3.528 \(\int \frac{x^3 \sinh ^{-1}(a x)^n}{\sqrt{1+a^2 x^2}} \, dx\)

Optimal. Leaf size=113 \[ \frac{3^{-n-1} \sinh ^{-1}(a x)^n \left (-\sinh ^{-1}(a x)\right )^{-n} \text{Gamma}\left (n+1,-3 \sinh ^{-1}(a x)\right )}{8 a^4}-\frac{3 \sinh ^{-1}(a x)^n \left (-\sinh ^{-1}(a x)\right )^{-n} \text{Gamma}\left (n+1,-\sinh ^{-1}(a x)\right )}{8 a^4}-\frac{3 \text{Gamma}\left (n+1,\sinh ^{-1}(a x)\right )}{8 a^4}+\frac{3^{-n-1} \text{Gamma}\left (n+1,3 \sinh ^{-1}(a x)\right )}{8 a^4} \]

[Out]

(3^(-1 - n)*ArcSinh[a*x]^n*Gamma[1 + n, -3*ArcSinh[a*x]])/(8*a^4*(-ArcSinh[a*x])^n) - (3*ArcSinh[a*x]^n*Gamma[
1 + n, -ArcSinh[a*x]])/(8*a^4*(-ArcSinh[a*x])^n) - (3*Gamma[1 + n, ArcSinh[a*x]])/(8*a^4) + (3^(-1 - n)*Gamma[
1 + n, 3*ArcSinh[a*x]])/(8*a^4)

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Rubi [A]  time = 0.252861, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {5779, 3312, 3308, 2181} \[ \frac{3^{-n-1} \sinh ^{-1}(a x)^n \left (-\sinh ^{-1}(a x)\right )^{-n} \text{Gamma}\left (n+1,-3 \sinh ^{-1}(a x)\right )}{8 a^4}-\frac{3 \sinh ^{-1}(a x)^n \left (-\sinh ^{-1}(a x)\right )^{-n} \text{Gamma}\left (n+1,-\sinh ^{-1}(a x)\right )}{8 a^4}-\frac{3 \text{Gamma}\left (n+1,\sinh ^{-1}(a x)\right )}{8 a^4}+\frac{3^{-n-1} \text{Gamma}\left (n+1,3 \sinh ^{-1}(a x)\right )}{8 a^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcSinh[a*x]^n)/Sqrt[1 + a^2*x^2],x]

[Out]

(3^(-1 - n)*ArcSinh[a*x]^n*Gamma[1 + n, -3*ArcSinh[a*x]])/(8*a^4*(-ArcSinh[a*x])^n) - (3*ArcSinh[a*x]^n*Gamma[
1 + n, -ArcSinh[a*x]])/(8*a^4*(-ArcSinh[a*x])^n) - (3*Gamma[1 + n, ArcSinh[a*x]])/(8*a^4) + (3^(-1 - n)*Gamma[
1 + n, 3*ArcSinh[a*x]])/(8*a^4)

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3308

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

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

Mathematica [A]  time = 0.189915, size = 100, normalized size = 0.88 \[ \frac{3^{-n-1} \left (-\sinh ^{-1}(a x)\right )^{-n} \left (\left (-\sinh ^{-1}(a x)\right )^n \left (\text{Gamma}\left (n+1,3 \sinh ^{-1}(a x)\right )-3^{n+2} \text{Gamma}\left (n+1,\sinh ^{-1}(a x)\right )\right )+\sinh ^{-1}(a x)^n \text{Gamma}\left (n+1,-3 \sinh ^{-1}(a x)\right )-3^{n+2} \sinh ^{-1}(a x)^n \text{Gamma}\left (n+1,-\sinh ^{-1}(a x)\right )\right )}{8 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcSinh[a*x]^n)/Sqrt[1 + a^2*x^2],x]

[Out]

(3^(-1 - n)*(ArcSinh[a*x]^n*Gamma[1 + n, -3*ArcSinh[a*x]] - 3^(2 + n)*ArcSinh[a*x]^n*Gamma[1 + n, -ArcSinh[a*x
]] + (-ArcSinh[a*x])^n*(-(3^(2 + n)*Gamma[1 + n, ArcSinh[a*x]]) + Gamma[1 + n, 3*ArcSinh[a*x]])))/(8*a^4*(-Arc
Sinh[a*x])^n)

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Maple [F]  time = 0.187, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{n}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x)

[Out]

int(x^3*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \operatorname{arsinh}\left (a x\right )^{n}}{\sqrt{a^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3*arcsinh(a*x)^n/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 )^{n}}{\sqrt{a^{2} x^{2} + 1}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(x^3*arcsinh(a*x)^n/sqrt(a^2*x^2 + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \operatorname{asinh}^{n}{\left (a x \right )}}{\sqrt{a^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asinh(a*x)**n/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**3*asinh(a*x)**n/sqrt(a**2*x**2 + 1), 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 )^{n}}{\sqrt{a^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^n/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3*arcsinh(a*x)^n/sqrt(a^2*x^2 + 1), x)