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

Optimal. Leaf size=73 \[ -\frac{\text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b}+\frac{\sqrt{1-(a+b x)^2} (a+b x)}{b \sin ^{-1}(a+b x)}-\frac{1-(a+b x)^2}{2 b \sin ^{-1}(a+b x)^2} \]

[Out]

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

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Rubi [A]  time = 0.112751, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4807, 4659, 4631, 3302} \[ -\frac{\text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b}+\frac{\sqrt{1-(a+b x)^2} (a+b x)}{b \sin ^{-1}(a+b x)}-\frac{1-(a+b x)^2}{2 b \sin ^{-1}(a+b x)^2} \]

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]

-(1 - (a + b*x)^2)/(2*b*ArcSin[a + b*x]^2) + ((a + b*x)*Sqrt[1 - (a + b*x)^2])/(b*ArcSin[a + b*x]) - CosIntegr
al[2*ArcSin[a + b*x]]/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 4659

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*
(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(b*(n + 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, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]

Rule 4631

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcSin
[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 0.270439, size = 88, normalized size = 1.21 \[ \frac{2 (a+b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)+a^2-2 \sin ^{-1}(a+b x)^2 \text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )+2 a b x+b^2 x^2-1}{2 b \sin ^{-1}(a+b x)^2} \]

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]

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

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Maple [A]  time = 0.05, size = 61, normalized size = 0.8 \begin{align*} -{\frac{4\,{\it Ci} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}-2\,\sin \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) +\cos \left ( 2\,\arcsin \left ( bx+a \right ) \right ) +1}{4\,b \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}} \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)^(1/2)/arcsin(b*x+a)^3,x)

[Out]

-1/4/b*(4*Ci(2*arcsin(b*x+a))*arcsin(b*x+a)^2-2*sin(2*arcsin(b*x+a))*arcsin(b*x+a)+cos(2*arcsin(b*x+a))+1)/arc
sin(b*x+a)^2

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)^(1/2)/arcsin(b*x+a)^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\arcsin \left (b x + a\right )^{3}}, x\right ) \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)^(1/2)/arcsin(b*x+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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((-b**2*x**2-2*a*b*x-a**2+1)**(1/2)/asin(b*x+a)**3,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 = 1.37681, size = 113, normalized size = 1.55 \begin{align*} -\frac{\operatorname{Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} + \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )}}{b \arcsin \left (b x + a\right )} + \frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{2 \, b \arcsin \left (b x + a\right )^{2}} \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)^(1/2)/arcsin(b*x+a)^3,x, algorithm="giac")

[Out]

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