Optimal. Leaf size=183 \[ -\frac{\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{8 b c^2}-\frac{3 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^2}-\frac{\sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^2}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{8 b c^2}+\frac{3 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^2}+\frac{\cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c^2} \]
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Rubi [A] time = 0.417174, antiderivative size = 179, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5779, 5448, 3303, 3298, 3301} \[ -\frac{\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b c^2}-\frac{3 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b c^2}-\frac{\sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b c^2}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{3 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b c^2}+\frac{\cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b c^2} \]
Antiderivative was successfully verified.
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Rule 5779
Rule 5448
Rule 3303
Rule 3298
Rule 3301
Rubi steps
\begin{align*} \int \frac{x \left (1+c^2 x^2\right )^{3/2}}{a+b \sinh ^{-1}(c x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh ^4(x) \sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{8 (a+b x)}+\frac{3 \sinh (3 x)}{16 (a+b x)}+\frac{\sinh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^2}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sinh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^2}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}\\ &=\frac{\cosh \left (\frac{a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^2}+\frac{\left (3 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}+\frac{\cosh \left (\frac{5 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}-\frac{\sinh \left (\frac{a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^2}-\frac{\left (3 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}-\frac{\sinh \left (\frac{5 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c^2}\\ &=-\frac{\text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac{a}{b}\right )}{8 b c^2}-\frac{3 \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{3 a}{b}\right )}{16 b c^2}-\frac{\text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{5 a}{b}\right )}{16 b c^2}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{3 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b c^2}+\frac{\cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b c^2}\\ \end{align*}
Mathematica [A] time = 0.484674, size = 136, normalized size = 0.74 \[ \frac{-2 \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-3 \sinh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac{5 a}{b}\right ) \text{Chi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+2 \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )+3 \cosh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac{5 a}{b}\right ) \text{Shi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )}{16 b c^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.175, size = 178, normalized size = 1. \begin{align*}{\frac{1}{32\,{c}^{2}b}{{\rm e}^{5\,{\frac{a}{b}}}}{\it Ei} \left ( 1,5\,{\it Arcsinh} \left ( cx \right ) +5\,{\frac{a}{b}} \right ) }+{\frac{3}{32\,{c}^{2}b}{{\rm e}^{3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,3\,{\it Arcsinh} \left ( cx \right ) +3\,{\frac{a}{b}} \right ) }+{\frac{1}{16\,{c}^{2}b}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( cx \right ) +{\frac{a}{b}} \right ) }-{\frac{1}{16\,{c}^{2}b}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ) }-{\frac{3}{32\,{c}^{2}b}{{\rm e}^{-3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-3\,{\it Arcsinh} \left ( cx \right ) -3\,{\frac{a}{b}} \right ) }-{\frac{1}{32\,{c}^{2}b}{{\rm e}^{-5\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-5\,{\it Arcsinh} \left ( cx \right ) -5\,{\frac{a}{b}} \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{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{b \operatorname{arsinh}\left (c x\right ) + a}\,{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{{\left (c^{2} x^{3} + x\right )} \sqrt{c^{2} x^{2} + 1}}{b \operatorname{arsinh}\left (c x\right ) + a}, 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 \left (c^{2} x^{2} + 1\right )^{\frac{3}{2}}}{a + b \operatorname{asinh}{\left (c 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{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{b \operatorname{arsinh}\left (c x\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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