\(\int \frac {(c-a^2 c x^2)^2}{\arccos (a x)^2} \, dx\) [23]

Optimal result

Integrand size = 20, antiderivative size = 77 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)^2} \, dx=\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \arccos (a x)}-\frac {5 c^2 \operatorname {CosIntegral}(\arccos (a x))}{8 a}+\frac {15 c^2 \operatorname {CosIntegral}(3 \arccos (a x))}{16 a}-\frac {5 c^2 \operatorname {CosIntegral}(5 \arccos (a x))}{16 a} \] Output:

c^2*(-a^2*x^2+1)^(5/2)/a/arccos(a*x)-5/8*c^2*Ci(arccos(a*x))/a+15/16*c^2*C 
i(3*arccos(a*x))/a-5/16*c^2*Ci(5*arccos(a*x))/a
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)^2} \, dx=\frac {c^2 \left (16 \left (1-a^2 x^2\right )^{5/2}-10 \arccos (a x) \operatorname {CosIntegral}(\arccos (a x))+15 \arccos (a x) \operatorname {CosIntegral}(3 \arccos (a x))-5 \arccos (a x) \operatorname {CosIntegral}(5 \arccos (a x))\right )}{16 a \arccos (a x)} \] Input:

Integrate[(c - a^2*c*x^2)^2/ArcCos[a*x]^2,x]
 

Output:

(c^2*(16*(1 - a^2*x^2)^(5/2) - 10*ArcCos[a*x]*CosIntegral[ArcCos[a*x]] + 1 
5*ArcCos[a*x]*CosIntegral[3*ArcCos[a*x]] - 5*ArcCos[a*x]*CosIntegral[5*Arc 
Cos[a*x]]))/(16*a*ArcCos[a*x])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5167, 5225, 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.

\(\displaystyle \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)^2} \, dx\)

\(\Big \downarrow \) 5167

\(\displaystyle 5 a c^2 \int \frac {x \left (1-a^2 x^2\right )^{3/2}}{\arccos (a x)}dx+\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \arccos (a x)}\)

\(\Big \downarrow \) 5225

\(\displaystyle \frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \arccos (a x)}-\frac {5 c^2 \int \frac {a x \left (1-a^2 x^2\right )^2}{\arccos (a x)}d\arccos (a x)}{a}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \arccos (a x)}-\frac {5 c^2 \int \left (\frac {a x}{8 \arccos (a x)}-\frac {3 \cos (3 \arccos (a x))}{16 \arccos (a x)}+\frac {\cos (5 \arccos (a x))}{16 \arccos (a x)}\right )d\arccos (a x)}{a}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \arccos (a x)}-\frac {5 c^2 \left (\frac {1}{8} \operatorname {CosIntegral}(\arccos (a x))-\frac {3}{16} \operatorname {CosIntegral}(3 \arccos (a x))+\frac {1}{16} \operatorname {CosIntegral}(5 \arccos (a x))\right )}{a}\)

Input:

Int[(c - a^2*c*x^2)^2/ArcCos[a*x]^2,x]
 

Output:

(c^2*(1 - a^2*x^2)^(5/2))/(a*ArcCos[a*x]) - (5*c^2*(CosIntegral[ArcCos[a*x 
]]/8 - (3*CosIntegral[3*ArcCos[a*x]])/16 + CosIntegral[5*ArcCos[a*x]]/16)) 
/a
 

Defintions of rubi rules used

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 5167

Int[((a_.) + ArcCos[(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*ArcCos[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*ArcCos[c*x])^(n + 
1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, - 
1]
 

rule 5225

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[(-(b*c^(m + 1))^(-1))*Simp[(d + e*x^2)^p/(1 - c 
^2*x^2)^p]   Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b]^(2*p + 1), x], 
 x, a + b*ArcCos[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 = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {c^{2} \left (15 \,\operatorname {Ci}\left (3 \arccos \left (a x \right )\right ) \arccos \left (a x \right )-5 \,\operatorname {Ci}\left (5 \arccos \left (a x \right )\right ) \arccos \left (a x \right )-10 \,\operatorname {Ci}\left (\arccos \left (a x \right )\right ) \arccos \left (a x \right )+10 \sqrt {-a^{2} x^{2}+1}-5 \sin \left (3 \arccos \left (a x \right )\right )+\sin \left (5 \arccos \left (a x \right )\right )\right )}{16 a \arccos \left (a x \right )}\)	\(83\)
default	\(\frac {c^{2} \left (15 \,\operatorname {Ci}\left (3 \arccos \left (a x \right )\right ) \arccos \left (a x \right )-5 \,\operatorname {Ci}\left (5 \arccos \left (a x \right )\right ) \arccos \left (a x \right )-10 \,\operatorname {Ci}\left (\arccos \left (a x \right )\right ) \arccos \left (a x \right )+10 \sqrt {-a^{2} x^{2}+1}-5 \sin \left (3 \arccos \left (a x \right )\right )+\sin \left (5 \arccos \left (a x \right )\right )\right )}{16 a \arccos \left (a x \right )}\)	\(83\)

Input:

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

Output:

1/16/a*c^2*(15*Ci(3*arccos(a*x))*arccos(a*x)-5*Ci(5*arccos(a*x))*arccos(a* 
x)-10*Ci(arccos(a*x))*arccos(a*x)+10*(-a^2*x^2+1)^(1/2)-5*sin(3*arccos(a*x 
))+sin(5*arccos(a*x)))/arccos(a*x)
 

Fricas [F]

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

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

Output:

integral((a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)/arccos(a*x)^2, x)
 

Sympy [F]

\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)^2} \, dx=c^{2} \left (\int \left (- \frac {2 a^{2} x^{2}}{\operatorname {acos}^{2}{\left (a x \right )}}\right )\, dx + \int \frac {a^{4} x^{4}}{\operatorname {acos}^{2}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {acos}^{2}{\left (a x \right )}}\, dx\right ) \] Input:

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

Output:

c**2*(Integral(-2*a**2*x**2/acos(a*x)**2, x) + Integral(a**4*x**4/acos(a*x 
)**2, x) + Integral(acos(a*x)**(-2), x))
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.61 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)^2} \, dx=\frac {\sqrt {-a^{2} x^{2} + 1} a^{3} c^{2} x^{4}}{\arccos \left (a x\right )} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a c^{2} x^{2}}{\arccos \left (a x\right )} - \frac {5 \, c^{2} \operatorname {Ci}\left (5 \, \arccos \left (a x\right )\right )}{16 \, a} + \frac {15 \, c^{2} \operatorname {Ci}\left (3 \, \arccos \left (a x\right )\right )}{16 \, a} - \frac {5 \, c^{2} \operatorname {Ci}\left (\arccos \left (a x\right )\right )}{8 \, a} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{2}}{a \arccos \left (a x\right )} \] Input:

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

Output:

sqrt(-a^2*x^2 + 1)*a^3*c^2*x^4/arccos(a*x) - 2*sqrt(-a^2*x^2 + 1)*a*c^2*x^ 
2/arccos(a*x) - 5/16*c^2*cos_integral(5*arccos(a*x))/a + 15/16*c^2*cos_int 
egral(3*arccos(a*x))/a - 5/8*c^2*cos_integral(arccos(a*x))/a + sqrt(-a^2*x 
^2 + 1)*c^2/(a*arccos(a*x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)^2} \, dx=c^{2} \left (\left (\int \frac {x^{4}}{\mathit {acos} \left (a x \right )^{2}}d x \right ) a^{4}-2 \left (\int \frac {x^{2}}{\mathit {acos} \left (a x \right )^{2}}d x \right ) a^{2}+\int \frac {1}{\mathit {acos} \left (a x \right )^{2}}d x \right ) \] Input:

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

Output:

c**2*(int(x**4/acos(a*x)**2,x)*a**4 - 2*int(x**2/acos(a*x)**2,x)*a**2 + in 
t(1/acos(a*x)**2,x))