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

Optimal. Leaf size=30 \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.211343, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5865, 5805, 6741, 12, 6742, 3301, 5448, 3298} \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcSinh[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.050618, size = 30, normalized size = 1. \[ \frac{\text{Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \text{Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcSinh[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.03, size = 27, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{{\it Shi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{2}}-a{\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/arcsinh(b*x+a),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/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}{\operatorname{arsinh}\left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/arcsinh(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/asinh(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/arcsinh(b*x + a), x)