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]

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

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Rubi [A]  time = 0.07, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 30

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

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 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 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 \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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.44, size = 63, normalized size = 1.00 \[ -\frac {b^{2} x^{2} + 2 \, a b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right ) - \arcsin \left (b x + a\right )^{2}}{4 \, b} \]

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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giac [A]  time = 0.37, size = 79, normalized size = 1.25 \[ \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )}{2 \, b} + \frac {\arcsin \left (b x + a\right )^{2}}{4 \, b} - \frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{4 \, b} - \frac {1}{8 \, b} \]

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

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maple [A]  time = 0.13, size = 96, normalized size = 1.52 \[ \frac {2 \arcsin \left (b x +a \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x b -b^{2} x^{2}+2 \arcsin \left (b x +a \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a -2 a b x +\arcsin \left (b x +a \right )^{2}-a^{2}}{4 b} \]

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

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maxima [B]  time = 0.42, size = 240, normalized size = 3.81 \[ -\frac {1}{4} \, {\left (x^{2} + \frac {2 \, a x}{b} - \frac {2 \, \arcsin \left (b x + a\right ) \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b^{2}} - \frac {\arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )^{2}}{b^{2}}\right )} b - \frac {1}{2} \, {\left (\frac {a^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b} - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x - \frac {{\left (a^{2} - 1\right )} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b}\right )} \arcsin \left (b x + a\right ) \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {asin}\left (a+b\,x\right )\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )} \operatorname {asin}{\left (a + b x \right )}\, dx \]

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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