3.62 \(\int \frac{x^2}{\sinh ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=81 \[ -\frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{8 a^3}+\frac{9 \text{Chi}\left (3 \sinh ^{-1}(a x)\right )}{8 a^3}-\frac{x^2 \sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac{x}{a^2 \sinh ^{-1}(a x)}-\frac{3 x^3}{2 \sinh ^{-1}(a x)} \]

[Out]

-(x^2*Sqrt[1 + a^2*x^2])/(2*a*ArcSinh[a*x]^2) - x/(a^2*ArcSinh[a*x]) - (3*x^3)/(2*ArcSinh[a*x]) - CoshIntegral
[ArcSinh[a*x]]/(8*a^3) + (9*CoshIntegral[3*ArcSinh[a*x]])/(8*a^3)

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Rubi [A]  time = 0.252573, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5667, 5774, 5669, 5448, 3301, 5657} \[ -\frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{8 a^3}+\frac{9 \text{Chi}\left (3 \sinh ^{-1}(a x)\right )}{8 a^3}-\frac{x^2 \sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac{x}{a^2 \sinh ^{-1}(a x)}-\frac{3 x^3}{2 \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSinh[a*x]^3,x]

[Out]

-(x^2*Sqrt[1 + a^2*x^2])/(2*a*ArcSinh[a*x]^2) - x/(a^2*ArcSinh[a*x]) - (3*x^3)/(2*ArcSinh[a*x]) - CoshIntegral
[ArcSinh[a*x]]/(8*a^3) + (9*CoshIntegral[3*ArcSinh[a*x]])/(8*a^3)

Rule 5667

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSinh[c*x])^(n +
1))/Sqrt[1 + c^2*x^2], x], x] - Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c
^2*x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x
)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -
1] && GtQ[d, 0]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 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 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 5657

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Cosh[a/b - x/b], x], x,
a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rubi steps

\begin{align*} \int \frac{x^2}{\sinh ^{-1}(a x)^3} \, dx &=-\frac{x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}+\frac{\int \frac{x}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2} \, dx}{a}+\frac{1}{2} (3 a) \int \frac{x^3}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2} \, dx\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{x}{a^2 \sinh ^{-1}(a x)}-\frac{3 x^3}{2 \sinh ^{-1}(a x)}+\frac{9}{2} \int \frac{x^2}{\sinh ^{-1}(a x)} \, dx+\frac{\int \frac{1}{\sinh ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{x}{a^2 \sinh ^{-1}(a x)}-\frac{3 x^3}{2 \sinh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^3}+\frac{9 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{x}{a^2 \sinh ^{-1}(a x)}-\frac{3 x^3}{2 \sinh ^{-1}(a x)}+\frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{a^3}+\frac{9 \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 x}+\frac{\cosh (3 x)}{4 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{x}{a^2 \sinh ^{-1}(a x)}-\frac{3 x^3}{2 \sinh ^{-1}(a x)}+\frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{a^3}-\frac{9 \operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^3}+\frac{9 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^3}\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac{x}{a^2 \sinh ^{-1}(a x)}-\frac{3 x^3}{2 \sinh ^{-1}(a x)}-\frac{\text{Chi}\left (\sinh ^{-1}(a x)\right )}{8 a^3}+\frac{9 \text{Chi}\left (3 \sinh ^{-1}(a x)\right )}{8 a^3}\\ \end{align*}

Mathematica [A]  time = 0.136761, size = 64, normalized size = 0.79 \[ -\frac{\frac{4 a x \left (a x \sqrt{a^2 x^2+1}+\left (3 a^2 x^2+2\right ) \sinh ^{-1}(a x)\right )}{\sinh ^{-1}(a x)^2}+\text{Chi}\left (\sinh ^{-1}(a x)\right )-9 \text{Chi}\left (3 \sinh ^{-1}(a x)\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSinh[a*x]^3,x]

[Out]

-((4*a*x*(a*x*Sqrt[1 + a^2*x^2] + (2 + 3*a^2*x^2)*ArcSinh[a*x]))/ArcSinh[a*x]^2 + CoshIntegral[ArcSinh[a*x]] -
9*CoshIntegral[3*ArcSinh[a*x]])/(8*a^3)

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Maple [A]  time = 0.029, size = 81, normalized size = 1. \begin{align*}{\frac{1}{{a}^{3}} \left ({\frac{1}{8\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{ax}{8\,{\it Arcsinh} \left ( ax \right ) }}-{\frac{{\it Chi} \left ({\it Arcsinh} \left ( ax \right ) \right ) }{8}}-{\frac{\cosh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{8\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}-{\frac{3\,\sinh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{8\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{9\,{\it Chi} \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arcsinh(a*x)^3,x)

[Out]

1/a^3*(1/8/arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)+1/8*a*x/arcsinh(a*x)-1/8*Chi(arcsinh(a*x))-1/8/arcsinh(a*x)^2*cosh
(3*arcsinh(a*x))-3/8/arcsinh(a*x)*sinh(3*arcsinh(a*x))+9/8*Chi(3*arcsinh(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arcsinh(a*x)^3,x, algorithm="maxima")

[Out]

-1/2*(a^8*x^9 + 3*a^6*x^7 + 3*a^4*x^5 + a^2*x^3 + (a^5*x^6 + a^3*x^4)*(a^2*x^2 + 1)^(3/2) + (3*a^6*x^7 + 5*a^4
*x^5 + 2*a^2*x^3)*(a^2*x^2 + 1) + (3*a^8*x^9 + 9*a^6*x^7 + 9*a^4*x^5 + 3*a^2*x^3 + (3*a^5*x^6 + 4*a^3*x^4 + a*
x^2)*(a^2*x^2 + 1)^(3/2) + (9*a^6*x^7 + 17*a^4*x^5 + 10*a^2*x^3 + 2*x)*(a^2*x^2 + 1) + (9*a^7*x^8 + 22*a^5*x^6
+ 18*a^3*x^4 + 5*a*x^2)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1)) + (3*a^7*x^8 + 7*a^5*x^6 + 5*a^3*x^4
+ a*x^2)*sqrt(a^2*x^2 + 1))/((a^8*x^6 + 3*a^6*x^4 + (a^2*x^2 + 1)^(3/2)*a^5*x^3 + 3*a^4*x^2 + 3*(a^6*x^4 + a^4
*x^2)*(a^2*x^2 + 1) + a^2 + 3*(a^7*x^5 + 2*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1))^2)
+ integrate(1/2*(9*a^10*x^10 + 36*a^8*x^8 + 54*a^6*x^6 + 36*a^4*x^4 + 9*a^2*x^2 + (9*a^6*x^6 + 4*a^4*x^4 - a^
2*x^2)*(a^2*x^2 + 1)^2 + (36*a^7*x^7 + 48*a^5*x^5 + 13*a^3*x^3 - 2*a*x)*(a^2*x^2 + 1)^(3/2) + (54*a^8*x^8 + 12
0*a^6*x^6 + 83*a^4*x^4 + 19*a^2*x^2 + 2)*(a^2*x^2 + 1) + (36*a^9*x^9 + 112*a^7*x^7 + 123*a^5*x^5 + 57*a^3*x^3
+ 10*a*x)*sqrt(a^2*x^2 + 1))/((a^10*x^8 + 4*a^8*x^6 + (a^2*x^2 + 1)^2*a^6*x^4 + 6*a^6*x^4 + 4*a^4*x^2 + 4*(a^7
*x^5 + a^5*x^3)*(a^2*x^2 + 1)^(3/2) + 6*(a^8*x^6 + 2*a^6*x^4 + a^4*x^2)*(a^2*x^2 + 1) + a^2 + 4*(a^9*x^7 + 3*a
^7*x^5 + 3*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1))), 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 (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arcsinh(a*x)^3,x, algorithm="fricas")

[Out]

integral(x^2/arcsinh(a*x)^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{asinh}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/asinh(a*x)**3,x)

[Out]

Integral(x**2/asinh(a*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{arsinh}\left (a x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arcsinh(a*x)^3,x, algorithm="giac")

[Out]

integrate(x^2/arcsinh(a*x)^3, x)