3.4.93 \(\int \frac {(1-c^2 x^2)^{3/2}}{(a+b \text {ArcSin}(c x))^2} \, dx\) [393]

Optimal. Leaf size=150 \[ -\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \text {ArcSin}(c x))}+\frac {\text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{b^2 c}+\frac {\text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{b^2 c}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c} \]

[Out]

-(-c^2*x^2+1)^2/b/c/(a+b*arcsin(c*x))-cos(2*a/b)*Si(2*(a+b*arcsin(c*x))/b)/b^2/c-1/2*cos(4*a/b)*Si(4*(a+b*arcs
in(c*x))/b)/b^2/c+Ci(2*(a+b*arcsin(c*x))/b)*sin(2*a/b)/b^2/c+1/2*Ci(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b^2/c

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Rubi [A]
time = 0.18, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4751, 4809, 4491, 3384, 3380, 3383} \begin {gather*} \frac {\sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{b^2 c}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{b^2 c}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b^2 c}-\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \text {ArcSin}(c x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - c^2*x^2)^2/(b*c*(a + b*ArcSin[c*x]))) + (CosIntegral[(2*(a + b*ArcSin[c*x]))/b]*Sin[(2*a)/b])/(b^2*c) +
(CosIntegral[(4*(a + b*ArcSin[c*x]))/b]*Sin[(4*a)/b])/(2*b^2*c) - (Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[
c*x]))/b])/(b^2*c) - (Cos[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/(2*b^2*c)

Rule 3380

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 3383

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 3384

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 4491

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 4751

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)/(b*(n + 1)))*Simp[(d + e*x^2)
^p/(1 - c^2*x^2)^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 4809

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

Rubi steps

\begin {align*} \int \frac {\left (1-c^2 x^2\right )^{3/2}}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {(4 c) \int \frac {x \left (1-c^2 x^2\right )}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {4 \text {Subst}\left (\int \frac {\cos ^3(x) \sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {4 \text {Subst}\left (\int \left (\frac {\sin (2 x)}{4 (a+b x)}+\frac {\sin (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\text {Subst}\left (\int \frac {\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}-\frac {\text {Subst}\left (\int \frac {\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {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 {\cos \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {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 {\sin \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}\\ &=-\frac {\left (1-c^2 x^2\right )^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 {\text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right ) \sin \left (\frac {4 a}{b}\right )}{2 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 {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c}\\ \end {align*}

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Mathematica [A]
time = 0.41, size = 122, normalized size = 0.81 \begin {gather*} \frac {-\frac {2 b \left (-1+c^2 x^2\right )^2}{a+b \text {ArcSin}(c x)}+2 \text {CosIntegral}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+\text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )-2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{2 b^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*(-1 + c^2*x^2)^2)/(a + b*ArcSin[c*x]) + 2*CosIntegral[2*(a/b + ArcSin[c*x])]*Sin[(2*a)/b] + CosIntegral
[4*(a/b + ArcSin[c*x])]*Sin[(4*a)/b] - 2*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] - Cos[(4*a)/b]*SinInt
egral[4*(a/b + ArcSin[c*x])])/(2*b^2*c)

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Maple [A]
time = 0.10, size = 250, normalized size = 1.67

method	result	size
default	\(-\frac {4 \arcsin \left (c x \right ) \sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (c x \right ) \cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b +8 \arcsin \left (c x \right ) \sinIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b -8 \arcsin \left (c x \right ) \cosineIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b +4 \sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -4 \cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a +8 \sinIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -8 \cosineIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +\cos \left (4 \arcsin \left (c x \right )\right ) b +4 \cos \left (2 \arcsin \left (c x \right )\right ) b +3 b}{8 c \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(250\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/c*(4*arcsin(c*x)*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*b-4*arcsin(c*x)*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*b+
8*arcsin(c*x)*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*b-8*arcsin(c*x)*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*b+4*Si(4*a
rcsin(c*x)+4*a/b)*cos(4*a/b)*a-4*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*a+8*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*a-8
*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*a+cos(4*arcsin(c*x))*b+4*cos(2*arcsin(c*x))*b+3*b)/(a+b*arcsin(c*x))/b^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^4*x^4 - 2*c^2*x^2 - (b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integrate(4*(c^3*x^3 - c*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 747 vs. \(2 (145) = 290\).
time = 0.53, size = 747, normalized size = 4.98 \begin {gather*} \frac {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \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 {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{4} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} + \frac {4 \, a \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \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 {4 \, a \cos \left (\frac {a}{b}\right )^{4} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\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 ) \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \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 ) \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 {4 \, b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\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 {4 \, a}{b} + 4 \, \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 ) \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 {4 \, a \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \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 )^{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 {{\left (c^{2} x^{2} - 1\right )}^{2} b}{b^{3} c \arcsin \left (c x\right ) + a b^{2} c} - \frac {b \arcsin \left (c x\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, {\left (b^{3} c \arcsin \left (c x\right ) + a b^{2} c\right )}} + \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 {a \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, {\left (b^{3} c \arcsin \left (c x\right ) + a b^{2} c\right )}} + \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*b*arcsin(c*x)*cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 4*b*ar
csin(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 4*a*cos(a/b)^3*cos_in
tegral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 4*a*cos(a/b)^4*sin_integral(4*a/b + 4*a
rcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 2*b*arcsin(c*x)*cos(a/b)*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a
/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 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) + 4*b*arcsin(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x)
+ a*b^2*c) - 2*b*arcsin(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_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 2*a*cos(a/b)*cos_integ
ral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 4*a*cos(a/b)^2*sin_integral(4*a/b + 4*arcs
in(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 2*a*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsin(c*x)
+ a*b^2*c) - (c^2*x^2 - 1)^2*b/(b^3*c*arcsin(c*x) + a*b^2*c) - 1/2*b*arcsin(c*x)*sin_integral(4*a/b + 4*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) - 1/2*a*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + a*sin_integral(2*a/b + 2
*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (1-c^2\,x^2\right )}^{3/2}}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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