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

Optimal. Leaf size=110 \[ \frac{(a+b x)^4}{16 b}-\frac{5 (a+b x)^2}{16 b}+\frac{\left (1-(a+b x)^2\right )^{3/2} (a+b x) \sin ^{-1}(a+b x)}{4 b}+\frac{3 \sqrt{1-(a+b x)^2} (a+b x) \sin ^{-1}(a+b x)}{8 b}+\frac{3 \sin ^{-1}(a+b x)^2}{16 b} \]

[Out]

(-5*(a + b*x)^2)/(16*b) + (a + b*x)^4/(16*b) + (3*(a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(8*b) + ((a
+ b*x)*(1 - (a + b*x)^2)^(3/2)*ArcSin[a + b*x])/(4*b) + (3*ArcSin[a + b*x]^2)/(16*b)

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Rubi [A]  time = 0.105756, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4807, 4649, 4647, 4641, 30, 14} \[ \frac{(a+b x)^4}{16 b}-\frac{5 (a+b x)^2}{16 b}+\frac{\left (1-(a+b x)^2\right )^{3/2} (a+b x) \sin ^{-1}(a+b x)}{4 b}+\frac{3 \sqrt{1-(a+b x)^2} (a+b x) \sin ^{-1}(a+b x)}{8 b}+\frac{3 \sin ^{-1}(a+b x)^2}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2 - 2*a*b*x - b^2*x^2)^(3/2)*ArcSin[a + b*x],x]

[Out]

(-5*(a + b*x)^2)/(16*b) + (a + b*x)^4/(16*b) + (3*(a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(8*b) + ((a
+ b*x)*(1 - (a + b*x)^2)^(3/2)*ArcSin[a + b*x])/(4*b) + (3*ArcSin[a + b*x]^2)/(16*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 4649

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \left (1-a^2-2 a b x-b^2 x^2\right )^{3/2} \sin ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (1-x^2\right )^{3/2} \sin ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)}{4 b}-\frac{\operatorname{Subst}\left (\int x \left (1-x^2\right ) \, dx,x,a+b x\right )}{4 b}+\frac{3 \operatorname{Subst}\left (\int \sqrt{1-x^2} \sin ^{-1}(x) \, dx,x,a+b x\right )}{4 b}\\ &=\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{8 b}+\frac{(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)}{4 b}-\frac{\operatorname{Subst}\left (\int \left (x-x^3\right ) \, dx,x,a+b x\right )}{4 b}-\frac{3 \operatorname{Subst}(\int x \, dx,x,a+b x)}{8 b}+\frac{3 \operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,a+b x\right )}{8 b}\\ &=-\frac{5 (a+b x)^2}{16 b}+\frac{(a+b x)^4}{16 b}+\frac{3 (a+b x) \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{8 b}+\frac{(a+b x) \left (1-(a+b x)^2\right )^{3/2} \sin ^{-1}(a+b x)}{4 b}+\frac{3 \sin ^{-1}(a+b x)^2}{16 b}\\ \end{align*}

Mathematica [A]  time = 0.0678455, size = 129, normalized size = 1.17 \[ \frac{1}{16} \left (-\frac{2 \sqrt{-a^2-2 a b x-b^2 x^2+1} \left (6 a^2 b x+2 a^3+6 a b^2 x^2-5 a+2 b^3 x^3-5 b x\right ) \sin ^{-1}(a+b x)}{b}+\left (6 a^2-5\right ) b x^2+2 a \left (2 a^2-5\right ) x+4 a b^2 x^3+\frac{3 \sin ^{-1}(a+b x)^2}{b}+b^3 x^4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - a^2 - 2*a*b*x - b^2*x^2)^(3/2)*ArcSin[a + b*x],x]

[Out]

(2*a*(-5 + 2*a^2)*x + (-5 + 6*a^2)*b*x^2 + 4*a*b^2*x^3 + b^3*x^4 - (2*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*(-5*a
+ 2*a^3 - 5*b*x + 6*a^2*b*x + 6*a*b^2*x^2 + 2*b^3*x^3)*ArcSin[a + b*x])/b + (3*ArcSin[a + b*x]^2)/b)/16

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Maple [B]  time = 0.061, size = 277, normalized size = 2.5 \begin{align*}{\frac{1}{16\,b} \left ( -4\,\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}\arcsin \left ( bx+a \right ){x}^{3}{b}^{3}+{x}^{4}{b}^{4}-12\,\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}\arcsin \left ( bx+a \right ){x}^{2}a{b}^{2}+4\,{x}^{3}a{b}^{3}-12\,\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}\arcsin \left ( bx+a \right ) x{a}^{2}b+6\,{x}^{2}{a}^{2}{b}^{2}-4\,\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}\arcsin \left ( bx+a \right ){a}^{3}+4\,x{a}^{3}b+10\,\arcsin \left ( bx+a \right ) \sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}xb-5\,{b}^{2}{x}^{2}+{a}^{4}+10\,\arcsin \left ( bx+a \right ) \sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}a-10\,xab+3\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}-5\,{a}^{2}+4 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(-4*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*arcsin(b*x+a)*x^3*b^3+x^4*b^4-12*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*arcsin
(b*x+a)*x^2*a*b^2+4*x^3*a*b^3-12*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*arcsin(b*x+a)*x*a^2*b+6*x^2*a^2*b^2-4*(-b^2*x^
2-2*a*b*x-a^2+1)^(1/2)*arcsin(b*x+a)*a^3+4*x*a^3*b+10*arcsin(b*x+a)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x*b-5*b^2*x
^2+a^4+10*arcsin(b*x+a)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a-10*x*a*b+3*arcsin(b*x+a)^2-5*a^2+4)/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((-b^2*x^2-2*a*b*x-a^2+1)^(3/2)*arcsin(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.35912, size = 285, normalized size = 2.59 \begin{align*} \frac{b^{4} x^{4} + 4 \, a b^{3} x^{3} +{\left (6 \, a^{2} - 5\right )} b^{2} x^{2} + 2 \,{\left (2 \, a^{3} - 5 \, a\right )} b x - 2 \,{\left (2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 2 \, a^{3} +{\left (6 \, a^{2} - 5\right )} b x - 5 \, a\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right ) + 3 \, \arcsin \left (b x + a\right )^{2}}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(b^4*x^4 + 4*a*b^3*x^3 + (6*a^2 - 5)*b^2*x^2 + 2*(2*a^3 - 5*a)*b*x - 2*(2*b^3*x^3 + 6*a*b^2*x^2 + 2*a^3 +
(6*a^2 - 5)*b*x - 5*a)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*arcsin(b*x + a) + 3*arcsin(b*x + a)^2)/b

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Sympy [A]  time = 7.82236, size = 298, normalized size = 2.71 \begin{align*} \begin{cases} \frac{a^{3} x}{4} - \frac{a^{3} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{4 b} + \frac{3 a^{2} b x^{2}}{8} - \frac{3 a^{2} x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{4} + \frac{a b^{2} x^{3}}{4} - \frac{3 a b x^{2} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{4} - \frac{5 a x}{8} + \frac{5 a \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{8 b} + \frac{b^{3} x^{4}}{16} - \frac{b^{2} x^{3} \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{4} - \frac{5 b x^{2}}{16} + \frac{5 x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{8} + \frac{3 \operatorname{asin}^{2}{\left (a + b x \right )}}{16 b} & \text{for}\: b \neq 0 \\x \left (1 - a^{2}\right )^{\frac{3}{2}} \operatorname{asin}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b**2*x**2-2*a*b*x-a**2+1)**(3/2)*asin(b*x+a),x)

[Out]

Piecewise((a**3*x/4 - a**3*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(4*b) + 3*a**2*b*x**2/8 - 3*a**
2*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/4 + a*b**2*x**3/4 - 3*a*b*x**2*sqrt(-a**2 - 2*a*b*x -
b**2*x**2 + 1)*asin(a + b*x)/4 - 5*a*x/8 + 5*a*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(8*b) + b**
3*x**4/16 - b**2*x**3*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/4 - 5*b*x**2/16 + 5*x*sqrt(-a**2 - 2
*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/8 + 3*asin(a + b*x)**2/(16*b), Ne(b, 0)), (x*(1 - a**2)**(3/2)*asin(a),
True))

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Giac [A]  time = 1.25101, size = 190, normalized size = 1.73 \begin{align*} \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}{\left (b x + a\right )} \arcsin \left (b x + a\right )}{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 )}{8 \, b} + \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{16 \, b} + \frac{3 \, \arcsin \left (b x + a\right )^{2}}{16 \, b} - \frac{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{16 \, b} - \frac{15}{128 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*(b*x + a)*arcsin(b*x + a)/b + 3/8*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*
(b*x + a)*arcsin(b*x + a)/b + 1/16*(b^2*x^2 + 2*a*b*x + a^2 - 1)^2/b + 3/16*arcsin(b*x + a)^2/b - 3/16*(b^2*x^
2 + 2*a*b*x + a^2 - 1)/b - 15/128/b