3.332 \(\int \frac{1}{\sqrt{1-a^2-2 a b x-b^2 x^2} \sin ^{-1}(a+b x)^3} \, dx\)

Optimal. Leaf size=15 \[ -\frac{1}{2 b \sin ^{-1}(a+b x)^2} \]

[Out]

-1/(2*b*ArcSin[a + b*x]^2)

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Rubi [A]  time = 0.0703002, antiderivative size = 15, 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} \[ -\frac{1}{2 b \sin ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*b*ArcSin[a + b*x]^2)

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

Mathematica [A]  time = 0.0145088, size = 15, normalized size = 1. \[ -\frac{1}{2 b \sin ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*b*ArcSin[a + b*x]^2)

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Maple [A]  time = 0.044, size = 14, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,b \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 82.6449, size = 45, normalized size = 3. \begin{align*} -\frac{1}{2 \, b \arctan \left (b x + a, \sqrt{b x + a + 1} \sqrt{-b x - a + 1}\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2)

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Fricas [A]  time = 2.06257, size = 38, normalized size = 2.53 \begin{align*} -\frac{1}{2 \, b \arcsin \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b*arcsin(b*x + a)^2)

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Sympy [A]  time = 2.26528, size = 29, normalized size = 1.93 \begin{align*} \begin{cases} - \frac{1}{2 b \operatorname{asin}^{2}{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x}{\sqrt{1 - a^{2}} \operatorname{asin}^{3}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-1/(2*b*asin(a + b*x)**2), Ne(b, 0)), (x/(sqrt(1 - a**2)*asin(a)**3), True))

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Giac [A]  time = 1.29891, size = 18, normalized size = 1.2 \begin{align*} -\frac{1}{2 \, b \arcsin \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b*arcsin(b*x + a)^2)