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

Optimal. Leaf size=75 \[ -\frac {\sin ^{-1}(a x)^n \left (-i \sin ^{-1}(a x)\right )^{-n} \Gamma \left (n+1,-i \sin ^{-1}(a x)\right )}{2 a^2}-\frac {\left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \Gamma \left (n+1,i \sin ^{-1}(a x)\right )}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4723, 3308, 2181} \[ -\frac {\sin ^{-1}(a x)^n \left (-i \sin ^{-1}(a x)\right )^{-n} \text {Gamma}\left (n+1,-i \sin ^{-1}(a x)\right )}{2 a^2}-\frac {\left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \text {Gamma}\left (n+1,i \sin ^{-1}(a x)\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcSin[a*x]^n*Gamma[1 + n, (-I)*ArcSin[a*x]])/(2*a^2*((-I)*ArcSin[a*x])^n) - (ArcSin[a*x]^n*Gamma[1 + n, I*A
rcSin[a*x]])/(2*a^2*(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 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 \sin ^{-1}(a x)^n}{\sqrt {1-a^2 x^2}} \, dx &=\frac {\operatorname {Subst}\left (\int x^n \sin (x) \, dx,x,\sin ^{-1}(a x)\right )}{a^2}\\ &=\frac {i \operatorname {Subst}\left (\int e^{-i x} x^n \, dx,x,\sin ^{-1}(a x)\right )}{2 a^2}-\frac {i \operatorname {Subst}\left (\int e^{i x} x^n \, dx,x,\sin ^{-1}(a x)\right )}{2 a^2}\\ &=-\frac {\left (-i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \Gamma \left (1+n,-i \sin ^{-1}(a x)\right )}{2 a^2}-\frac {\left (i \sin ^{-1}(a x)\right )^{-n} \sin ^{-1}(a x)^n \Gamma \left (1+n,i \sin ^{-1}(a x)\right )}{2 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 70, normalized size = 0.93 \[ -\frac {\sin ^{-1}(a x)^n \left (\sin ^{-1}(a x)^2\right )^{-n} \left (\left (-i \sin ^{-1}(a x)\right )^n \Gamma \left (n+1,i \sin ^{-1}(a x)\right )+\left (i \sin ^{-1}(a x)\right )^n \Gamma \left (n+1,-i \sin ^{-1}(a x)\right )\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x \arcsin \left (a x\right )^{n}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {x \arcsin \left (a x \right )^{n}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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