\(\int \frac {(d-c^2 d x^2)^2}{a+b \arccos (c x)} \, dx\) [28]

Optimal result

Integrand size = 24, antiderivative size = 201 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{a+b \arccos (c x)} \, dx=\frac {5 d^2 \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{8 b c}-\frac {5 d^2 \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16 b c}+\frac {d^2 \operatorname {CosIntegral}\left (\frac {5 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{16 b c}-\frac {5 d^2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{8 b c}+\frac {5 d^2 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{16 b c}-\frac {d^2 \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arccos (c x))}{b}\right )}{16 b c} \] Output:

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

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.69 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{a+b \arccos (c x)} \, dx=\frac {d^2 \left (10 \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right ) \sin \left (\frac {a}{b}\right )-5 \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+\operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {5 a}{b}\right )-10 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )+5 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )-\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arccos (c x)\right )\right )\right )}{16 b c} \] Input:

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

Output:

(d^2*(10*CosIntegral[a/b + ArcCos[c*x]]*Sin[a/b] - 5*CosIntegral[3*(a/b + 
ArcCos[c*x])]*Sin[(3*a)/b] + CosIntegral[5*(a/b + ArcCos[c*x])]*Sin[(5*a)/ 
b] - 10*Cos[a/b]*SinIntegral[a/b + ArcCos[c*x]] + 5*Cos[(3*a)/b]*SinIntegr 
al[3*(a/b + ArcCos[c*x])] - Cos[(5*a)/b]*SinIntegral[5*(a/b + ArcCos[c*x]) 
]))/(16*b*c)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5169, 25, 3042, 3793, 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.

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

\(\Big \downarrow \) 5169

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((d^2*((-5*CosIntegral[(a + b*ArcCos[c*x])/b]*Sin[a/b])/8 + (5*CosIntegra 
l[(3*(a + b*ArcCos[c*x]))/b]*Sin[(3*a)/b])/16 - (CosIntegral[(5*(a + b*Arc 
Cos[c*x]))/b]*Sin[(5*a)/b])/16 + (5*Cos[a/b]*SinIntegral[(a + b*ArcCos[c*x 
])/b])/8 - (5*Cos[(3*a)/b]*SinIntegral[(3*(a + b*ArcCos[c*x]))/b])/16 + (C 
os[(5*a)/b]*SinIntegral[(5*(a + b*ArcCos[c*x]))/b])/16))/(b*c))
 

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 5169

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[(-(b*c)^(-1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Subst[ 
Int[x^n*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, 0]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(-\frac {d^{2} \left (\operatorname {Si}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )-\operatorname {Ci}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )-5 \,\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )+5 \,\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )+10 \,\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-10 \,\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )\right )}{16 c b}\)	\(142\)
default	\(-\frac {d^{2} \left (\operatorname {Si}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )-\operatorname {Ci}\left (5 \arccos \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )-5 \,\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )+5 \,\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )+10 \,\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-10 \,\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )\right )}{16 c b}\)	\(142\)

Input:

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

Output:

-1/16/c*d^2*(Si(5*arccos(c*x)+5*a/b)*cos(5*a/b)-Ci(5*arccos(c*x)+5*a/b)*si 
n(5*a/b)-5*Si(3*arccos(c*x)+3*a/b)*cos(3*a/b)+5*Ci(3*arccos(c*x)+3*a/b)*si 
n(3*a/b)+10*Si(arccos(c*x)+a/b)*cos(a/b)-10*Ci(arccos(c*x)+a/b)*sin(a/b))/ 
b
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

d**2*(Integral(-2*c**2*x**2/(a + b*acos(c*x)), x) + Integral(c**4*x**4/(a 
+ b*acos(c*x)), x) + Integral(1/(a + b*acos(c*x)), x))
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (189) = 378\).

Time = 0.16 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.97 \[ \int \frac {\left (d-c^2 d x^2\right )^2}{a+b \arccos (c x)} \, dx=\frac {d^{2} \cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {5 \, a}{b} + 5 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c} - \frac {d^{2} \cos \left (\frac {a}{b}\right )^{5} \operatorname {Si}\left (\frac {5 \, a}{b} + 5 \, \arccos \left (c x\right )\right )}{b c} - \frac {3 \, d^{2} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {5 \, a}{b} + 5 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c} - \frac {5 \, d^{2} \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c} + \frac {5 \, d^{2} \cos \left (\frac {a}{b}\right )^{3} \operatorname {Si}\left (\frac {5 \, a}{b} + 5 \, \arccos \left (c x\right )\right )}{4 \, b c} + \frac {5 \, d^{2} \cos \left (\frac {a}{b}\right )^{3} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, b c} + \frac {d^{2} \operatorname {Ci}\left (\frac {5 \, a}{b} + 5 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{16 \, b c} + \frac {5 \, d^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{16 \, b c} + \frac {5 \, d^{2} \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{8 \, b c} - \frac {5 \, d^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {5 \, a}{b} + 5 \, \arccos \left (c x\right )\right )}{16 \, b c} - \frac {15 \, d^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{16 \, b c} - \frac {5 \, d^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{8 \, b c} \] Input:

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

Output:

d^2*cos(a/b)^4*cos_integral(5*a/b + 5*arccos(c*x))*sin(a/b)/(b*c) - d^2*co 
s(a/b)^5*sin_integral(5*a/b + 5*arccos(c*x))/(b*c) - 3/4*d^2*cos(a/b)^2*co 
s_integral(5*a/b + 5*arccos(c*x))*sin(a/b)/(b*c) - 5/4*d^2*cos(a/b)^2*cos_ 
integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b*c) + 5/4*d^2*cos(a/b)^3*sin_in 
tegral(5*a/b + 5*arccos(c*x))/(b*c) + 5/4*d^2*cos(a/b)^3*sin_integral(3*a/ 
b + 3*arccos(c*x))/(b*c) + 1/16*d^2*cos_integral(5*a/b + 5*arccos(c*x))*si 
n(a/b)/(b*c) + 5/16*d^2*cos_integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b*c) 
 + 5/8*d^2*cos_integral(a/b + arccos(c*x))*sin(a/b)/(b*c) - 5/16*d^2*cos(a 
/b)*sin_integral(5*a/b + 5*arccos(c*x))/(b*c) - 15/16*d^2*cos(a/b)*sin_int 
egral(3*a/b + 3*arccos(c*x))/(b*c) - 5/8*d^2*cos(a/b)*sin_integral(a/b + a 
rccos(c*x))/(b*c)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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