\(\int \frac {x^2 (1-c^2 x^2)^{3/2}}{(a+b \arcsin (c x))^2} \, dx\) [359]

Optimal result

Integrand size = 28, antiderivative size = 220 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=-\frac {x^2 \left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}+\frac {\operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{16 b^2 c^3}-\frac {\operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{4 b^2 c^3}-\frac {3 \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {6 a}{b}\right )}{16 b^2 c^3}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{4 b^2 c^3}+\frac {3 \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^3} \] Output:

-x^2*(-c^2*x^2+1)^2/b/c/(a+b*arcsin(c*x))+1/16*Ci(2*(a+b*arcsin(c*x))/b)*s 
in(2*a/b)/b^2/c^3-1/4*Ci(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b^2/c^3-3/16*Ci 
(6*(a+b*arcsin(c*x))/b)*sin(6*a/b)/b^2/c^3-1/16*cos(2*a/b)*Si(2*(a+b*arcsi 
n(c*x))/b)/b^2/c^3+1/4*cos(4*a/b)*Si(4*(a+b*arcsin(c*x))/b)/b^2/c^3+3/16*c 
os(6*a/b)*Si(6*(a+b*arcsin(c*x))/b)/b^2/c^3
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.39 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=-\frac {16 b c^2 x^2-32 b c^4 x^4+16 b c^6 x^6-(a+b \arcsin (c x)) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+4 (a+b \arcsin (c x)) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )+3 a \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {6 a}{b}\right )+3 b \arcsin (c x) \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {6 a}{b}\right )+a \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+b \arcsin (c x) \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-4 a \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-4 b \arcsin (c x) \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-3 a \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-3 b \arcsin (c x) \cos \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{16 b^2 c^3 (a+b \arcsin (c x))} \] Input:

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

Output:

-1/16*(16*b*c^2*x^2 - 32*b*c^4*x^4 + 16*b*c^6*x^6 - (a + b*ArcSin[c*x])*Co 
sIntegral[2*(a/b + ArcSin[c*x])]*Sin[(2*a)/b] + 4*(a + b*ArcSin[c*x])*CosI 
ntegral[4*(a/b + ArcSin[c*x])]*Sin[(4*a)/b] + 3*a*CosIntegral[6*(a/b + Arc 
Sin[c*x])]*Sin[(6*a)/b] + 3*b*ArcSin[c*x]*CosIntegral[6*(a/b + ArcSin[c*x] 
)]*Sin[(6*a)/b] + a*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + b*Ar 
cSin[c*x]*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] - 4*a*Cos[(4*a)/ 
b]*SinIntegral[4*(a/b + ArcSin[c*x])] - 4*b*ArcSin[c*x]*Cos[(4*a)/b]*SinIn 
tegral[4*(a/b + ArcSin[c*x])] - 3*a*Cos[(6*a)/b]*SinIntegral[6*(a/b + ArcS 
in[c*x])] - 3*b*ArcSin[c*x]*Cos[(6*a)/b]*SinIntegral[6*(a/b + ArcSin[c*x]) 
])/(b^2*c^3*(a + b*ArcSin[c*x]))
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5214, 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.

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

\(\Big \downarrow \) 5214

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

\(\Big \downarrow \) 5224

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 5214

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_. 
)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^m*Sqrt[1 - c^2*x^2]*(d + e*x^2)^p* 
((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Simp[f*(m/(b*c*(n + 1)) 
)*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p 
- 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x] + Simp[c*((m + 2*p + 1)/(b*f*(n 
+ 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f*x)^(m + 1)*(1 - c^2*x^2 
)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f 
}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1 
, 0] && IGtQ[m, -3]
 

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 = 0.40 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.65

method	result	size
default	\(-\frac {2 \arcsin \left (c x \right ) \operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b -2 \arcsin \left (c x \right ) \operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b -6 \arcsin \left (c x \right ) \operatorname {Si}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) b +6 \arcsin \left (c x \right ) \operatorname {Ci}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) b -8 \arcsin \left (c x \right ) \operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b +8 \arcsin \left (c x \right ) \sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) b +2 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -2 \,\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a -6 \,\operatorname {Si}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right ) a +6 \,\operatorname {Ci}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right ) a -8 \,\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a +8 \sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) a +\cos \left (2 \arcsin \left (c x \right )\right ) b -\cos \left (6 \arcsin \left (c x \right )\right ) b -2 \cos \left (4 \arcsin \left (c x \right )\right ) b +2 b}{32 c^{3} \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(364\)

Input:

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

Output:

-1/32/c^3*(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-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-8*arcsi 
n(c*x)*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*b+8*arcsin(c*x)*sin(4*a/b)*Ci(4* 
arcsin(c*x)+4*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-6*Si(6*arcsin(c*x)+6*a/b)*cos(6*a/b)*a+6*Ci(6*arc 
sin(c*x)+6*a/b)*sin(6*a/b)*a-8*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*a+8*sin( 
4*a/b)*Ci(4*arcsin(c*x)+4*a/b)*a+cos(2*arcsin(c*x))*b-cos(6*arcsin(c*x))*b 
-2*cos(4*arcsin(c*x))*b+2*b)/(a+b*arcsin(c*x))/b^2
 

Fricas [F]

\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {x^{2} \left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

-(c^4*x^6 - 2*c^2*x^4 + x^2 - (b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x 
+ 1)) + a*b*c)*integrate(2*(3*c^4*x^5 - 4*c^2*x^3 + x)/(b^2*c*arctan2(c*x, 
 sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c), x))/(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. 1553 vs. \(2 (207) = 414\).

Time = 0.24 (sec) , antiderivative size = 1553, normalized size of antiderivative = 7.06 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\text {Too large to display} \] Input:

integrate(x^2*(-c^2*x^2+1)^(3/2)/(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^3*arcsin(c*x) + a*b^2*c^3) + 6*b*arcsin(c*x)*cos(a/b)^6*sin_integral 
(6*a/b + 6*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 6*a*cos(a/b)^5 
*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2 
*c^3) + 6*a*cos(a/b)^6*sin_integral(6*a/b + 6*arcsin(c*x))/(b^3*c^3*arcsin 
(c*x) + a*b^2*c^3) + 6*b*arcsin(c*x)*cos(a/b)^3*cos_integral(6*a/b + 6*arc 
sin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 2*b*arcsin(c*x)*cos 
(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) 
+ a*b^2*c^3) - 9*b*arcsin(c*x)*cos(a/b)^4*sin_integral(6*a/b + 6*arcsin(c* 
x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 2*b*arcsin(c*x)*cos(a/b)^4*sin_int 
egral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 6*a*cos(a 
/b)^3*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + 
a*b^2*c^3) - 2*a*cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/( 
b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 9*a*cos(a/b)^4*sin_integral(6*a/b + 6*a 
rcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 2*a*cos(a/b)^4*sin_integra 
l(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 9/8*b*arcsin( 
c*x)*cos(a/b)*cos_integral(6*a/b + 6*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin 
(c*x) + a*b^2*c^3) + b*arcsin(c*x)*cos(a/b)*cos_integral(4*a/b + 4*arcsin( 
c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/8*b*arcsin(c*x)*cos(a 
/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) +...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {x^2\,{\left (1-c^2\,x^2\right )}^{3/2}}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=-\left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{4}}{\mathit {asin} \left (c x \right )^{2} b^{2}+2 \mathit {asin} \left (c x \right ) a b +a^{2}}d x \right ) c^{2}+\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{2}}{\mathit {asin} \left (c x \right )^{2} b^{2}+2 \mathit {asin} \left (c x \right ) a b +a^{2}}d x \] Input:

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

Output:

 - 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**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)