Optimal. Leaf size=150 \[ \frac{\text{PolyLog}\left (2,-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{\text{PolyLog}\left (2,-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{2} x \log \left (\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac{1}{2} x \log \left (\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+x \tanh ^{-1}(d \tanh (a+b x)+c) \]
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Rubi [A] time = 0.22185, 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 = {6235, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{\text{PolyLog}\left (2,-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{2} x \log \left (\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac{1}{2} x \log \left (\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+x \tanh ^{-1}(d \tanh (a+b x)+c) \]
Antiderivative was successfully verified.
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Rule 6235
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \tanh ^{-1}(c+d \tanh (a+b x)) \, dx &=x \tanh ^{-1}(c+d \tanh (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 \tanh ^{-1}(c+d \tanh (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 \tanh ^{-1}(c+d \tanh (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 \tanh ^{-1}(c+d \tanh (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 = 6.08505, size = 131, normalized size = 0.87 \[ \frac{\text{PolyLog}\left (2,-\frac{(c+d-1) e^{2 (a+b x)}}{c-d-1}\right )-\text{PolyLog}\left (2,-\frac{(c+d+1) e^{2 (a+b x)}}{c-d+1}\right )+2 b x \left (\log \left (\frac{(c+d-1) e^{2 (a+b x)}}{c-d-1}+1\right )-\log \left (\frac{(c+d+1) e^{2 (a+b x)}}{c-d+1}+1\right )\right )}{4 b}+x \tanh ^{-1}(d \tanh (a+b x)+c) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.158, size = 306, normalized size = 2. \begin{align*} -{\frac{{\it Artanh} \left ( c+d\tanh \left ( bx+a \right ) \right ) \ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{2\,b}}+{\frac{{\it Artanh} \left ( c+d\tanh \left ( bx+a \right ) \right ) \ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{2\,b}}+{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c-1}{c-d-1}} \right ) }+{\frac{\ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c-1}{c-d-1}} \right ) }-{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c+1}{1+c-d}} \right ) }-{\frac{\ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c+1}{1+c-d}} \right ) }-{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c-1}{c+d-1}} \right ) }-{\frac{\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +c-1}{c+d-1}} \right ) }+{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +c+1}{1+c+d}} \right ) }+{\frac{\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{4\,b}\ln \left ({\frac{d\tanh \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 = 2.0972, 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{artanh}\left (d \tanh \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 = 2.10806, size = 1602, normalized size = 10.68 \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{artanh}\left (d \tanh \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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