Optimal. Leaf size=63 \[ \frac{a^3 x}{b^2 \left (a^2-b^2\right )}-\frac{a^2 \log (a \sinh (x)+b \cosh (x))}{b \left (a^2-b^2\right )}-\frac{a x}{b^2}+\frac{\log (\sinh (x))}{b} \]
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Rubi [A] time = 0.0920795, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3541, 3475, 3484, 3530} \[ \frac{a^3 x}{b^2 \left (a^2-b^2\right )}-\frac{a^2 \log (a \sinh (x)+b \cosh (x))}{b \left (a^2-b^2\right )}-\frac{a x}{b^2}+\frac{\log (\sinh (x))}{b} \]
Antiderivative was successfully verified.
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Rule 3541
Rule 3475
Rule 3484
Rule 3530
Rubi steps
\begin{align*} \int \frac{\coth ^2(x)}{a+b \coth (x)} \, dx &=-\frac{a x}{b^2}+\frac{a^2 \int \frac{1}{a+b \coth (x)} \, dx}{b^2}+\frac{\int \coth (x) \, dx}{b}\\ &=-\frac{a x}{b^2}+\frac{a^3 x}{b^2 \left (a^2-b^2\right )}+\frac{\log (\sinh (x))}{b}-\frac{\left (i a^2\right ) \int \frac{-i b-i a \coth (x)}{a+b \coth (x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{a x}{b^2}+\frac{a^3 x}{b^2 \left (a^2-b^2\right )}+\frac{\log (\sinh (x))}{b}-\frac{a^2 \log (b \cosh (x)+a \sinh (x))}{b \left (a^2-b^2\right )}\\ \end{align*}
Mathematica [A] time = 0.0760584, size = 49, normalized size = 0.78 \[ \frac{-a^2 \log (a \sinh (x)+b \cosh (x))+a^2 \log (\sinh (x))+a b x-b^2 \log (\sinh (x))}{a^2 b-b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.019, size = 60, normalized size = 1. \begin{align*}{\frac{\ln \left ( 1+{\rm coth} \left (x\right ) \right ) }{2\,a-2\,b}}-{\frac{\ln \left ({\rm coth} \left (x\right )-1 \right ) }{2\,b+2\,a}}-{\frac{{a}^{2}\ln \left ( a+b{\rm coth} \left (x\right ) \right ) }{ \left ( a+b \right ) \left ( a-b \right ) b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.27629, size = 85, normalized size = 1.35 \begin{align*} -\frac{a^{2} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{2} b - b^{3}} + \frac{x}{a + b} + \frac{\log \left (e^{\left (-x\right )} + 1\right )}{b} + \frac{\log \left (e^{\left (-x\right )} - 1\right )}{b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.80317, size = 186, normalized size = 2.95 \begin{align*} -\frac{a^{2} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) -{\left (a b + b^{2}\right )} x -{\left (a^{2} - b^{2}\right )} \log \left (\frac{2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b - b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 2.94032, size = 372, normalized size = 5.9 \begin{align*} \begin{cases} \tilde{\infty } \left (x - \log{\left (\tanh{\left (x \right )} + 1 \right )} + \log{\left (\tanh{\left (x \right )} \right )}\right ) & \text{for}\: a = 0 \wedge b = 0 \\\frac{3 x \tanh{\left (x \right )}}{2 b \tanh{\left (x \right )} - 2 b} - \frac{3 x}{2 b \tanh{\left (x \right )} - 2 b} - \frac{2 \log{\left (\tanh{\left (x \right )} + 1 \right )} \tanh{\left (x \right )}}{2 b \tanh{\left (x \right )} - 2 b} + \frac{2 \log{\left (\tanh{\left (x \right )} + 1 \right )}}{2 b \tanh{\left (x \right )} - 2 b} + \frac{2 \log{\left (\tanh{\left (x \right )} \right )} \tanh{\left (x \right )}}{2 b \tanh{\left (x \right )} - 2 b} - \frac{2 \log{\left (\tanh{\left (x \right )} \right )}}{2 b \tanh{\left (x \right )} - 2 b} - \frac{1}{2 b \tanh{\left (x \right )} - 2 b} & \text{for}\: a = - b \\\frac{x \tanh{\left (x \right )}}{2 b \tanh{\left (x \right )} + 2 b} + \frac{x}{2 b \tanh{\left (x \right )} + 2 b} - \frac{2 \log{\left (\tanh{\left (x \right )} + 1 \right )} \tanh{\left (x \right )}}{2 b \tanh{\left (x \right )} + 2 b} - \frac{2 \log{\left (\tanh{\left (x \right )} + 1 \right )}}{2 b \tanh{\left (x \right )} + 2 b} + \frac{2 \log{\left (\tanh{\left (x \right )} \right )} \tanh{\left (x \right )}}{2 b \tanh{\left (x \right )} + 2 b} + \frac{2 \log{\left (\tanh{\left (x \right )} \right )}}{2 b \tanh{\left (x \right )} + 2 b} + \frac{1}{2 b \tanh{\left (x \right )} + 2 b} & \text{for}\: a = b \\\frac{x - \frac{1}{\tanh{\left (x \right )}}}{a} & \text{for}\: b = 0 \\\frac{x - \log{\left (\tanh{\left (x \right )} + 1 \right )} + \log{\left (\tanh{\left (x \right )} \right )}}{b} & \text{for}\: a = 0 \\- \frac{a^{2} \log{\left (\tanh{\left (x \right )} + \frac{b}{a} \right )}}{a^{2} b - b^{3}} + \frac{a^{2} \log{\left (\tanh{\left (x \right )} \right )}}{a^{2} b - b^{3}} + \frac{a b x}{a^{2} b - b^{3}} - \frac{b^{2} x}{a^{2} b - b^{3}} + \frac{b^{2} \log{\left (\tanh{\left (x \right )} + 1 \right )}}{a^{2} b - b^{3}} - \frac{b^{2} \log{\left (\tanh{\left (x \right )} \right )}}{a^{2} b - b^{3}} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.14789, size = 80, normalized size = 1.27 \begin{align*} -\frac{a^{2} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - a + b \right |}\right )}{a^{2} b - b^{3}} + \frac{x}{a - b} + \frac{\log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right )}{b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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