Optimal. Leaf size=156 \[ -\frac{e \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}+\frac{e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}-\frac{e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e \sqrt{(c+d x)^2+1} (c+d x)}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \]
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Rubi [A] time = 0.331282, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {5865, 12, 5667, 5774, 5669, 5448, 3303, 3298, 3301, 5675} \[ -\frac{e \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{b^3 d}+\frac{e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{b^3 d}-\frac{e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e \sqrt{(c+d x)^2+1} (c+d x)}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 12
Rule 5667
Rule 5774
Rule 5669
Rule 5448
Rule 3303
Rule 3298
Rule 3301
Rule 5675
Rubi steps
\begin{align*} \int \frac{c e+d e x}{\left (a+b \sinh ^{-1}(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e x}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e \operatorname{Subst}\left (\int \frac{x}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e (c+d x) \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac{e \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}+\frac{e \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}\\ &=-\frac{e (c+d x) \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{(2 e) \operatorname{Subst}\left (\int \frac{x}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}\\ &=-\frac{e (c+d x) \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{(2 e) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e (c+d x) \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{(2 e) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 (a+b x)} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e (c+d x) \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e (c+d x) \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (e \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 )}{b^2 d}-\frac{\left (e \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 )}{b^2 d}\\ &=-\frac{e (c+d x) \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e (c+d x)^2}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right ) \sinh \left (\frac{2 a}{b}\right )}{b^3 d}+\frac{e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c+d x)\right )}{b^3 d}\\ \end{align*}
Mathematica [A] time = 0.31807, size = 120, normalized size = 0.77 \[ \frac{e \left (-\frac{b^2 (c+d x) \sqrt{(c+d x)^2+1}}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}-2 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+2 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\frac{b \left (-2 (c+d x)^2-1\right )}{a+b \sinh ^{-1}(c+d x)}\right )}{2 b^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.061, size = 239, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -{\frac{e \left ( 2\,b{\it Arcsinh} \left ( dx+c \right ) +2\,a-b \right ) }{8\,{b}^{2} \left ({b}^{2} \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}+2\,ab{\it Arcsinh} \left ( dx+c \right ) +{a}^{2} \right ) } \left ( 2\, \left ( dx+c \right ) ^{2}-2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}+1 \right ) }+{\frac{e}{2\,{b}^{3}}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) }-{\frac{e}{8\,b \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}} \left ( 2\, \left ( dx+c \right ) ^{2}+1+2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{e}{4\,{b}^{2} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) } \left ( 2\, \left ( dx+c \right ) ^{2}+1+2\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) }-{\frac{e}{2\,{b}^{3}}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( dx+c \right ) -2\,{\frac{a}{b}} \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 e x + c e}{b^{3} \operatorname{arsinh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{arsinh}\left (d x + c\right ) + a^{3}}, 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 \left (\int \frac{c}{a^{3} + 3 a^{2} b \operatorname{asinh}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asinh}^{3}{\left (c + d x \right )}}\, dx + \int \frac{d x}{a^{3} + 3 a^{2} b \operatorname{asinh}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asinh}^{3}{\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{d e x + c e}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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