\(\int \frac {x^3 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx\) [383]

Optimal result

Integrand size = 28, antiderivative size = 214 \[ \int \frac {x^3 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx=-\frac {x^3 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}+\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{16 b^2 c^4}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arccos (c x))}{b}\right )}{16 b^2 c^4}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{16 b^2 c^4}-\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arccos (c x))}{b}\right )}{16 b^2 c^4} \] Output:

-x^3*(-c^2*x^2+1)/b/c/(a+b*arccos(c*x))+1/8*cos(a/b)*Ci((a+b*arccos(c*x))/ 
b)/b^2/c^4+3/16*cos(3*a/b)*Ci(3*(a+b*arccos(c*x))/b)/b^2/c^4-5/16*cos(5*a/ 
b)*Ci(5*(a+b*arccos(c*x))/b)/b^2/c^4+1/8*sin(a/b)*Si((a+b*arccos(c*x))/b)/ 
b^2/c^4+3/16*sin(3*a/b)*Si(3*(a+b*arccos(c*x))/b)/b^2/c^4-5/16*sin(5*a/b)* 
Si(5*(a+b*arccos(c*x))/b)/b^2/c^4
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx=\frac {\frac {16 b c^3 x^3}{a+b \arccos (c x)}-\frac {16 b c^5 x^5}{a+b \arccos (c x)}-2 \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right ) \sin \left (\frac {a}{b}\right )+3 \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+5 \operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {5 a}{b}\right )+2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )-3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )-5 \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{16 b^2 c^4} \] Input:

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

Output:

((16*b*c^3*x^3)/(a + b*ArcCos[c*x]) - (16*b*c^5*x^5)/(a + b*ArcCos[c*x]) - 
 2*CosIntegral[a/b + ArcCos[c*x]]*Sin[a/b] + 3*CosIntegral[3*(a/b + ArcCos 
[c*x])]*Sin[(3*a)/b] + 5*CosIntegral[5*(a/b + ArcCos[c*x])]*Sin[(5*a)/b] + 
 2*Cos[a/b]*SinIntegral[a/b + ArcCos[c*x]] - 3*Cos[(3*a)/b]*SinIntegral[3* 
(a/b + ArcCos[c*x])] - 5*Cos[(5*a)/b]*SinIntegral[5*(a/b + ArcCos[c*x])])/ 
(16*b^2*c^4)
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.36, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5215, 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.

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

\(\Big \downarrow \) 5215

\(\displaystyle \frac {5 c \int \frac {x^4}{a+b \arccos (c x)}dx}{b}-\frac {3 \int \frac {x^2}{a+b \arccos (c x)}dx}{b c}+\frac {x^3 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 5147

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4906

\(\displaystyle \frac {5 \int \left (\frac {\sin \left (\frac {5 a}{b}-\frac {5 (a+b \arccos (c x))}{b}\right )}{16 (a+b \arccos (c x))}+\frac {3 \sin \left (\frac {3 a}{b}-\frac {3 (a+b \arccos (c x))}{b}\right )}{16 (a+b \arccos (c x))}+\frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{8 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b^2 c^4}-\frac {3 \int \left (\frac {\sin \left (\frac {3 a}{b}-\frac {3 (a+b \arccos (c x))}{b}\right )}{4 (a+b \arccos (c x))}+\frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{4 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b^2 c^4}+\frac {x^3 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \left (-\frac {1}{4} \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )-\frac {1}{4} \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )+\frac {1}{4} \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )+\frac {1}{4} \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )\right )}{b^2 c^4}-\frac {5 \left (-\frac {1}{8} \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )-\frac {3}{16} \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )-\frac {1}{16} \sin \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arccos (c x))}{b}\right )+\frac {1}{8} \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )+\frac {3}{16} \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )+\frac {1}{16} \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arccos (c x))}{b}\right )\right )}{b^2 c^4}+\frac {x^3 \left (1-c^2 x^2\right )}{b c (a+b \arccos (c x))}\)

Input:

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

Output:

(x^3*(1 - c^2*x^2))/(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) - (5*(-1/8*(Cos 
Integral[(a + b*ArcCos[c*x])/b]*Sin[a/b]) - (3*CosIntegral[(3*(a + b*ArcCo 
s[c*x]))/b]*Sin[(3*a)/b])/16 - (CosIntegral[(5*(a + b*ArcCos[c*x]))/b]*Sin 
[(5*a)/b])/16 + (Cos[a/b]*SinIntegral[(a + b*ArcCos[c*x])/b])/8 + (3*Cos[( 
3*a)/b]*SinIntegral[(3*(a + b*ArcCos[c*x]))/b])/16 + (Cos[(5*a)/b]*SinInte 
gral[(5*(a + b*ArcCos[c*x]))/b])/16))/(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 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 = 0.18 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {2 \arccos \left (c x \right ) \operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -2 \arccos \left (c x \right ) \operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b -5 \arccos \left (c x \right ) \operatorname {Si}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) b +5 \arccos \left (c x \right ) \operatorname {Ci}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) b -3 \arccos \left (c x \right ) \operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b +3 \arccos \left (c x \right ) \operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b +2 \,\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -2 \,\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a -5 \,\operatorname {Si}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a +5 \,\operatorname {Ci}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a -3 \,\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a +3 \,\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a +2 c x b -\cos \left (5 \arccos \left (c x \right )\right ) b -\cos \left (3 \arccos \left (c x \right )\right ) b}{16 c^{4} \left (a +b \arccos \left (c x \right )\right ) b^{2}}\)	\(341\)

Input:

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

Output:

1/16/c^4*(2*arccos(c*x)*Si(arccos(c*x)+a/b)*cos(a/b)*b-2*arccos(c*x)*Ci(ar 
ccos(c*x)+a/b)*sin(a/b)*b-5*arccos(c*x)*Si(5*arccos(c*x)+5*a/b)*cos(5*a/b) 
*b+5*arccos(c*x)*Ci(5*arccos(c*x)+5*a/b)*sin(5*a/b)*b-3*arccos(c*x)*Si(3*a 
rccos(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+2*Si(arccos(c*x)+a/b)*cos(a/b)*a-2*Ci(arccos(c*x)+a/b)*sin(a/b)*a- 
5*Si(5*arccos(c*x)+5*a/b)*cos(5*a/b)*a+5*Ci(5*arccos(c*x)+5*a/b)*sin(5*a/b 
)*a-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+2*c*x*b-cos(5*arccos(c*x))*b-cos(3*arccos(c*x))*b)/(a+b*arccos(c*x 
))/b^2
 

Fricas [F]

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

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

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

Output:

-(c^2*x^5 - x^3 - (b^2*c*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) + a*b* 
c)*integrate((5*c^2*x^4 - 3*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. \[ \int \frac {x^3 \sqrt {1-c^2 x^2}}{(a+b \arccos (c x))^2} \, dx=\int \frac {x^3\,\sqrt {1-c^2\,x^2}}{{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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