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

Optimal. Leaf size=63 \[ -\frac{(a+b x)^2}{4 b}+\frac{\sqrt{1-(a+b x)^2} (a+b x) \sin ^{-1}(a+b x)}{2 b}+\frac{\sin ^{-1}(a+b x)^2}{4 b} \]

[Out]

-(a + b*x)^2/(4*b) + ((a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(2*b) + ArcSin[a + b*x]^2/(4*b)

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Rubi [A]  time = 0.070168, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {4807, 4647, 4641, 30} \[ -\frac{(a+b x)^2}{4 b}+\frac{\sqrt{1-(a+b x)^2} (a+b x) \sin ^{-1}(a+b x)}{2 b}+\frac{\sin ^{-1}(a+b x)^2}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*x)^2/(4*b) + ((a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(2*b) + ArcSin[a + b*x]^2/(4*b)

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 4647

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \sqrt{1-a^2-2 a b x-b^2 x^2} \sin ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{1-x^2} \sin ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{2 b}-\frac{\operatorname{Subst}(\int x \, dx,x,a+b x)}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac{(a+b x)^2}{4 b}+\frac{(a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{2 b}+\frac{\sin ^{-1}(a+b x)^2}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0597914, size = 64, normalized size = 1.02 \[ \frac{2 (a+b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)-b x (2 a+b x)+\sin ^{-1}(a+b x)^2}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*x*(2*a + b*x)) + 2*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x] + ArcSin[a + b*x]^2)/(4*b)

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Maple [A]  time = 0.054, size = 96, normalized size = 1.5 \begin{align*}{\frac{1}{4\,b} \left ( 2\,\arcsin \left ( bx+a \right ) \sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}xb-{b}^{2}{x}^{2}+2\,\arcsin \left ( bx+a \right ) \sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}a-2\,xab+ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}-{a}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*arcsin(b*x+a)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x*b-b^2*x^2+2*arcsin(b*x+a)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)
*a-2*x*a*b+arcsin(b*x+a)^2-a^2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.46247, size = 153, normalized size = 2.43 \begin{align*} -\frac{b^{2} x^{2} + 2 \, a b x - 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )} \arcsin \left (b x + a\right ) - \arcsin \left (b x + a\right )^{2}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^2*x^2 + 2*a*b*x - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)*arcsin(b*x + a) - arcsin(b*x + a)^2)/
b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{- \left (a + b x - 1\right ) \left (a + b x + 1\right )} \operatorname{asin}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20567, size = 107, normalized size = 1.7 \begin{align*} \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )} \arcsin \left (b x + a\right )}{2 \, b} + \frac{\arcsin \left (b x + a\right )^{2}}{4 \, b} - \frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{4 \, b} - \frac{1}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)*arcsin(b*x + a)/b + 1/4*arcsin(b*x + a)^2/b - 1/4*(b^2*x^2 +
2*a*b*x + a^2 - 1)/b - 1/8/b