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

Optimal. Leaf size=19 \[ \frac{\sin ^{-1}(a+b x)^{n+1}}{b (n+1)} \]

[Out]

ArcSin[a + b*x]^(1 + n)/(b*(1 + n))

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Rubi [A]  time = 0.0749411, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {4807, 4641} \[ \frac{\sin ^{-1}(a+b x)^{n+1}}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^n/Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2],x]

[Out]

ArcSin[a + b*x]^(1 + n)/(b*(1 + n))

Rule 4807

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> Di
st[1/d, Subst[Int[(-(C/d^2) + (C*x^2)/d^2)^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rubi steps

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

Mathematica [A]  time = 0.0281393, size = 19, normalized size = 1. \[ \frac{\sin ^{-1}(a+b x)^{n+1}}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^n/Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2],x]

[Out]

ArcSin[a + b*x]^(1 + n)/(b*(1 + n))

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Maple [A]  time = 0.06, size = 20, normalized size = 1.1 \begin{align*}{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{1+n}}{b \left ( 1+n \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(b*x+a)^(1+n)/b/(1+n)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 3.09604, size = 61, normalized size = 3.21 \begin{align*} \frac{\arcsin \left (b x + a\right )^{n} \arcsin \left (b x + a\right )}{b n + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(b*x + a)^n*arcsin(b*x + a)/(b*n + b)

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Sympy [A]  time = 1.47999, size = 60, normalized size = 3.16 \begin{align*} \begin{cases} \frac{x}{\sqrt{1 - a^{2}} \operatorname{asin}{\left (a \right )}} & \text{for}\: b = 0 \wedge n = -1 \\\frac{x \operatorname{asin}^{n}{\left (a \right )}}{\sqrt{1 - a^{2}}} & \text{for}\: b = 0 \\\frac{\log{\left (\operatorname{asin}{\left (a + b x \right )} \right )}}{b} & \text{for}\: n = -1 \\\frac{\operatorname{asin}{\left (a + b x \right )} \operatorname{asin}^{n}{\left (a + b x \right )}}{b n + b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(sqrt(1 - a**2)*asin(a)), Eq(b, 0) & Eq(n, -1)), (x*asin(a)**n/sqrt(1 - a**2), Eq(b, 0)), (log(as
in(a + b*x))/b, Eq(n, -1)), (asin(a + b*x)*asin(a + b*x)**n/(b*n + b), True))

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Giac [A]  time = 1.32728, size = 26, normalized size = 1.37 \begin{align*} \frac{\arcsin \left (b x + a\right )^{n + 1}}{b{\left (n + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsin(b*x + a)^(n + 1)/(b*(n + 1))