Optimal. Leaf size=30 \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \]
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Rubi [A] time = 0.211343, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5865, 5805, 6741, 12, 6742, 3301, 5448, 3298} \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 5805
Rule 6741
Rule 12
Rule 6742
Rule 3301
Rule 5448
Rule 3298
Rubi steps
\begin{align*} \int \frac{x}{\sinh ^{-1}(a+b x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-\frac{a}{b}+\frac{x}{b}}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cosh (x) \left (-\frac{a}{b}+\frac{\sinh (x)}{b}\right )}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cosh (x) (-a+\sinh (x))}{b x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cosh (x) (-a+\sinh (x))}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a \cosh (x)}{x}+\frac{\cosh (x) \sinh (x)}{x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}\\ \end{align*}
Mathematica [A] time = 0.050618, size = 30, normalized size = 1. \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.03, size = 27, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{{\it Shi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{2}}-a{\it Chi} \left ({\it Arcsinh} \left ( bx+a \right ) \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{arsinh}\left (b x + a\right )}\,{d x} \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{x}{\operatorname{arsinh}\left (b x + a\right )}, 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*} \int \frac{x}{\operatorname{asinh}{\left (a + b x \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 \frac{x}{\operatorname{arsinh}\left (b x + a\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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