3.81 \(\int \frac{x^2}{\sinh ^{-1}(a+b x)} \, dx\)

Optimal. Leaf size=60 \[ \frac{a^2 \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}-\frac{\text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac{\text{Chi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}-\frac{a \text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3} \]

[Out]

-CoshIntegral[ArcSinh[a + b*x]]/(4*b^3) + (a^2*CoshIntegral[ArcSinh[a + b*x]])/b^3 + CoshIntegral[3*ArcSinh[a
+ b*x]]/(4*b^3) - (a*SinhIntegral[2*ArcSinh[a + b*x]])/b^3

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Rubi [A]  time = 0.527069, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5865, 5805, 6741, 12, 6742, 3301, 5448, 3298} \[ \frac{a^2 \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}-\frac{\text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac{\text{Chi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}-\frac{a \text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSinh[a + b*x],x]

[Out]

-CoshIntegral[ArcSinh[a + b*x]]/(4*b^3) + (a^2*CoshIntegral[ArcSinh[a + b*x]])/b^3 + CoshIntegral[3*ArcSinh[a
+ b*x]]/(4*b^3) - (a*SinhIntegral[2*ArcSinh[a + b*x]])/b^3

Rule 5865

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

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 6741

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6742

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

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.135993, size = 44, normalized size = 0.73 \[ \frac{\left (4 a^2-1\right ) \text{Chi}\left (\sinh ^{-1}(a+b x)\right )+\text{Chi}\left (3 \sinh ^{-1}(a+b x)\right )-4 a \text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSinh[a + b*x],x]

[Out]

((-1 + 4*a^2)*CoshIntegral[ArcSinh[a + b*x]] + CoshIntegral[3*ArcSinh[a + b*x]] - 4*a*SinhIntegral[2*ArcSinh[a
+ b*x]])/(4*b^3)

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Maple [A]  time = 0.04, size = 49, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{3}} \left ( -a{\it Shi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) -{\frac{{\it Chi} \left ({\it Arcsinh} \left ( bx+a \right ) \right ) }{4}}+{\frac{{\it Chi} \left ( 3\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{4}}+{a}^{2}{\it Chi} \left ({\it Arcsinh} \left ( bx+a \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(-a*Shi(2*arcsinh(b*x+a))-1/4*Chi(arcsinh(b*x+a))+1/4*Chi(3*arcsinh(b*x+a))+a^2*Chi(arcsinh(b*x+a)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/arcsinh(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/arcsinh(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/asinh(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/arcsinh(b*x + a), x)