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

Optimal. Leaf size=86 \[ \frac{\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b^2 c}-\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b^2 c}-\frac{1-c^2 x^2}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

-((1 - c^2*x^2)/(b*c*(a + b*ArcSin[c*x]))) + (CosIntegral[(2*(a + b*ArcSin[c*x]))/b]*Sin[(2*a)/b])/(b^2*c) - (
Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/(b^2*c)

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Rubi [A]  time = 0.162483, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4659, 4635, 4406, 12, 3303, 3299, 3302} \[ \frac{\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c}-\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c}-\frac{1-c^2 x^2}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - c^2*x^2)/(b*c*(a + b*ArcSin[c*x]))) + (CosIntegral[(2*a)/b + 2*ArcSin[c*x]]*Sin[(2*a)/b])/(b^2*c) - (Co
s[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]])/(b^2*c)

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 4635

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
in[x]^m*Cos[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

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

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

Mathematica [A]  time = 0.198103, size = 72, normalized size = 0.84 \[ \frac{\frac{b \left (c^2 x^2-1\right )}{a+b \sin ^{-1}(c x)}+\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )}{b^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*(-1 + c^2*x^2))/(a + b*ArcSin[c*x]) + CosIntegral[2*(a/b + ArcSin[c*x])]*Sin[(2*a)/b] - Cos[(2*a)/b]*SinIn
tegral[2*(a/b + ArcSin[c*x])])/(b^2*c)

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Maple [A]  time = 0.046, size = 134, normalized size = 1.6 \begin{align*} -{\frac{1}{2\,{b}^{2}c \left ( a+b\arcsin \left ( cx \right ) \right ) } \left ( 2\,\arcsin \left ( cx \right ){\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) b-2\,\arcsin \left ( cx \right ){\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) b+2\,{\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) a-2\,{\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) a+\cos \left ( 2\,\arcsin \left ( cx \right ) \right ) b+b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*x^2 - 2*(b^2*c^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c^2)*integrate(x/(b^2*arctan2(c*x, sqrt
(c*x + 1)*sqrt(-c*x + 1)) + a*b), x) - 1)/(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.55515, size = 392, normalized size = 4.56 \begin{align*} \frac{2 \, b \arcsin \left (c x\right ) \cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right ) \sin \left (\frac{a}{b}\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac{2 \, b \arcsin \left (c x\right ) \cos \left (\frac{a}{b}\right )^{2} \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac{2 \, a \cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right ) \sin \left (\frac{a}{b}\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac{2 \, a \cos \left (\frac{a}{b}\right )^{2} \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac{b \arcsin \left (c x\right ) \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac{{\left (c^{2} x^{2} - 1\right )} b}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac{a \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*arcsin(c*x)*cos(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 2*b*arcs
in(c*x)*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 2*a*cos(a/b)*cos_integr
al(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 2*a*cos(a/b)^2*sin_integral(2*a/b + 2*arcsi
n(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + b*arcsin(c*x)*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsin(c*x) +
a*b^2*c) + (c^2*x^2 - 1)*b/(b^3*c*arcsin(c*x) + a*b^2*c) + a*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsi
n(c*x) + a*b^2*c)