3.20 \(\int \frac{(d+e x)^2}{a+b \sinh ^{-1}(c x)} \, dx\)

Optimal. Leaf size=245 \[ -\frac{d e \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b c^2}-\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{d e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b c^2}+\frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}-\frac{e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{d^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c}-\frac{d^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c} \]

[Out]

(d^2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(b*c) - (e^2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(4*b
*c^3) + (e^2*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(4*b*c^3) - (d*e*CoshIntegral[(2*a)/b + 2*A
rcSinh[c*x]]*Sinh[(2*a)/b])/(b*c^2) - (d^2*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(b*c) + (e^2*Sinh[a/b]*
SinhIntegral[a/b + ArcSinh[c*x]])/(4*b*c^3) + (d*e*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcSinh[c*x]])/(b*c^
2) - (e^2*Sinh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(4*b*c^3)

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Rubi [A]  time = 0.70211, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5805, 6742, 3303, 3298, 3301, 5448} \[ -\frac{d e \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b c^2}-\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{d e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b c^2}+\frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}-\frac{e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{d^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c}-\frac{d^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + b*ArcSinh[c*x]),x]

[Out]

(d^2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(b*c) - (e^2*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(4*b
*c^3) + (e^2*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(4*b*c^3) - (d*e*CoshIntegral[(2*a)/b + 2*A
rcSinh[c*x]]*Sinh[(2*a)/b])/(b*c^2) - (d^2*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(b*c) + (e^2*Sinh[a/b]*
SinhIntegral[a/b + ArcSinh[c*x]])/(4*b*c^3) + (d*e*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcSinh[c*x]])/(b*c^
2) - (e^2*Sinh[(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(4*b*c^3)

Rule 5805

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst
[Int[(a + b*x)^n*Cosh[x]*(c*d + e*Sinh[x])^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ
[m, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{a+b \sinh ^{-1}(c x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh (x) (c d+e \sinh (x))^2}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{c^2 d^2 \cosh (x)}{a+b x}+\frac{e^2 \cosh (x) \sinh ^2(x)}{a+b x}+\frac{c d e \sinh (2 x)}{a+b x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=\frac{d^2 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac{(d e) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^2}+\frac{e^2 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=\frac{e^2 \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 (a+b x)}+\frac{\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}+\frac{\left (d^2 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac{\left (d 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 x)\right )}{c^2}-\frac{\left (d^2 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c}-\frac{\left (d 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 x)\right )}{c^2}\\ &=\frac{d^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c}-\frac{d e \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{2 a}{b}\right )}{b c^2}-\frac{d^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c}+\frac{d e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b c^2}-\frac{e^2 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}+\frac{e^2 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}\\ &=\frac{d^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c}-\frac{d e \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{2 a}{b}\right )}{b c^2}-\frac{d^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c}+\frac{d e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b c^2}-\frac{\left (e^2 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}+\frac{\left (e^2 \cosh \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 )}{4 c^3}+\frac{\left (e^2 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^3}-\frac{\left (e^2 \sinh \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 )}{4 c^3}\\ &=\frac{d^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c}-\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}-\frac{d e \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{2 a}{b}\right )}{b c^2}-\frac{d^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{b c}+\frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{d e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{b c^2}-\frac{e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b c^3}\\ \end{align*}

Mathematica [A]  time = 0.399842, size = 188, normalized size = 0.77 \[ \frac{\cosh \left (\frac{a}{b}\right ) \left (4 c^2 d^2-e^2\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-4 c^2 d^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-4 c d e \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+4 c d e \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )}{4 b c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(a + b*ArcSinh[c*x]),x]

[Out]

((4*c^2*d^2 - e^2)*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]] + e^2*Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcSin
h[c*x])] - 4*c*d*e*CoshIntegral[2*(a/b + ArcSinh[c*x])]*Sinh[(2*a)/b] - 4*c^2*d^2*Sinh[a/b]*SinhIntegral[a/b +
ArcSinh[c*x]] + e^2*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] + 4*c*d*e*Cosh[(2*a)/b]*SinhIntegral[2*(a/b +
ArcSinh[c*x])] - e^2*Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])])/(4*b*c^3)

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Maple [A]  time = 0.128, size = 254, normalized size = 1. \begin{align*}{\frac{1}{c} \left ( -{\frac{{e}^{2}}{8\,{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{{e}^{2}}{8\,{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{{d}^{2}}{2\,b}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( cx \right ) +{\frac{a}{b}} \right ) }+{\frac{{e}^{2}}{8\,{c}^{2}b}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\it Arcsinh} \left ( cx \right ) +{\frac{a}{b}} \right ) }-{\frac{{d}^{2}}{2\,b}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ) }+{\frac{{e}^{2}}{8\,{c}^{2}b}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\it Arcsinh} \left ( cx \right ) -{\frac{a}{b}} \right ) }-{\frac{de}{2\,bc}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( cx \right ) -2\,{\frac{a}{b}} \right ) }+{\frac{de}{2\,bc}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( cx \right ) +2\,{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(a+b*arcsinh(c*x)),x)

[Out]

1/c*(-1/8/c^2*e^2/b*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)-1/8/c^2*e^2/b*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b
)-1/2/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)*d^2+1/8/c^2/b*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)*e^2-1/2/b*exp(-a/b)*Ei(1
,-arcsinh(c*x)-a/b)*d^2+1/8/c^2/b*exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)*e^2-1/2/c*d*e/b*exp(-2*a/b)*Ei(1,-2*arcsin
h(c*x)-2*a/b)+1/2/c*d*e/b*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{2}}{b \operatorname{arsinh}\left (c x\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(a+b*arcsinh(c*x)),x, algorithm="maxima")

[Out]

integrate((e*x + d)^2/(b*arcsinh(c*x) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{e^{2} x^{2} + 2 \, d e x + d^{2}}{b \operatorname{arsinh}\left (c x\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral((e^2*x^2 + 2*d*e*x + d^2)/(b*arcsinh(c*x) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{2}}{a + b \operatorname{asinh}{\left (c x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(a+b*asinh(c*x)),x)

[Out]

Integral((d + e*x)**2/(a + b*asinh(c*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{2}}{b \operatorname{arsinh}\left (c x\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

integrate((e*x + d)^2/(b*arcsinh(c*x) + a), x)