Optimal. Leaf size=150 \[ \frac{\text{PolyLog}\left (2,\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{\text{PolyLog}\left (2,\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{2} x \log \left (1-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )-\frac{1}{2} x \log \left (1-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )+x \coth ^{-1}(d \coth (a+b x)+c) \]
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Rubi [A] time = 0.234753, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {6238, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{\text{PolyLog}\left (2,\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{2} x \log \left (1-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )-\frac{1}{2} x \log \left (1-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )+x \coth ^{-1}(d \coth (a+b x)+c) \]
Antiderivative was successfully verified.
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Rule 6238
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \coth ^{-1}(c+d \coth (a+b x)) \, dx &=x \coth ^{-1}(c+d \coth (a+b x))-(b (1-c-d)) \int \frac{e^{2 a+2 b x} x}{1-c+d+(-1+c+d) e^{2 a+2 b x}} \, dx+(b (1+c+d)) \int \frac{e^{2 a+2 b x} x}{1+c-d+(-1-c-d) e^{2 a+2 b x}} \, dx\\ &=x \coth ^{-1}(c+d \coth (a+b x))+\frac{1}{2} x \log \left (1-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{2} x \log \left (1-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{1}{2} \int \log \left (1+\frac{(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx-\frac{1}{2} \int \log \left (1+\frac{(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx\\ &=x \coth ^{-1}(c+d \coth (a+b x))+\frac{1}{2} x \log \left (1-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{2} x \log \left (1-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{(-1-c-d) x}{1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}-\frac{\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{(-1+c+d) x}{1-c+d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=x \coth ^{-1}(c+d \coth (a+b x))+\frac{1}{2} x \log \left (1-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{2} x \log \left (1-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{\text{Li}_2\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{\text{Li}_2\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}\\ \end{align*}
Mathematica [A] time = 1.12623, size = 131, normalized size = 0.87 \[ x \coth ^{-1}(d \coth (a+b x)+c)-\frac{-\text{PolyLog}\left (2,\frac{(c+d-1) e^{2 (a+b x)}}{c-d-1}\right )+\text{PolyLog}\left (2,\frac{(c+d+1) e^{2 (a+b x)}}{c-d+1}\right )-2 b x \left (\log \left (1-\frac{(c+d-1) e^{2 (a+b x)}}{c-d-1}\right )-\log \left (1-\frac{(c+d+1) e^{2 (a+b x)}}{c-d+1}\right )\right )}{4 b} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.171, size = 306, normalized size = 2. \begin{align*} -{\frac{{\rm arccoth} \left (c+d{\rm coth} \left (bx+a\right )\right )\ln \left ( d{\rm coth} \left (bx+a\right )-d \right ) }{2\,b}}+{\frac{{\rm arccoth} \left (c+d{\rm coth} \left (bx+a\right )\right )\ln \left ( d{\rm coth} \left (bx+a\right )+d \right ) }{2\,b}}+{\frac{1}{4\,b}{\it dilog} \left ({\frac{d{\rm coth} \left (bx+a\right )+c+1}{1+c+d}} \right ) }+{\frac{\ln \left ( d{\rm coth} \left (bx+a\right )-d \right ) }{4\,b}\ln \left ({\frac{d{\rm coth} \left (bx+a\right )+c+1}{1+c+d}} \right ) }-{\frac{1}{4\,b}{\it dilog} \left ({\frac{d{\rm coth} \left (bx+a\right )+c-1}{c+d-1}} \right ) }-{\frac{\ln \left ( d{\rm coth} \left (bx+a\right )-d \right ) }{4\,b}\ln \left ({\frac{d{\rm coth} \left (bx+a\right )+c-1}{c+d-1}} \right ) }+{\frac{1}{4\,b}{\it dilog} \left ({\frac{d{\rm coth} \left (bx+a\right )+c-1}{c-d-1}} \right ) }+{\frac{\ln \left ( d{\rm coth} \left (bx+a\right )+d \right ) }{4\,b}\ln \left ({\frac{d{\rm coth} \left (bx+a\right )+c-1}{c-d-1}} \right ) }-{\frac{1}{4\,b}{\it dilog} \left ({\frac{d{\rm coth} \left (bx+a\right )+c+1}{1+c-d}} \right ) }-{\frac{\ln \left ( d{\rm coth} \left (bx+a\right )+d \right ) }{4\,b}\ln \left ({\frac{d{\rm coth} \left (bx+a\right )+c+1}{1+c-d}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 2.27015, size = 192, normalized size = 1.28 \begin{align*} -\frac{1}{4} \, b d{\left (\frac{2 \, b x \log \left (-\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) +{\rm Li}_2\left (\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right )}{b^{2} d} - \frac{2 \, b x \log \left (-\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) +{\rm Li}_2\left (\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right )}{b^{2} d}\right )} + x \operatorname{arcoth}\left (d \coth \left (b x + a\right ) + c\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.16709, size = 1585, normalized size = 10.57 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 \operatorname{arcoth}\left (d \coth \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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