pi ^{2} a^{3} b - 16 \, a^{5} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right ) + 8 \,{\left (4 \,{\left (3 \, \pi ^{2} a b^{3} - 4 \, a^{3} b^{3}\right )} x^{3} + 8 \,{\left (3 \, \pi ^{2} a^{2} b^{2} - 4 \, a^{4} b^{2}\right )} x^{2} +{\left (3 \, \pi ^{4} a b + 8 \, \pi ^{2} a^{3} b - 16 \, a^{5} b\right )} x\right )} \log \left (x\right )\right )}}{4 \,{\left (\pi ^{6} b^{2} + 12 \, \pi ^{4} a^{2} b^{2} + 48 \, \pi ^{2} a^{4} b^{2} + 64 \, a^{6} b^{2}\right )} x^{3} + 8 \,{\left (\pi ^{6} a b + 12 \, \pi ^{4} a^{3} b + 48 \, \pi ^{2} a^{5} b + 64 \, a^{7} b\right )} x^{2} +{\left (\pi ^{8} + 16 \, \pi ^{6} a^{2} + 96 \, \pi ^{4} a^{4} + 256 \, \pi ^{2} a^{6} + 256 \, a^{8}\right )} x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \operatorname{acoth}^{2}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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