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 = 0.21, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, 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 12
Rule 3298
Rule 3301
Rule 5448
Rule 5805
Rule 5865
Rule 6741
Rule 6742
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*}
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Mathematica [A] time = 0.05, size = 30, normalized size = 1.00 \[ \frac {\text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac {a \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{\operatorname {arsinh}\left (b x + a\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\operatorname {arsinh}\left (b x + a\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 27, normalized size = 0.90 \[ \frac {\frac {\Shi \left (2 \arcsinh \left (b x +a \right )\right )}{2}-a \Chi \left (\arcsinh \left (b x +a \right )\right )}{b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\operatorname {arsinh}\left (b x + a\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x}{\mathrm {asinh}\left (a+b\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\operatorname {asinh}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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