\(\int \frac {x (1-c^2 x^2)^{5/2}}{a+b \arccos (c x)} \, dx\) [337]

Optimal result

Integrand size = 26, antiderivative size = 245 \[ \int \frac {x \left (1-c^2 x^2\right )^{5/2}}{a+b \arccos (c x)} \, dx=-\frac {5 \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{64 b c^2}-\frac {9 \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{64 b c^2}-\frac {5 \operatorname {CosIntegral}\left (\frac {5 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{64 b c^2}-\frac {\operatorname {CosIntegral}\left (\frac {7 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {7 a}{b}\right )}{64 b c^2}+\frac {5 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{64 b c^2}+\frac {9 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{64 b c^2}+\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arccos (c x))}{b}\right )}{64 b c^2}+\frac {\cos \left (\frac {7 a}{b}\right ) \text {Si}\left (\frac {7 (a+b \arccos (c x))}{b}\right )}{64 b c^2} \] Output:

-5/64*Ci((a+b*arccos(c*x))/b)*sin(a/b)/b/c^2-9/64*Ci(3*(a+b*arccos(c*x))/b 
)*sin(3*a/b)/b/c^2-5/64*Ci(5*(a+b*arccos(c*x))/b)*sin(5*a/b)/b/c^2-1/64*Ci 
(7*(a+b*arccos(c*x))/b)*sin(7*a/b)/b/c^2+5/64*cos(a/b)*Si((a+b*arccos(c*x) 
)/b)/b/c^2+9/64*cos(3*a/b)*Si(3*(a+b*arccos(c*x))/b)/b/c^2+5/64*cos(5*a/b) 
*Si(5*(a+b*arccos(c*x))/b)/b/c^2+1/64*cos(7*a/b)*Si(7*(a+b*arccos(c*x))/b) 
/b/c^2
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.73 \[ \int \frac {x \left (1-c^2 x^2\right )^{5/2}}{a+b \arccos (c x)} \, dx=\frac {-5 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right )+9 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )-5 \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arccos (c x)\right )\right )+\cos \left (\frac {7 a}{b}\right ) \operatorname {CosIntegral}\left (7 \left (\frac {a}{b}+\arccos (c x)\right )\right )-5 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )+9 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )-5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arccos (c x)\right )\right )+\sin \left (\frac {7 a}{b}\right ) \text {Si}\left (7 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{64 b c^2} \] Input:

Integrate[(x*(1 - c^2*x^2)^(5/2))/(a + b*ArcCos[c*x]),x]
 

Output:

(-5*Cos[a/b]*CosIntegral[a/b + ArcCos[c*x]] + 9*Cos[(3*a)/b]*CosIntegral[3 
*(a/b + ArcCos[c*x])] - 5*Cos[(5*a)/b]*CosIntegral[5*(a/b + ArcCos[c*x])] 
+ Cos[(7*a)/b]*CosIntegral[7*(a/b + ArcCos[c*x])] - 5*Sin[a/b]*SinIntegral 
[a/b + ArcCos[c*x]] + 9*Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcCos[c*x])] - 
5*Sin[(5*a)/b]*SinIntegral[5*(a/b + ArcCos[c*x])] + Sin[(7*a)/b]*SinIntegr 
al[7*(a/b + ArcCos[c*x])])/(64*b*c^2)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, 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.

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

\(\Big \downarrow \) 5225

\(\displaystyle -\frac {\int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^6\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c^2}\)

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {\int \left (-\frac {\cos \left (\frac {7 a}{b}-\frac {7 (a+b \arccos (c x))}{b}\right )}{64 (a+b \arccos (c x))}+\frac {5 \cos \left (\frac {5 a}{b}-\frac {5 (a+b \arccos (c x))}{b}\right )}{64 (a+b \arccos (c x))}-\frac {9 \cos \left (\frac {3 a}{b}-\frac {3 (a+b \arccos (c x))}{b}\right )}{64 (a+b \arccos (c x))}+\frac {5 \cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{64 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b c^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {5}{64} \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )-\frac {9}{64} \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )+\frac {5}{64} \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arccos (c x))}{b}\right )-\frac {1}{64} \cos \left (\frac {7 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {7 (a+b \arccos (c x))}{b}\right )+\frac {5}{64} \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )-\frac {9}{64} \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )+\frac {5}{64} \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arccos (c x))}{b}\right )-\frac {1}{64} \sin \left (\frac {7 a}{b}\right ) \text {Si}\left (\frac {7 (a+b \arccos (c x))}{b}\right )}{b c^2}\)

Input:

Int[(x*(1 - c^2*x^2)^(5/2))/(a + b*ArcCos[c*x]),x]
 

Output:

-(((5*Cos[a/b]*CosIntegral[(a + b*ArcCos[c*x])/b])/64 - (9*Cos[(3*a)/b]*Co 
sIntegral[(3*(a + b*ArcCos[c*x]))/b])/64 + (5*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 + (5*Sin[a/b]*SinIntegral[(a + b*ArcCos[c*x])/b])/64 - (9*Sin 
[(3*a)/b]*SinIntegral[(3*(a + b*ArcCos[c*x]))/b])/64 + (5*Sin[(5*a)/b]*Sin 
Integral[(5*(a + b*ArcCos[c*x]))/b])/64 - (Sin[(7*a)/b]*SinIntegral[(7*(a 
+ b*ArcCos[c*x]))/b])/64)/(b*c^2))
 

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 = 0.14 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {5 \,\operatorname {Si}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )+5 \,\operatorname {Ci}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )-\operatorname {Si}\left (7 \arccos \left (c x \right )+\frac {7 a}{b}\right ) \sin \left (\frac {7 a}{b}\right )-\operatorname {Ci}\left (7 \arccos \left (c x \right )+\frac {7 a}{b}\right ) \cos \left (\frac {7 a}{b}\right )-9 \,\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )-9 \,\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )+5 \,\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+5 \,\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{64 c^{2} b}\)	\(186\)

Input:

int(x*(-c^2*x^2+1)^(5/2)/(a+b*arccos(c*x)),x,method=_RETURNVERBOSE)
 

Output:

-1/64/c^2*(5*Si(5*arccos(c*x)+5*a/b)*sin(5*a/b)+5*Ci(5*arccos(c*x)+5*a/b)* 
cos(5*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)-9*Si(3*arccos(c*x)+3*a/b)*sin(3*a/b)-9*Ci(3*arccos(c*x)+3*a/b)*cos( 
3*a/b)+5*Si(arccos(c*x)+a/b)*sin(a/b)+5*Ci(arccos(c*x)+a/b)*cos(a/b))/b
 

Fricas [F]

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

integrate(x*(-c^2*x^2+1)^(5/2)/(a+b*arccos(c*x)),x, algorithm="fricas")
 

Output:

integral((c^4*x^5 - 2*c^2*x^3 + x)*sqrt(-c^2*x^2 + 1)/(b*arccos(c*x) + a), 
 x)
 

Sympy [F]

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

integrate(x*(-c**2*x**2+1)**(5/2)/(a+b*acos(c*x)),x)
 

Output:

Integral(x*(-(c*x - 1)*(c*x + 1))**(5/2)/(a + b*acos(c*x)), x)
 

Maxima [F]

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

integrate(x*(-c^2*x^2+1)^(5/2)/(a+b*arccos(c*x)),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (229) = 458\).

Time = 0.17 (sec) , antiderivative size = 613, normalized size of antiderivative = 2.50 \[ \int \frac {x \left (1-c^2 x^2\right )^{5/2}}{a+b \arccos (c x)} \, dx =\text {Too large to display} \] Input:

integrate(x*(-c^2*x^2+1)^(5/2)/(a+b*arccos(c*x)),x, algorithm="giac")
 

Output:

cos(a/b)^7*cos_integral(7*a/b + 7*arccos(c*x))/(b*c^2) + cos(a/b)^6*sin(a/ 
b)*sin_integral(7*a/b + 7*arccos(c*x))/(b*c^2) - 7/4*cos(a/b)^5*cos_integr 
al(7*a/b + 7*arccos(c*x))/(b*c^2) - 5/4*cos(a/b)^5*cos_integral(5*a/b + 5* 
arccos(c*x))/(b*c^2) - 5/4*cos(a/b)^4*sin(a/b)*sin_integral(7*a/b + 7*arcc 
os(c*x))/(b*c^2) - 5/4*cos(a/b)^4*sin(a/b)*sin_integral(5*a/b + 5*arccos(c 
*x))/(b*c^2) + 7/8*cos(a/b)^3*cos_integral(7*a/b + 7*arccos(c*x))/(b*c^2) 
+ 25/16*cos(a/b)^3*cos_integral(5*a/b + 5*arccos(c*x))/(b*c^2) + 9/16*cos( 
a/b)^3*cos_integral(3*a/b + 3*arccos(c*x))/(b*c^2) + 3/8*cos(a/b)^2*sin(a/ 
b)*sin_integral(7*a/b + 7*arccos(c*x))/(b*c^2) + 15/16*cos(a/b)^2*sin(a/b) 
*sin_integral(5*a/b + 5*arccos(c*x))/(b*c^2) + 9/16*cos(a/b)^2*sin(a/b)*si 
n_integral(3*a/b + 3*arccos(c*x))/(b*c^2) - 7/64*cos(a/b)*cos_integral(7*a 
/b + 7*arccos(c*x))/(b*c^2) - 25/64*cos(a/b)*cos_integral(5*a/b + 5*arccos 
(c*x))/(b*c^2) - 27/64*cos(a/b)*cos_integral(3*a/b + 3*arccos(c*x))/(b*c^2 
) - 5/64*cos(a/b)*cos_integral(a/b + arccos(c*x))/(b*c^2) - 1/64*sin(a/b)* 
sin_integral(7*a/b + 7*arccos(c*x))/(b*c^2) - 5/64*sin(a/b)*sin_integral(5 
*a/b + 5*arccos(c*x))/(b*c^2) - 9/64*sin(a/b)*sin_integral(3*a/b + 3*arcco 
s(c*x))/(b*c^2) - 5/64*sin(a/b)*sin_integral(a/b + arccos(c*x))/(b*c^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt( - c**2*x**2 + 1)*x**5)/(acos(c*x)*b + a),x)*c**4 - 2*int((sqrt( 
 - c**2*x**2 + 1)*x**3)/(acos(c*x)*b + a),x)*c**2 + int((sqrt( - c**2*x**2 
 + 1)*x)/(acos(c*x)*b + a),x)