Optimal. Leaf size=145 \[ \frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{4 b d}-\frac{e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 b d}-\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{4 b d}+\frac{e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 b d} \]
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Rubi [A] time = 0.340267, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5865, 12, 5669, 5448, 3303, 3298, 3301} \[ \frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{4 b d}-\frac{e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{8 b d}-\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{4 b d}+\frac{e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{8 b d} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 12
Rule 5669
Rule 5448
Rule 3303
Rule 3298
Rule 3301
Rubi steps
\begin{align*} \int \frac{(c e+d e x)^3}{a+b \sinh ^{-1}(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \left (-\frac{\sinh (2 x)}{4 (a+b x)}+\frac{\sinh (4 x)}{8 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d}-\frac{e^3 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}\\ &=-\frac{\left (e^3 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}+\frac{\left (e^3 \cosh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d}+\frac{\left (e^3 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 d}-\frac{\left (e^3 \sinh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d}\\ &=\frac{e^3 \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{2 a}{b}\right )}{4 b d}-\frac{e^3 \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{4 a}{b}\right )}{8 b d}-\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{4 b d}+\frac{e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c+d x)\right )}{8 b d}\\ \end{align*}
Mathematica [A] time = 0.212484, size = 109, normalized size = 0.75 \[ \frac{e^3 \left (2 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )-2 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )}{8 b d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.079, size = 134, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{e}^{3}}{16\,b}{{\rm e}^{4\,{\frac{a}{b}}}}{\it Ei} \left ( 1,4\,{\it Arcsinh} \left ( dx+c \right ) +4\,{\frac{a}{b}} \right ) }-{\frac{{e}^{3}}{8\,b}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) }+{\frac{{e}^{3}}{8\,b}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( dx+c \right ) -2\,{\frac{a}{b}} \right ) }-{\frac{{e}^{3}}{16\,b}{{\rm e}^{-4\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-4\,{\it Arcsinh} \left ( dx+c \right ) -4\,{\frac{a}{b}} \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{{\left (d e x + c e\right )}^{3}}{b \operatorname{arsinh}\left (d x + c\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{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}{b \operatorname{arsinh}\left (d x + c\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*} e^{3} \left (\int \frac{c^{3}}{a + b \operatorname{asinh}{\left (c + d x \right )}}\, dx + \int \frac{d^{3} x^{3}}{a + b \operatorname{asinh}{\left (c + d x \right )}}\, dx + \int \frac{3 c d^{2} x^{2}}{a + b \operatorname{asinh}{\left (c + d x \right )}}\, dx + \int \frac{3 c^{2} d x}{a + b \operatorname{asinh}{\left (c + d x \right )}}\, dx\right ) \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 (d e x + c e\right )}^{3}}{b \operatorname{arsinh}\left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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