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

Optimal. Leaf size=111 \[ -\frac {(a+b x) \sqrt {1-(a+b x)^2}}{4 b}+\frac {\sin ^{-1}(a+b x)^3}{6 b}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b}-\frac {(a+b x)^2 \sin ^{-1}(a+b x)}{2 b}+\frac {\sin ^{-1}(a+b x)}{4 b} \]

[Out]

1/4*arcsin(b*x+a)/b-1/2*(b*x+a)^2*arcsin(b*x+a)/b+1/6*arcsin(b*x+a)^3/b-1/4*(b*x+a)*(1-(b*x+a)^2)^(1/2)/b+1/2*
(b*x+a)*arcsin(b*x+a)^2*(1-(b*x+a)^2)^(1/2)/b

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Rubi [A]  time = 0.13, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4807, 4647, 4641, 4627, 321, 216} \[ -\frac {(a+b x) \sqrt {1-(a+b x)^2}}{4 b}+\frac {\sin ^{-1}(a+b x)^3}{6 b}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b}-\frac {(a+b x)^2 \sin ^{-1}(a+b x)}{2 b}+\frac {\sin ^{-1}(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*x)*Sqrt[1 - (a + b*x)^2])/(4*b) + ArcSin[a + b*x]/(4*b) - ((a + b*x)^2*ArcSin[a + b*x])/(2*b) + ((a +
b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x]^2)/(2*b) + ArcSin[a + b*x]^3/(6*b)

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && 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)^2 \, dx &=\frac {\operatorname {Subst}\left (\int \sqrt {1-x^2} \sin ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)^2}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{2 b}-\frac {\operatorname {Subst}\left (\int x \sin ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=-\frac {(a+b x)^2 \sin ^{-1}(a+b x)}{2 b}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b}+\frac {\sin ^{-1}(a+b x)^3}{6 b}+\frac {\operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{4 b}-\frac {(a+b x)^2 \sin ^{-1}(a+b x)}{2 b}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b}+\frac {\sin ^{-1}(a+b x)^3}{6 b}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{4 b}\\ &=-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{4 b}+\frac {\sin ^{-1}(a+b x)}{4 b}-\frac {(a+b x)^2 \sin ^{-1}(a+b x)}{2 b}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)^2}{2 b}+\frac {\sin ^{-1}(a+b x)^3}{6 b}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 116, normalized size = 1.05 \[ \frac {-3 (a+b x) \sqrt {-a^2-2 a b x-b^2 x^2+1}+6 (a+b x) \sqrt {-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)^2-3 \left (2 a^2+4 a b x+2 b^2 x^2-1\right ) \sin ^{-1}(a+b x)+2 \sin ^{-1}(a+b x)^3}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] - 3*(-1 + 2*a^2 + 4*a*b*x + 2*b^2*x^2)*ArcSin[a + b*x] + 6*(a
+ b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x]^2 + 2*ArcSin[a + b*x]^3)/(12*b)

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fricas [A]  time = 0.55, size = 91, normalized size = 0.82 \[ \frac {2 \, \arcsin \left (b x + a\right )^{3} - 3 \, {\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} - 1\right )} \arcsin \left (b x + a\right ) + 3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (2 \, {\left (b x + a\right )} \arcsin \left (b x + a\right )^{2} - b x - a\right )}}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*arcsin(b*x + a)^3 - 3*(2*b^2*x^2 + 4*a*b*x + 2*a^2 - 1)*arcsin(b*x + a) + 3*sqrt(-b^2*x^2 - 2*a*b*x -
a^2 + 1)*(2*(b*x + a)*arcsin(b*x + a)^2 - b*x - a))/b

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giac [A]  time = 0.40, size = 125, normalized size = 1.13 \[ \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{2 \, b} + \frac {\arcsin \left (b x + a\right )^{3}}{6 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \arcsin \left (b x + a\right )}{2 \, b} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{4 \, b} - \frac {\arcsin \left (b x + a\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^2*(-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)^2/b + 1/6*arcsin(b*x + a)^3/b - 1/2*(b^2*x^2
+ 2*a*b*x + a^2 - 1)*arcsin(b*x + a)/b - 1/4*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/b - 1/4*arcsin(b*x +
a)/b

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maple [A]  time = 0.14, size = 179, normalized size = 1.61 \[ \frac {6 \arcsin \left (b x +a \right )^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x b -6 \arcsin \left (b x +a \right ) x^{2} b^{2}+6 \arcsin \left (b x +a \right )^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a -12 \arcsin \left (b x +a \right ) x a b +2 \arcsin \left (b x +a \right )^{3}-6 \arcsin \left (b x +a \right ) a^{2}-3 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x b -3 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +3 \arcsin \left (b x +a \right )}{12 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*arcsin(b*x+a)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x*b-6*arcsin(b*x+a)*x^2*b^2+6*arcsin(b*x+a)^2*(-b^2*x^2
-2*a*b*x-a^2+1)^(1/2)*a-12*arcsin(b*x+a)*x*a*b+2*arcsin(b*x+a)^3-6*arcsin(b*x+a)*a^2-3*(-b^2*x^2-2*a*b*x-a^2+1
)^(1/2)*x*b-3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a+3*arcsin(b*x+a))/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*arcsin(b*x + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {asin}\left (a+b\,x\right )}^2\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asin(a + b*x)^2*(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 \[ \int \sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )} \operatorname {asin}^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(b*x+a)**2*(-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)**2, x)

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