Optimal. Leaf size=69 \[ -\frac{\text{PolyLog}\left (2,-(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{2} x \log \left ((d+1) e^{2 a+2 b x}+1\right )+x \tanh ^{-1}(d \tanh (a+b x)+d+1)+\frac{b x^2}{2} \]
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Rubi [A] time = 0.136717, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6231, 2184, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{2} x \log \left ((d+1) e^{2 a+2 b x}+1\right )+x \tanh ^{-1}(d \tanh (a+b x)+d+1)+\frac{b x^2}{2} \]
Antiderivative was successfully verified.
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Rule 6231
Rule 2184
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \tanh ^{-1}(1+d+d \tanh (a+b x)) \, dx &=x \tanh ^{-1}(1+d+d \tanh (a+b x))+b \int \frac{x}{1+(1+d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^2}{2}+x \tanh ^{-1}(1+d+d \tanh (a+b x))-(b (1+d)) \int \frac{e^{2 a+2 b x} x}{1+(1+d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^2}{2}+x \tanh ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{2} x \log \left (1+(1+d) e^{2 a+2 b x}\right )+\frac{1}{2} \int \log \left (1+(1+d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac{b x^2}{2}+x \tanh ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{2} x \log \left (1+(1+d) e^{2 a+2 b x}\right )+\frac{\operatorname{Subst}\left (\int \frac{\log (1+(1+d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=\frac{b x^2}{2}+x \tanh ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{2} x \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac{\text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}\\ \end{align*}
Mathematica [B] time = 4.3156, size = 201, normalized size = 2.91 \[ \frac{-2 \text{PolyLog}\left (2,-\sqrt{-d-1} e^{a+b x}\right )-2 \text{PolyLog}\left (2,\sqrt{-d-1} e^{a+b x}\right )-2 \log \left (e^{a+b x}\right ) \log \left (1-\sqrt{-d-1} e^{a+b x}\right )-2 \log \left (e^{a+b x}\right ) \log \left (\sqrt{-d-1} e^{a+b x}+1\right )+2 \log \left (e^{a+b x}\right ) \log \left ((d+1) e^{a+b x}+e^{-a-b x}\right )-2 b x \log (d \sinh (a+b x)+(d+2) \cosh (a+b x))+\log ^2\left (e^{a+b x}\right )+b^2 x^2}{4 b}+x \tanh ^{-1}(d \tanh (a+b x)+d+1) \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.116, size = 247, normalized size = 3.6 \begin{align*}{\frac{{\it Artanh} \left ( 1+d+d\tanh \left ( bx+a \right ) \right ) \ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{2\,b}}-{\frac{{\it Artanh} \left ( 1+d+d\tanh \left ( bx+a \right ) \right ) \ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{2\,b}}+{\frac{ \left ( \ln \left ( d\tanh \left ( bx+a \right ) +d \right ) \right ) ^{2}}{8\,b}}-{\frac{1}{4\,b}{\it dilog} \left ( 1+{\frac{d\tanh \left ( bx+a \right ) }{2}}+{\frac{d}{2}} \right ) }-{\frac{\ln \left ( d\tanh \left ( bx+a \right ) +d \right ) }{4\,b}\ln \left ( 1+{\frac{d\tanh \left ( bx+a \right ) }{2}}+{\frac{d}{2}} \right ) }-{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +d}{2\,d}} \right ) }-{\frac{\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +d}{2\,d}} \right ) }+{\frac{1}{4\,b}{\it dilog} \left ({\frac{d\tanh \left ( bx+a \right ) +d+2}{2\,d+2}} \right ) }+{\frac{\ln \left ( d\tanh \left ( bx+a \right ) -d \right ) }{4\,b}\ln \left ({\frac{d\tanh \left ( bx+a \right ) +d+2}{2\,d+2}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 3.26336, size = 97, normalized size = 1.41 \begin{align*} \frac{1}{4} \, b d{\left (\frac{2 \, x^{2}}{d} - \frac{2 \, b x \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{2} d}\right )} + x \operatorname{artanh}\left (d \tanh \left (b x + a\right ) + d + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.05621, size = 709, normalized size = 10.28 \begin{align*} \frac{b^{2} x^{2} + b x \log \left (-\frac{{\left (d + 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) + a \log \left (2 \,{\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \,{\left (d + 1\right )} \sinh \left (b x + a\right ) + \sqrt{-4 \, d - 4}\right ) + a \log \left (2 \,{\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \,{\left (d + 1\right )} \sinh \left (b x + a\right ) - \sqrt{-4 \, d - 4}\right ) -{\left (b x + a\right )} \log \left (\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) -{\left (b x + a\right )} \log \left (-\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) -{\rm Li}_2\left (\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) -{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{2 \, b} \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 \operatorname{atanh}{\left (d \tanh{\left (a + b x \right )} + d + 1 \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 \operatorname{artanh}\left (d \tanh \left (b x + a\right ) + d + 1\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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