\(\int \frac {x^2 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx\) [384]

Optimal result

Integrand size = 28, antiderivative size = 94 \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx=-\frac {x^2 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}-\frac {\operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c^3}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )}{2 b^2 c^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx=\frac {-\frac {2 b c^2 x^2 \left (-1+c^2 x^2\right )}{a+b \arccos (c x)}+\operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{2 b^2 c^3} \] Input:

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

Output:

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(193\) vs. \(2(94)=188\).

Time = 1.07 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5215, 5147, 25, 4906, 27, 2009, 3042, 3784, 25, 3042, 3780, 3783}

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.

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

\(\Big \downarrow \) 5215

\(\displaystyle \frac {4 c \int \frac {x^3}{a+b \arccos (c x)}dx}{b}-\frac {2 \int \frac {x}{a+b \arccos (c x)}dx}{b c}+\frac {x^2 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 5147

\(\displaystyle -\frac {4 \int -\frac {\cos ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}+\frac {2 \int -\frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}+\frac {x^2 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4 \int \frac {\cos ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}-\frac {2 \int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}+\frac {x^2 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arccos (c x))}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))-\sin \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arccos (c x))}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}-\frac {4 \left (-\frac {1}{4} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right )+\frac {1}{4} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )+\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )\right )}{b^2 c^3}+\frac {x^2 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 3780

\(\displaystyle \frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\sin \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arccos (c x))}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b^2 c^3}-\frac {4 \left (-\frac {1}{4} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right )+\frac {1}{4} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )+\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )\right )}{b^2 c^3}+\frac {x^2 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 3783

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

Input:

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

Output:

(x^2*(1 - c^2*x^2))/(b*c*(a + b*ArcCos[c*x])) + (-(CosIntegral[(2*(a + b*A 
rcCos[c*x]))/b]*Sin[(2*a)/b]) + Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcCos[ 
c*x]))/b])/(b^2*c^3) - (4*(-1/4*(CosIntegral[(2*(a + b*ArcCos[c*x]))/b]*Si 
n[(2*a)/b]) - (CosIntegral[(4*(a + b*ArcCos[c*x]))/b]*Sin[(4*a)/b])/8 + (C 
os[(2*a)/b]*SinIntegral[(2*(a + b*ArcCos[c*x]))/b])/4 + (Cos[(4*a)/b]*SinI 
ntegral[(4*(a + b*ArcCos[c*x]))/b])/8))/(b^2*c^3)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3780

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

rule 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3784

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* 
e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[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 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 5147

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

rule 5215

Int[((a_.) + ArcCos[(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*ArcCos[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*ArcCos[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*ArcCos[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]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.45

method	result	size
default	\(-\frac {4 \arccos \left (c x \right ) \operatorname {Si}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -4 \arccos \left (c x \right ) \operatorname {Ci}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b +4 \,\operatorname {Si}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -4 \,\operatorname {Ci}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a +\cos \left (4 \arccos \left (c x \right )\right ) b -b}{8 c^{3} \left (a +b \arccos \left (c x \right )\right ) b^{2}}\)	\(136\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx=\int \frac {x^{2} \sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (88) = 176\).

Time = 0.23 (sec) , antiderivative size = 556, normalized size of antiderivative = 5.91 \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-b*c^4*x^4/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + 4*b*arccos(c*x)*cos(a/b)^3* 
cos_integral(4*a/b + 4*arccos(c*x))*sin(a/b)/(b^3*c^3*arccos(c*x) + a*b^2* 
c^3) - 4*b*arccos(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arccos(c*x))/(b^3 
*c^3*arccos(c*x) + a*b^2*c^3) + 4*a*cos(a/b)^3*cos_integral(4*a/b + 4*arcc 
os(c*x))*sin(a/b)/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 4*a*cos(a/b)^4*sin_i 
ntegral(4*a/b + 4*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + b*c^2*x 
^2/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 2*b*arccos(c*x)*cos(a/b)*cos_integr 
al(4*a/b + 4*arccos(c*x))*sin(a/b)/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + 4*b 
*arccos(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arccos(c*x))/(b^3*c^3*arcco 
s(c*x) + a*b^2*c^3) - 2*a*cos(a/b)*cos_integral(4*a/b + 4*arccos(c*x))*sin 
(a/b)/(b^3*c^3*arccos(c*x) + a*b^2*c^3) + 4*a*cos(a/b)^2*sin_integral(4*a/ 
b + 4*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 1/2*b*arccos(c*x)*s 
in_integral(4*a/b + 4*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3) - 1/2 
*a*sin_integral(4*a/b + 4*arccos(c*x))/(b^3*c^3*arccos(c*x) + a*b^2*c^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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