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

Optimal. Leaf size=82 \[ \frac {\cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b c}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b c}+\frac {\log \left (a+b \sin ^{-1}(c x)\right )}{2 b c} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4661, 3312, 3303, 3299, 3302} \[ \frac {\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b c}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b c}+\frac {\log \left (a+b \sin ^{-1}(c x)\right )}{2 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(2*a)/b]*CosIntegral[(2*a)/b + 2*ArcSin[c*x]])/(2*b*c) + Log[a + b*ArcSin[c*x]]/(2*b*c) + (Sin[(2*a)/b]*S
inIntegral[(2*a)/b + 2*ArcSin[c*x]])/(2*b*c)

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]

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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 62, normalized size = 0.76 \[ \frac {\cos \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\log \left (a+b \sin ^{-1}(c x)\right )}{2 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.50, size = 102, normalized size = 1.24 \[ \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c} + \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c} - \frac {\operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{2 \, b c} + \frac {\log \left (b \arcsin \left (c x\right ) + a\right )}{2 \, b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(a/b)^2*cos_integral(2*a/b + 2*arcsin(c*x))/(b*c) + cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(c*x))/(
b*c) - 1/2*cos_integral(2*a/b + 2*arcsin(c*x))/(b*c) + 1/2*log(b*arcsin(c*x) + a)/(b*c)

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maple [A]  time = 0.08, size = 77, normalized size = 0.94 \[ \frac {\Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{2 c b}+\frac {\Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{2 c b}+\frac {\ln \left (a +b \arcsin \left (c x \right )\right )}{2 b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-c^{2} x^{2} + 1}}{b \arcsin \left (c x\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {1-c^2\,x^2}}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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