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

Optimal. Leaf size=47 \[ \frac{\text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{2 b}+\frac{\text{CosIntegral}\left (4 \sin ^{-1}(a+b x)\right )}{8 b}+\frac{3 \log \left (\sin ^{-1}(a+b x)\right )}{8 b} \]

[Out]

CosIntegral[2*ArcSin[a + b*x]]/(2*b) + CosIntegral[4*ArcSin[a + b*x]]/(8*b) + (3*Log[ArcSin[a + b*x]])/(8*b)

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Rubi [A]  time = 0.138157, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4807, 4661, 3312, 3302} \[ \frac{\text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{2 b}+\frac{\text{CosIntegral}\left (4 \sin ^{-1}(a+b x)\right )}{8 b}+\frac{3 \log \left (\sin ^{-1}(a+b x)\right )}{8 b} \]

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]

CosIntegral[2*ArcSin[a + b*x]]/(2*b) + CosIntegral[4*ArcSin[a + b*x]]/(8*b) + (3*Log[ArcSin[a + b*x]])/(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 4661

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[(
a + b*x)^n*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && I
GtQ[2*p, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 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{\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 \frac{\left (1-x^2\right )^{3/2}}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\cos ^4(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{3}{8 x}+\frac{\cos (2 x)}{2 x}+\frac{\cos (4 x)}{8 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac{3 \log \left (\sin ^{-1}(a+b x)\right )}{8 b}+\frac{\operatorname{Subst}\left (\int \frac{\cos (4 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b}+\frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b}\\ &=\frac{\text{Ci}\left (2 \sin ^{-1}(a+b x)\right )}{2 b}+\frac{\text{Ci}\left (4 \sin ^{-1}(a+b x)\right )}{8 b}+\frac{3 \log \left (\sin ^{-1}(a+b x)\right )}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.314244, size = 37, normalized size = 0.79 \[ \frac{4 \text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )+\text{CosIntegral}\left (4 \sin ^{-1}(a+b x)\right )+3 \log \left (\sin ^{-1}(a+b x)\right )}{8 b} \]

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]

(4*CosIntegral[2*ArcSin[a + b*x]] + CosIntegral[4*ArcSin[a + b*x]] + 3*Log[ArcSin[a + b*x]])/(8*b)

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Maple [A]  time = 0.052, size = 42, normalized size = 0.9 \begin{align*}{\frac{{\it Ci} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) }{2\,b}}+{\frac{{\it Ci} \left ( 4\,\arcsin \left ( bx+a \right ) \right ) }{8\,b}}+{\frac{3\,\ln \left ( \arcsin \left ( bx+a \right ) \right ) }{8\,b}} \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]

1/2*Ci(2*arcsin(b*x+a))/b+1/8*Ci(4*arcsin(b*x+a))/b+3/8*ln(arcsin(b*x+a))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}{\arcsin \left (b x + a\right )}\,{d x} \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]

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

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac{3}{2}}}{\operatorname{asin}{\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)**(3/2)/asin(b*x+a),x)

[Out]

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

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Giac [A]  time = 1.2108, size = 55, normalized size = 1.17 \begin{align*} \frac{\operatorname{Ci}\left (4 \, \arcsin \left (b x + a\right )\right )}{8 \, b} + \frac{\operatorname{Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{2 \, b} + \frac{3 \, \log \left (\arcsin \left (b x + a\right )\right )}{8 \, 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]

1/8*cos_integral(4*arcsin(b*x + a))/b + 1/2*cos_integral(2*arcsin(b*x + a))/b + 3/8*log(arcsin(b*x + a))/b