\(\int \frac {x^3 (1-c^2 x^2)^{3/2}}{a+b \arccos (c x)} \, dx\) [327]

Optimal result

Integrand size = 28, antiderivative size = 245 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{3/2}}{a+b \arccos (c x)} \, dx=-\frac {3 \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{64 b c^4}-\frac {3 \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{64 b c^4}+\frac {\operatorname {CosIntegral}\left (\frac {5 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{64 b c^4}+\frac {\operatorname {CosIntegral}\left (\frac {7 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {7 a}{b}\right )}{64 b c^4}+\frac {3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{64 b c^4}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{64 b c^4}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arccos (c x))}{b}\right )}{64 b c^4}-\frac {\cos \left (\frac {7 a}{b}\right ) \text {Si}\left (\frac {7 (a+b \arccos (c x))}{b}\right )}{64 b c^4} \] Output:

-3/64*Ci((a+b*arccos(c*x))/b)*sin(a/b)/b/c^4-3/64*Ci(3*(a+b*arccos(c*x))/b 
)*sin(3*a/b)/b/c^4+1/64*Ci(5*(a+b*arccos(c*x))/b)*sin(5*a/b)/b/c^4+1/64*Ci 
(7*(a+b*arccos(c*x))/b)*sin(7*a/b)/b/c^4+3/64*cos(a/b)*Si((a+b*arccos(c*x) 
)/b)/b/c^4+3/64*cos(3*a/b)*Si(3*(a+b*arccos(c*x))/b)/b/c^4-1/64*cos(5*a/b) 
*Si(5*(a+b*arccos(c*x))/b)/b/c^4-1/64*cos(7*a/b)*Si(7*(a+b*arccos(c*x))/b) 
/b/c^4
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{3/2}}{a+b \arccos (c x)} \, dx=\frac {-3 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right )+3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )+\cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arccos (c x)\right )\right )-\cos \left (\frac {7 a}{b}\right ) \operatorname {CosIntegral}\left (7 \left (\frac {a}{b}+\arccos (c x)\right )\right )-3 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )+3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )+\sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arccos (c x)\right )\right )-\sin \left (\frac {7 a}{b}\right ) \text {Si}\left (7 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{64 b c^4} \] Input:

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

Output:

(-3*Cos[a/b]*CosIntegral[a/b + ArcCos[c*x]] + 3*Cos[(3*a)/b]*CosIntegral[3 
*(a/b + ArcCos[c*x])] + Cos[(5*a)/b]*CosIntegral[5*(a/b + ArcCos[c*x])] - 
Cos[(7*a)/b]*CosIntegral[7*(a/b + ArcCos[c*x])] - 3*Sin[a/b]*SinIntegral[a 
/b + ArcCos[c*x]] + 3*Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcCos[c*x])] + Si 
n[(5*a)/b]*SinIntegral[5*(a/b + ArcCos[c*x])] - Sin[(7*a)/b]*SinIntegral[7 
*(a/b + ArcCos[c*x])])/(64*b*c^4)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {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.

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

\(\Big \downarrow \) 5225

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

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {\int \left (\frac {\cos \left (\frac {7 a}{b}-\frac {7 (a+b \arccos (c x))}{b}\right )}{64 (a+b \arccos (c x))}-\frac {\cos \left (\frac {5 a}{b}-\frac {5 (a+b \arccos (c x))}{b}\right )}{64 (a+b \arccos (c x))}-\frac {3 \cos \left (\frac {3 a}{b}-\frac {3 (a+b \arccos (c x))}{b}\right )}{64 (a+b \arccos (c x))}+\frac {3 \cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{64 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b c^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {3}{64} \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )-\frac {3}{64} \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )-\frac {1}{64} \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arccos (c x))}{b}\right )+\frac {1}{64} \cos \left (\frac {7 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {7 (a+b \arccos (c x))}{b}\right )+\frac {3}{64} \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )-\frac {3}{64} \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )-\frac {1}{64} \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arccos (c x))}{b}\right )+\frac {1}{64} \sin \left (\frac {7 a}{b}\right ) \text {Si}\left (\frac {7 (a+b \arccos (c x))}{b}\right )}{b c^4}\)

Input:

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

Output:

-(((3*Cos[a/b]*CosIntegral[(a + b*ArcCos[c*x])/b])/64 - (3*Cos[(3*a)/b]*Co 
sIntegral[(3*(a + b*ArcCos[c*x]))/b])/64 - (Cos[(5*a)/b]*CosIntegral[(5*(a 
 + b*ArcCos[c*x]))/b])/64 + (Cos[(7*a)/b]*CosIntegral[(7*(a + b*ArcCos[c*x 
]))/b])/64 + (3*Sin[a/b]*SinIntegral[(a + b*ArcCos[c*x])/b])/64 - (3*Sin[( 
3*a)/b]*SinIntegral[(3*(a + b*ArcCos[c*x]))/b])/64 - (Sin[(5*a)/b]*SinInte 
gral[(5*(a + b*ArcCos[c*x]))/b])/64 + (Sin[(7*a)/b]*SinIntegral[(7*(a + b* 
ArcCos[c*x]))/b])/64)/(b*c^4))
 

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 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 = 0.17 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.75

method	result	size
default	\(-\frac {3 \,\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+3 \,\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\operatorname {Si}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )-\operatorname {Ci}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )-3 \,\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )-3 \,\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )+\operatorname {Si}\left (7 \arccos \left (c x \right )+\frac {7 a}{b}\right ) \sin \left (\frac {7 a}{b}\right )+\operatorname {Ci}\left (7 \arccos \left (c x \right )+\frac {7 a}{b}\right ) \cos \left (\frac {7 a}{b}\right )}{64 c^{4} b}\)	\(184\)

Input:

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

Output:

-1/64/c^4*(3*Si(arccos(c*x)+a/b)*sin(a/b)+3*Ci(arccos(c*x)+a/b)*cos(a/b)-S 
i(5*arccos(c*x)+5*a/b)*sin(5*a/b)-Ci(5*arccos(c*x)+5*a/b)*cos(5*a/b)-3*Si( 
3*arccos(c*x)+3*a/b)*sin(3*a/b)-3*Ci(3*arccos(c*x)+3*a/b)*cos(3*a/b)+Si(7* 
arccos(c*x)+7*a/b)*sin(7*a/b)+Ci(7*arccos(c*x)+7*a/b)*cos(7*a/b))/b
 

Fricas [F]

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

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

Output:

integral(-(c^2*x^5 - x^3)*sqrt(-c^2*x^2 + 1)/(b*arccos(c*x) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 615 vs. \(2 (229) = 458\).

Time = 0.16 (sec) , antiderivative size = 615, normalized size of antiderivative = 2.51 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{3/2}}{a+b \arccos (c x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-cos(a/b)^7*cos_integral(7*a/b + 7*arccos(c*x))/(b*c^4) - cos(a/b)^6*sin(a 
/b)*sin_integral(7*a/b + 7*arccos(c*x))/(b*c^4) + 7/4*cos(a/b)^5*cos_integ 
ral(7*a/b + 7*arccos(c*x))/(b*c^4) + 1/4*cos(a/b)^5*cos_integral(5*a/b + 5 
*arccos(c*x))/(b*c^4) + 5/4*cos(a/b)^4*sin(a/b)*sin_integral(7*a/b + 7*arc 
cos(c*x))/(b*c^4) + 1/4*cos(a/b)^4*sin(a/b)*sin_integral(5*a/b + 5*arccos( 
c*x))/(b*c^4) - 7/8*cos(a/b)^3*cos_integral(7*a/b + 7*arccos(c*x))/(b*c^4) 
 - 5/16*cos(a/b)^3*cos_integral(5*a/b + 5*arccos(c*x))/(b*c^4) + 3/16*cos( 
a/b)^3*cos_integral(3*a/b + 3*arccos(c*x))/(b*c^4) - 3/8*cos(a/b)^2*sin(a/ 
b)*sin_integral(7*a/b + 7*arccos(c*x))/(b*c^4) - 3/16*cos(a/b)^2*sin(a/b)* 
sin_integral(5*a/b + 5*arccos(c*x))/(b*c^4) + 3/16*cos(a/b)^2*sin(a/b)*sin 
_integral(3*a/b + 3*arccos(c*x))/(b*c^4) + 7/64*cos(a/b)*cos_integral(7*a/ 
b + 7*arccos(c*x))/(b*c^4) + 5/64*cos(a/b)*cos_integral(5*a/b + 5*arccos(c 
*x))/(b*c^4) - 9/64*cos(a/b)*cos_integral(3*a/b + 3*arccos(c*x))/(b*c^4) - 
 3/64*cos(a/b)*cos_integral(a/b + arccos(c*x))/(b*c^4) + 1/64*sin(a/b)*sin 
_integral(7*a/b + 7*arccos(c*x))/(b*c^4) + 1/64*sin(a/b)*sin_integral(5*a/ 
b + 5*arccos(c*x))/(b*c^4) - 3/64*sin(a/b)*sin_integral(3*a/b + 3*arccos(c 
*x))/(b*c^4) - 3/64*sin(a/b)*sin_integral(a/b + arccos(c*x))/(b*c^4)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 \left (1-c^2 x^2\right )^{3/2}}{a+b \arccos (c x)} \, dx=-\left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{5}}{\mathit {acos} \left (c x \right ) b +a}d x \right ) c^{2}+\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{3}}{\mathit {acos} \left (c x \right ) b +a}d x \] Input:

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

Output:

 - int((sqrt( - c**2*x**2 + 1)*x**5)/(acos(c*x)*b + a),x)*c**2 + int((sqrt 
( - c**2*x**2 + 1)*x**3)/(acos(c*x)*b + a),x)