Optimal. Leaf size=60 \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d} \]
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Rubi [A] time = 0.095525, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5865, 12, 5659, 3716, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 12
Rule 5659
Rule 3716
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{\sinh ^{-1}(a+b x)}{\frac{a d}{b}+d x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b \sinh ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 d}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}-\frac{\operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}-\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}+\frac{\text{Li}_2\left (e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}\\ \end{align*}
Mathematica [A] time = 0.0165718, size = 52, normalized size = 0.87 \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a+b x)}\right )-\sinh ^{-1}(a+b x) \left (\sinh ^{-1}(a+b x)-2 \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )\right )}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.037, size = 125, normalized size = 2.1 \begin{align*} -{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}{2\,d}}+{\frac{{\it Arcsinh} \left ( bx+a \right ) }{d}\ln \left ( 1+bx+a+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) }+{\frac{1}{d}{\it polylog} \left ( 2,-bx-a-\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) }+{\frac{{\it Arcsinh} \left ( bx+a \right ) }{d}\ln \left ( 1-bx-a-\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) }+{\frac{1}{d}{\it polylog} \left ( 2,bx+a+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arsinh}\left (b x + a\right )}{b d x + a d}, x\right ) \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*} \frac{b \int \frac{\operatorname{asinh}{\left (a + b x \right )}}{a + b x}\, dx}{d} \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{\operatorname{arsinh}\left (b x + a\right )}{d x + \frac{a d}{b}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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