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

Optimal. Leaf size=109 \[ \frac {\sin ^{-1}(a x)^{n+1}}{2 a^3 (n+1)}+\frac {i 2^{-n-3} \sin ^{-1}(a x)^n \left (-i \sin ^{-1}(a x)\right )^{-n} \Gamma \left (n+1,-2 i \sin ^{-1}(a x)\right )}{a^3}-\frac {i 2^{-n-3} \left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \Gamma \left (n+1,2 i \sin ^{-1}(a x)\right )}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4723, 3312, 3307, 2181} \[ \frac {i 2^{-n-3} \sin ^{-1}(a x)^n \left (-i \sin ^{-1}(a x)\right )^{-n} \text {Gamma}\left (n+1,-2 i \sin ^{-1}(a x)\right )}{a^3}-\frac {i 2^{-n-3} \left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \text {Gamma}\left (n+1,2 i \sin ^{-1}(a x)\right )}{a^3}+\frac {\sin ^{-1}(a x)^{n+1}}{2 a^3 (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcSin[a*x]^(1 + n)/(2*a^3*(1 + n)) + (I*2^(-3 - n)*ArcSin[a*x]^n*Gamma[1 + n, (-2*I)*ArcSin[a*x]])/(a^3*((-I)
*ArcSin[a*x])^n) - (I*2^(-3 - n)*ArcSin[a*x]^n*Gamma[1 + n, (2*I)*ArcSin[a*x]])/(a^3*(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 3307

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

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

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Mathematica [A]  time = 0.27, size = 109, normalized size = 1.00 \[ \frac {2^{-n-3} \sin ^{-1}(a x)^n \left (\sin ^{-1}(a x)^2\right )^{-n} \left (2^{n+2} \sin ^{-1}(a x) \left (\sin ^{-1}(a x)^2\right )^n-i (n+1) \left (-i \sin ^{-1}(a x)\right )^n \Gamma \left (n+1,2 i \sin ^{-1}(a x)\right )+i (n+1) \left (i \sin ^{-1}(a x)\right )^n \Gamma \left (n+1,-2 i \sin ^{-1}(a x)\right )\right )}{a^3 (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x^{2} \arcsin \left (a x\right )^{n}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \arcsin \left (a x \right )^{n}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*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^2*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 \[ \int \frac {x^2\,{\mathrm {asin}\left (a\,x\right )}^n}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^2*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 \[ \int \frac {x^{2} \operatorname {asin}^{n}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*asin(a*x)**n/sqrt(-(a*x - 1)*(a*x + 1)), x)

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