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

Optimal. Leaf size=135 \[ \frac{3 (a+b x)^2}{8 b}+\frac{\sin ^{-1}(a+b x)^4}{8 b}+\frac{(a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^3}{2 b}-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)^2}{4 b}+\frac{3 \sin ^{-1}(a+b x)^2}{8 b}-\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{4 b} \]

[Out]

(3*(a + b*x)^2)/(8*b) - (3*(a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(4*b) + (3*ArcSin[a + b*x]^2)/(8*b
) - (3*(a + b*x)^2*ArcSin[a + b*x]^2)/(4*b) + ((a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x]^3)/(2*b) + ArcS
in[a + b*x]^4/(8*b)

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Rubi [A]  time = 0.193401, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4807, 4647, 4641, 4627, 4707, 30} \[ \frac{3 (a+b x)^2}{8 b}+\frac{\sin ^{-1}(a+b x)^4}{8 b}+\frac{(a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^3}{2 b}-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)^2}{4 b}+\frac{3 \sin ^{-1}(a+b x)^2}{8 b}-\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x)^2)/(8*b) - (3*(a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(4*b) + (3*ArcSin[a + b*x]^2)/(8*b
) - (3*(a + b*x)^2*ArcSin[a + b*x]^2)/(4*b) + ((a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x]^3)/(2*b) + ArcS
in[a + b*x]^4/(8*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 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 4707

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

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)^3 \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{1-x^2} \sin ^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^3}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)^3}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{2 b}-\frac{3 \operatorname{Subst}\left (\int x \sin ^{-1}(x)^2 \, dx,x,a+b x\right )}{2 b}\\ &=-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)^2}{4 b}+\frac{(a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^3}{2 b}+\frac{\sin ^{-1}(a+b x)^4}{8 b}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2 \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{4 b}-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)^2}{4 b}+\frac{(a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^3}{2 b}+\frac{\sin ^{-1}(a+b x)^4}{8 b}+\frac{3 \operatorname{Subst}(\int x \, dx,x,a+b x)}{4 b}+\frac{3 \operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{4 b}\\ &=\frac{3 (a+b x)^2}{8 b}-\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{4 b}+\frac{3 \sin ^{-1}(a+b x)^2}{8 b}-\frac{3 (a+b x)^2 \sin ^{-1}(a+b x)^2}{4 b}+\frac{(a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)^3}{2 b}+\frac{\sin ^{-1}(a+b x)^4}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.123856, size = 133, normalized size = 0.99 \[ \frac{4 (a+b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)^3-3 \left (2 a^2+4 a b x+2 b^2 x^2-1\right ) \sin ^{-1}(a+b x)^2-6 (a+b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)+3 b x (2 a+b x)+\sin ^{-1}(a+b x)^4}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*x*(2*a + b*x) - 6*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x] - 3*(-1 + 2*a^2 + 4*a*b*x +
2*b^2*x^2)*ArcSin[a + b*x]^2 + 4*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x]^3 + ArcSin[a + b
*x]^4)/(8*b)

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Maple [A]  time = 0.085, size = 215, normalized size = 1.6 \begin{align*}{\frac{1}{8\,b} \left ( 4\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}xb-6\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}{x}^{2}{b}^{2}+4\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}a-12\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}xab+ \left ( \arcsin \left ( bx+a \right ) \right ) ^{4}-6\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}{a}^{2}-6\,\arcsin \left ( bx+a \right ) \sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}xb+3\,{b}^{2}{x}^{2}-6\,\arcsin \left ( bx+a \right ) \sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}a+6\,xab+3\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}+3\,{a}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*arcsin(b*x+a)^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x*b-6*arcsin(b*x+a)^2*x^2*b^2+4*arcsin(b*x+a)^3*(-b^2*x^
2-2*a*b*x-a^2+1)^(1/2)*a-12*arcsin(b*x+a)^2*x*a*b+arcsin(b*x+a)^4-6*arcsin(b*x+a)^2*a^2-6*arcsin(b*x+a)*(-b^2*
x^2-2*a*b*x-a^2+1)^(1/2)*x*b+3*b^2*x^2-6*arcsin(b*x+a)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a+6*x*a*b+3*arcsin(b*x+a
)^2+3*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)^3*(-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 = 2.31098, size = 278, normalized size = 2.06 \begin{align*} \frac{3 \, b^{2} x^{2} + \arcsin \left (b x + a\right )^{4} + 6 \, a b x - 3 \,{\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} - 1\right )} \arcsin \left (b x + a\right )^{2} + 2 \, \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 )^{3} - 3 \,{\left (b x + a\right )} \arcsin \left (b x + a\right )\right )}}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*b^2*x^2 + arcsin(b*x + a)^4 + 6*a*b*x - 3*(2*b^2*x^2 + 4*a*b*x + 2*a^2 - 1)*arcsin(b*x + a)^2 + 2*sqrt(
-b^2*x^2 - 2*a*b*x - a^2 + 1)*(2*(b*x + a)*arcsin(b*x + a)^3 - 3*(b*x + a)*arcsin(b*x + a)))/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}^{3}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(b*x+a)**3*(-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)**3, x)

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Giac [A]  time = 1.29283, size = 219, normalized size = 1.62 \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 )^{3}}{2 \, b} + \frac{\arcsin \left (b x + a\right )^{4}}{8 \, b} - \frac{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \arcsin \left (b x + a\right )^{2}}{4 \, b} - \frac{3 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )} \arcsin \left (b x + a\right )}{4 \, b} - \frac{3 \, \arcsin \left (b x + a\right )^{2}}{8 \, b} + \frac{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{8 \, b} + \frac{3}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3*(-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)^3/b + 1/8*arcsin(b*x + a)^4/b - 3/4*(b^2*x^2
+ 2*a*b*x + a^2 - 1)*arcsin(b*x + a)^2/b - 3/4*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)*arcsin(b*x + a)/b
- 3/8*arcsin(b*x + a)^2/b + 3/8*(b^2*x^2 + 2*a*b*x + a^2 - 1)/b + 3/16/b