\(\int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx\) [528]

Optimal result

Integrand size = 23, antiderivative size = 113 \[ \int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx=\frac {3^{-1-n} (-\text {arcsinh}(a x))^{-n} \text {arcsinh}(a x)^n \Gamma (1+n,-3 \text {arcsinh}(a x))}{8 a^4}-\frac {3 (-\text {arcsinh}(a x))^{-n} \text {arcsinh}(a x)^n \Gamma (1+n,-\text {arcsinh}(a x))}{8 a^4}-\frac {3 \Gamma (1+n,\text {arcsinh}(a x))}{8 a^4}+\frac {3^{-1-n} \Gamma (1+n,3 \text {arcsinh}(a x))}{8 a^4} \]

[Out]

1/8*3^(-1-n)*arcsinh(a*x)^n*GAMMA(1+n,-3*arcsinh(a*x))/a^4/((-arcsinh(a*x))^n)-3/8*arcsinh(a*x)^n*GAMMA(1+n,-a
rcsinh(a*x))/a^4/((-arcsinh(a*x))^n)-3/8*GAMMA(1+n,arcsinh(a*x))/a^4+1/8*3^(-1-n)*GAMMA(1+n,3*arcsinh(a*x))/a^
4

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5819, 3393, 3389, 2212} \[ \int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx=\frac {3^{-n-1} \text {arcsinh}(a x)^n (-\text {arcsinh}(a x))^{-n} \Gamma (n+1,-3 \text {arcsinh}(a x))}{8 a^4}-\frac {3 \text {arcsinh}(a x)^n (-\text {arcsinh}(a x))^{-n} \Gamma (n+1,-\text {arcsinh}(a x))}{8 a^4}-\frac {3 \Gamma (n+1,\text {arcsinh}(a x))}{8 a^4}+\frac {3^{-n-1} \Gamma (n+1,3 \text {arcsinh}(a x))}{8 a^4} \]

[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 2212

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

Rule 3389

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 3393

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 5819

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

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx=\frac {3^{-1-n} (-\text {arcsinh}(a x))^{-n} \left (\text {arcsinh}(a x)^n \Gamma (1+n,-3 \text {arcsinh}(a x))-3^{2+n} \text {arcsinh}(a x)^n \Gamma (1+n,-\text {arcsinh}(a x))+(-\text {arcsinh}(a x))^n \left (-3^{2+n} \Gamma (1+n,\text {arcsinh}(a x))+\Gamma (1+n,3 \text {arcsinh}(a x))\right )\right )}{8 a^4} \]

[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)

Maple [F]

\[\int \frac {x^{3} \operatorname {arcsinh}\left (a x \right )^{n}}{\sqrt {a^{2} x^{2}+1}}d x\]

[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)

Fricas [F]

\[ \int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx=\int { \frac {x^{3} \operatorname {arsinh}\left (a x\right )^{n}}{\sqrt {a^{2} x^{2} + 1}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx=\int \frac {x^{3} \operatorname {asinh}^{n}{\left (a x \right )}}{\sqrt {a^{2} x^{2} + 1}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx=\int { \frac {x^{3} \operatorname {arsinh}\left (a x\right )^{n}}{\sqrt {a^{2} x^{2} + 1}} \,d x } \]

[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)

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \text {arcsinh}(a x)^n}{\sqrt {1+a^2 x^2}} \, dx=\int \frac {x^3\,{\mathrm {asinh}\left (a\,x\right )}^n}{\sqrt {a^2\,x^2+1}} \,d x \]

[In]

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

[Out]

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