\(\int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))^2} \, dx\) [412]

Optimal result

Integrand size = 28, antiderivative size = 142 \[ \int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))^2} \, dx=-\frac {x^3}{b c (a+b \arccos (c x))}+\frac {3 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b^2 c^4}-\frac {3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^4}+\frac {3 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b^2 c^4}-\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b^2 c^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.79 \[ \int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))^2} \, dx=\frac {x^3}{b c (a+b \arccos (c x))}+\frac {3 \left (-\operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right ) \sin \left (\frac {a}{b}\right )-\operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )\right )}{4 b^2 c^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5223, 5147, 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.

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

\(\Big \downarrow \) 5223

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

\(\Big \downarrow \) 5147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 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 5223

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {3 \arccos \left (c x \right ) \operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b -3 \arccos \left (c x \right ) \operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b +3 \arccos \left (c x \right ) \operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -3 \arccos \left (c x \right ) \operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +3 \,\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a -3 \,\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a +3 \,\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -3 \,\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +3 c x b +\cos \left (3 \arccos \left (c x \right )\right ) b}{4 c^{4} \left (a +b \arccos \left (c x \right )\right ) b^{2}}\)	\(226\)

Input:

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

Output:

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

Fricas [F]

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

integrate(x^3/(-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^3/(a^2*c^2*x^2 + (b^2*c^2*x^2 - b^2)*arccos 
(c*x)^2 - a^2 + 2*(a*b*c^2*x^2 - a*b)*arccos(c*x)), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

(x^3 - 3*(b^2*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b*c)*integr 
ate(x^2/(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 [F(-2)]

Exception generated. \[ \int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \arccos (c x))^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**3/(sqrt( - c**2*x**2 + 1)*acos(c*x)**2*b**2 + 2*sqrt( - c**2*x**2 + 
 1)*acos(c*x)*a*b + sqrt( - c**2*x**2 + 1)*a**2),x)