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

Optimal. Leaf size=13 \[ -\frac {1}{b \sin ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 13, normalized size of antiderivative = 1.00, 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} \[ -\frac {1}{b \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*ArcSin[a + b*x]))

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 13, normalized size = 1.00 \[ -\frac {1}{b \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*ArcSin[a + b*x]))

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fricas [A]  time = 0.42, size = 13, normalized size = 1.00 \[ -\frac {1}{b \arcsin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(b*arcsin(b*x + a))

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giac [A]  time = 0.32, size = 13, normalized size = 1.00 \[ -\frac {1}{b \arcsin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(b*arcsin(b*x + a))

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maple [A]  time = 0.10, size = 14, normalized size = 1.08 \[ -\frac {1}{b \arcsin \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.88, size = 33, normalized size = 2.54 \[ -\frac {1}{b \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.27, size = 13, normalized size = 1.00 \[ -\frac {1}{b\,\mathrm {asin}\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(b*asin(a + b*x))

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sympy [A]  time = 1.29, size = 26, normalized size = 2.00 \[ \begin {cases} - \frac {1}{b \operatorname {asin}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {1 - a^{2}} \operatorname {asin}^{2}{\relax (a )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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