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

Optimal. Leaf size=82 \[ -\frac{\text{Shi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Shi}\left (4 \sinh ^{-1}(a x)\right )}{a^4}-\frac{x^3 \sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac{2 x^4}{\sinh ^{-1}(a x)} \]

[Out]

-(x^3*Sqrt[1 + a^2*x^2])/(2*a*ArcSinh[a*x]^2) - (3*x^2)/(2*a^2*ArcSinh[a*x]) - (2*x^4)/ArcSinh[a*x] - SinhInte
gral[2*ArcSinh[a*x]]/(2*a^4) + SinhIntegral[4*ArcSinh[a*x]]/a^4

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Rubi [A]  time = 0.304009, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5667, 5774, 5669, 5448, 3298, 12} \[ -\frac{\text{Shi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Shi}\left (4 \sinh ^{-1}(a x)\right )}{a^4}-\frac{x^3 \sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac{2 x^4}{\sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^3*Sqrt[1 + a^2*x^2])/(2*a*ArcSinh[a*x]^2) - (3*x^2)/(2*a^2*ArcSinh[a*x]) - (2*x^4)/ArcSinh[a*x] - SinhInte
gral[2*ArcSinh[a*x]]/(2*a^4) + SinhIntegral[4*ArcSinh[a*x]]/a^4

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 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 12

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

Rubi steps

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

Mathematica [A]  time = 0.203097, size = 69, normalized size = 0.84 \[ -\frac{\frac{a^2 x^2 \left (a x \sqrt{a^2 x^2+1}+\left (4 a^2 x^2+3\right ) \sinh ^{-1}(a x)\right )}{\sinh ^{-1}(a x)^2}+\text{Shi}\left (2 \sinh ^{-1}(a x)\right )-2 \text{Shi}\left (4 \sinh ^{-1}(a x)\right )}{2 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 82, normalized size = 1. \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{\sinh \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{8\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}+{\frac{\cosh \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{4\,{\it Arcsinh} \left ( ax \right ) }}-{\frac{{\it Shi} \left ( 2\,{\it Arcsinh} \left ( ax \right ) \right ) }{2}}-{\frac{\sinh \left ( 4\,{\it Arcsinh} \left ( ax \right ) \right ) }{16\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 4\,{\it Arcsinh} \left ( ax \right ) \right ) }{4\,{\it Arcsinh} \left ( ax \right ) }}+{\it Shi} \left ( 4\,{\it Arcsinh} \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(1/8/arcsinh(a*x)^2*sinh(2*arcsinh(a*x))+1/4/arcsinh(a*x)*cosh(2*arcsinh(a*x))-1/2*Shi(2*arcsinh(a*x))-1
/16/arcsinh(a*x)^2*sinh(4*arcsinh(a*x))-1/4/arcsinh(a*x)*cosh(4*arcsinh(a*x))+Shi(4*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^3/arcsinh(a*x)^3,x, algorithm="maxima")

[Out]

-1/2*(a^8*x^10 + 3*a^6*x^8 + 3*a^4*x^6 + a^2*x^4 + (a^5*x^7 + a^3*x^5)*(a^2*x^2 + 1)^(3/2) + (3*a^6*x^8 + 5*a^
4*x^6 + 2*a^2*x^4)*(a^2*x^2 + 1) + (4*a^8*x^10 + 12*a^6*x^8 + 12*a^4*x^6 + 4*a^2*x^4 + 2*(2*a^5*x^7 + 3*a^3*x^
5 + a*x^3)*(a^2*x^2 + 1)^(3/2) + 3*(4*a^6*x^8 + 8*a^4*x^6 + 5*a^2*x^4 + x^2)*(a^2*x^2 + 1) + (12*a^7*x^9 + 30*
a^5*x^7 + 25*a^3*x^5 + 7*a*x^3)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1)) + (3*a^7*x^9 + 7*a^5*x^7 + 5*a
^3*x^5 + a*x^3)*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*(16*a^10*x^11 + 64*a^8*x^9 + 96*a^6*x^7 + 64*a^4*x^5 + 16*a^2*x^3 + 4*(4*a^6*x^7 + 3*a
^4*x^5)*(a^2*x^2 + 1)^2 + (64*a^7*x^8 + 100*a^5*x^6 + 42*a^3*x^4 + 3*a*x^2)*(a^2*x^2 + 1)^(3/2) + 6*(16*a^8*x^
9 + 38*a^6*x^7 + 30*a^4*x^5 + 9*a^2*x^3 + x)*(a^2*x^2 + 1) + (64*a^9*x^10 + 204*a^7*x^8 + 234*a^5*x^6 + 115*a^
3*x^4 + 21*a*x^2)*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^{3}}{\operatorname{arsinh}\left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3/arcsinh(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/asinh(a*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/arcsinh(a*x)^3, x)