\(\int \frac {(1-c^2 x^2)^{5/2}}{(a+b \arcsin (c x))^2} \, dx\) [368]

Optimal result

Integrand size = 25, antiderivative size = 217 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx=-\frac {\left (1-c^2 x^2\right )^3}{b c (a+b \arcsin (c x))}+\frac {15 \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{16 b^2 c}+\frac {3 \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{4 b^2 c}+\frac {3 \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {6 a}{b}\right )}{16 b^2 c}-\frac {15 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c}-\frac {3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{4 b^2 c}-\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c} \] Output:

-(-c^2*x^2+1)^3/b/c/(a+b*arcsin(c*x))+15/16*Ci(2*(a+b*arcsin(c*x))/b)*sin( 
2*a/b)/b^2/c+3/4*Ci(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b^2/c+3/16*Ci(6*(a+b 
*arcsin(c*x))/b)*sin(6*a/b)/b^2/c-15/16*cos(2*a/b)*Si(2*(a+b*arcsin(c*x))/ 
b)/b^2/c-3/4*cos(4*a/b)*Si(4*(a+b*arcsin(c*x))/b)/b^2/c-3/16*cos(6*a/b)*Si 
(6*(a+b*arcsin(c*x))/b)/b^2/c
 

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.43 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx=-\frac {16 b-48 b c^2 x^2+48 b c^4 x^4-16 b c^6 x^6-15 (a+b \arcsin (c x)) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )-12 (a+b \arcsin (c x)) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )-3 a \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {6 a}{b}\right )-3 b \arcsin (c x) \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {6 a}{b}\right )+15 a \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+15 b \arcsin (c x) \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+12 a \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+12 b \arcsin (c x) \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+3 a \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+3 b \arcsin (c x) \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{16 b^2 c (a+b \arcsin (c x))} \] Input:

Integrate[(1 - c^2*x^2)^(5/2)/(a + b*ArcSin[c*x])^2,x]
 

Output:

-1/16*(16*b - 48*b*c^2*x^2 + 48*b*c^4*x^4 - 16*b*c^6*x^6 - 15*(a + b*ArcSi 
n[c*x])*CosIntegral[2*(a/b + ArcSin[c*x])]*Sin[(2*a)/b] - 12*(a + b*ArcSin 
[c*x])*CosIntegral[4*(a/b + ArcSin[c*x])]*Sin[(4*a)/b] - 3*a*CosIntegral[6 
*(a/b + ArcSin[c*x])]*Sin[(6*a)/b] - 3*b*ArcSin[c*x]*CosIntegral[6*(a/b + 
ArcSin[c*x])]*Sin[(6*a)/b] + 15*a*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin 
[c*x])] + 15*b*ArcSin[c*x]*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] 
 + 12*a*Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] + 12*b*ArcSin[c*x] 
*Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] + 3*a*Cos[(6*a)/b]*SinInt 
egral[6*(a/b + ArcSin[c*x])] + 3*b*ArcSin[c*x]*Cos[(6*a)/b]*SinIntegral[6* 
(a/b + ArcSin[c*x])])/(b^2*c*(a + b*ArcSin[c*x]))
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5166, 5224, 25, 4906, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx\)

\(\Big \downarrow \) 5166

\(\displaystyle -\frac {6 c \int \frac {x \left (1-c^2 x^2\right )^2}{a+b \arcsin (c x)}dx}{b}-\frac {\left (1-c^2 x^2\right )^3}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 5224

\(\displaystyle -\frac {6 \int -\frac {\cos ^5\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b^2 c}-\frac {\left (1-c^2 x^2\right )^3}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {6 \int \frac {\cos ^5\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b^2 c}-\frac {\left (1-c^2 x^2\right )^3}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {6 \int \left (\frac {\sin \left (\frac {6 a}{b}-\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}+\frac {\sin \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 (a+b \arcsin (c x))}+\frac {5 \sin \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b^2 c}-\frac {\left (1-c^2 x^2\right )^3}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {6 \left (-\frac {5}{32} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )-\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} \sin \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )+\frac {5}{32} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )+\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )+\frac {1}{32} \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )\right )}{b^2 c}-\frac {\left (1-c^2 x^2\right )^3}{b c (a+b \arcsin (c x))}\)

Input:

Int[(1 - c^2*x^2)^(5/2)/(a + b*ArcSin[c*x])^2,x]
 

Output:

-((1 - c^2*x^2)^3/(b*c*(a + b*ArcSin[c*x]))) - (6*((-5*CosIntegral[(2*(a + 
 b*ArcSin[c*x]))/b]*Sin[(2*a)/b])/32 - (CosIntegral[(4*(a + b*ArcSin[c*x]) 
)/b]*Sin[(4*a)/b])/8 - (CosIntegral[(6*(a + b*ArcSin[c*x]))/b]*Sin[(6*a)/b 
])/32 + (5*Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/32 + (Cos[ 
(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/8 + (Cos[(6*a)/b]*SinInte 
gral[(6*(a + b*ArcSin[c*x]))/b])/32))/(b^2*c)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG 
tQ[p, 0]
 

rule 5166

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] + Simp[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 5224

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[(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]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.68

method	result	size
default	\(-\frac {6 \arcsin \left (c x \right ) \operatorname {Si}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) b -6 \arcsin \left (c x \right ) \operatorname {Ci}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) b +24 \arcsin \left (c x \right ) \operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -24 \arcsin \left (c x \right ) \sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) b +30 \arcsin \left (c x \right ) \operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b -30 \arcsin \left (c x \right ) \operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b +6 \,\operatorname {Si}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) a -6 \,\operatorname {Ci}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) a +24 \,\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -24 \sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) a +30 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -30 \,\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +\cos \left (6 \arcsin \left (c x \right )\right ) b +6 \cos \left (4 \arcsin \left (c x \right )\right ) b +15 \cos \left (2 \arcsin \left (c x \right )\right ) b +10 b}{32 c \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(364\)

Input:

int((-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x))^2,x,method=_RETURNVERBOSE)
 

Output:

-1/32/c*(6*arcsin(c*x)*Si(6*arcsin(c*x)+6*a/b)*cos(6*a/b)*b-6*arcsin(c*x)* 
Ci(6*arcsin(c*x)+6*a/b)*sin(6*a/b)*b+24*arcsin(c*x)*Si(4*arcsin(c*x)+4*a/b 
)*cos(4*a/b)*b-24*arcsin(c*x)*sin(4*a/b)*Ci(4*arcsin(c*x)+4*a/b)*b+30*arcs 
in(c*x)*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*b-30*arcsin(c*x)*Ci(2*arcsin(c* 
x)+2*a/b)*sin(2*a/b)*b+6*Si(6*arcsin(c*x)+6*a/b)*cos(6*a/b)*a-6*Ci(6*arcsi 
n(c*x)+6*a/b)*sin(6*a/b)*a+24*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*a-24*sin( 
4*a/b)*Ci(4*arcsin(c*x)+4*a/b)*a+30*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*a-3 
0*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*a+cos(6*arcsin(c*x))*b+6*cos(4*arcsin 
(c*x))*b+15*cos(2*arcsin(c*x))*b+10*b)/(a+b*arcsin(c*x))/b^2
 

Fricas [F]

\[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x))^2,x, algorithm="fricas")
 

Output:

integral((c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(-c^2*x^2 + 1)/(b^2*arcsin(c*x)^2 + 
 2*a*b*arcsin(c*x) + a^2), x)
 

Sympy [F]

\[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \] Input:

integrate((-c**2*x**2+1)**(5/2)/(a+b*asin(c*x))**2,x)
 

Output:

Integral((-(c*x - 1)*(c*x + 1))**(5/2)/(a + b*asin(c*x))**2, x)
 

Maxima [F]

\[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x))^2,x, algorithm="maxima")
 

Output:

(c^6*x^6 - 3*c^4*x^4 + 3*c^2*x^2 - (b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt( 
-c*x + 1)) + a*b*c)*integrate(6*(c^5*x^5 - 2*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)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1394 vs. \(2 (203) = 406\).

Time = 0.23 (sec) , antiderivative size = 1394, normalized size of antiderivative = 6.42 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx=\text {Too large to display} \] Input:

integrate((-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x))^2,x, algorithm="giac")
 

Output:

6*b*arcsin(c*x)*cos(a/b)^5*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b 
^3*c*arcsin(c*x) + a*b^2*c) - 6*b*arcsin(c*x)*cos(a/b)^6*sin_integral(6*a/ 
b + 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 6*a*cos(a/b)^5*cos_inte 
gral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 6*a*c 
os(a/b)^6*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c 
) - 6*b*arcsin(c*x)*cos(a/b)^3*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b 
)/(b^3*c*arcsin(c*x) + a*b^2*c) + 6*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) + 9*b*arcsin 
(c*x)*cos(a/b)^4*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + 
a*b^2*c) - 6*b*arcsin(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) - 6*a*cos(a/b)^3*cos_integral(6*a/b + 6*arcs 
in(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 6*a*cos(a/b)^3*cos_integ 
ral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 9*a*co 
s(a/b)^4*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) 
 - 6*a*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + 
 a*b^2*c) + 9/8*b*arcsin(c*x)*cos(a/b)*cos_integral(6*a/b + 6*arcsin(c*x)) 
*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 3*b*arcsin(c*x)*cos(a/b)*cos_int 
egral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 15/8 
*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) - 27/8*b*arcsin(c*x)*cos(a/b)^2*sin_integral(6...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \] Input:

int((1 - c^2*x^2)^(5/2)/(a + b*asin(c*x))^2,x)
 

Output:

int((1 - c^2*x^2)^(5/2)/(a + b*asin(c*x))^2, x)
 

Reduce [F]

\[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {\sqrt {-c^{2} x^{2}+1}}{\mathit {asin} \left (c x \right )^{2} b^{2}+2 \mathit {asin} \left (c x \right ) a b +a^{2}}d x +\left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{4}}{\mathit {asin} \left (c x \right )^{2} b^{2}+2 \mathit {asin} \left (c x \right ) a b +a^{2}}d x \right ) c^{4}-2 \left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{2}}{\mathit {asin} \left (c x \right )^{2} b^{2}+2 \mathit {asin} \left (c x \right ) a b +a^{2}}d x \right ) c^{2} \] Input:

int((-c^2*x^2+1)^(5/2)/(a+b*asin(c*x))^2,x)
 

Output:

int(sqrt( - c**2*x**2 + 1)/(asin(c*x)**2*b**2 + 2*asin(c*x)*a*b + a**2),x) 
 + int((sqrt( - c**2*x**2 + 1)*x**4)/(asin(c*x)**2*b**2 + 2*asin(c*x)*a*b 
+ a**2),x)*c**4 - 2*int((sqrt( - c**2*x**2 + 1)*x**2)/(asin(c*x)**2*b**2 + 
 2*asin(c*x)*a*b + a**2),x)*c**2