3.498 \(\int \frac {x^3 \sin ^{-1}(a x)^n}{\sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=163 \[ -\frac {3 \sin ^{-1}(a x)^n \left (-i \sin ^{-1}(a x)\right )^{-n} \Gamma \left (n+1,-i \sin ^{-1}(a x)\right )}{8 a^4}+\frac {3^{-n-1} \sin ^{-1}(a x)^n \left (-i \sin ^{-1}(a x)\right )^{-n} \Gamma \left (n+1,-3 i \sin ^{-1}(a x)\right )}{8 a^4}-\frac {3 \left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \Gamma \left (n+1,i \sin ^{-1}(a x)\right )}{8 a^4}+\frac {3^{-n-1} \left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \Gamma \left (n+1,3 i \sin ^{-1}(a x)\right )}{8 a^4} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4723, 3312, 3308, 2181} \[ -\frac {3 \sin ^{-1}(a x)^n \left (-i \sin ^{-1}(a x)\right )^{-n} \text {Gamma}\left (n+1,-i \sin ^{-1}(a x)\right )}{8 a^4}+\frac {3^{-n-1} \sin ^{-1}(a x)^n \left (-i \sin ^{-1}(a x)\right )^{-n} \text {Gamma}\left (n+1,-3 i \sin ^{-1}(a x)\right )}{8 a^4}-\frac {3 \left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \text {Gamma}\left (n+1,i \sin ^{-1}(a x)\right )}{8 a^4}+\frac {3^{-n-1} \left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \text {Gamma}\left (n+1,3 i \sin ^{-1}(a x)\right )}{8 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 4723

Int[((a_.) + ArcSin[(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*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 153, normalized size = 0.94 \[ -\frac {3^{-n-1} \sin ^{-1}(a x)^n \left (\sin ^{-1}(a x)^2\right )^{-2 n} \left (\left (-i \sin ^{-1}(a x)\right )^n \left (3^{n+2} \left (\sin ^{-1}(a x)^2\right )^n \Gamma \left (n+1,i \sin ^{-1}(a x)\right )-\left (\sin ^{-1}(a x)^2\right )^n \Gamma \left (n+1,3 i \sin ^{-1}(a x)\right )-\left (i \sin ^{-1}(a x)\right )^{2 n} \Gamma \left (n+1,-3 i \sin ^{-1}(a x)\right )\right )+3^{n+2} \left (i \sin ^{-1}(a x)\right )^n \left (\sin ^{-1}(a x)^2\right )^n \Gamma \left (n+1,-i \sin ^{-1}(a x)\right )\right )}{8 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(3^(-1 - n)*ArcSin[a*x]^n*(3^(2 + n)*(I*ArcSin[a*x])^n*(ArcSin[a*x]^2)^n*Gamma[1 + n, (-I)*ArcSin[a*x]] +
((-I)*ArcSin[a*x])^n*(3^(2 + n)*(ArcSin[a*x]^2)^n*Gamma[1 + n, I*ArcSin[a*x]] - (I*ArcSin[a*x])^(2*n)*Gamma[1
+ n, (-3*I)*ArcSin[a*x]] - (ArcSin[a*x]^2)^n*Gamma[1 + n, (3*I)*ArcSin[a*x]])))/(a^4*(ArcSin[a*x]^2)^(2*n))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x^{3} \arcsin \left (a x\right )^{n}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \arcsin \left (a x \right )^{n}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,{\mathrm {asin}\left (a\,x\right )}^n}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*asin(a*x)^n)/(1 - a^2*x^2)^(1/2),x)

[Out]

int((x^3*asin(a*x)^n)/(1 - a^2*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \operatorname {asin}^{n}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*asin(a*x)**n/sqrt(-(a*x - 1)*(a*x + 1)), x)

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