\(\int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [471]

Optimal result

Integrand size = 21, antiderivative size = 179 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {x}{b^3}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^3 (a+b)^{5/2} d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

x/b^3-a*(2*a^4-5*a^2*b^2+6*b^4)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b 
)^(1/2))/(a-b)^(5/2)/b^3/(a+b)^(5/2)/d-1/2*a^2*cos(d*x+c)*sin(d*x+c)/b/(a^ 
2-b^2)/d/(a+b*cos(d*x+c))^2-1/2*a^2*(2*a^2-5*b^2)*sin(d*x+c)/b^2/(a^2-b^2) 
^2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 1.56 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {2 (c+d x)+\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}-\frac {a^2 b \left (2 a^3-5 a b^2+3 b \left (a^2-2 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}}{2 b^3 d} \] Input:

Integrate[Cos[c + d*x]^3/(a + b*Cos[c + d*x])^3,x]
 

Output:

(2*(c + d*x) + (2*a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTanh[((a - b)*Tan[(c + 
d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) - (a^2*b*(2*a^3 - 5*a*b^2 + 
 3*b*(a^2 - 2*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b 
*Cos[c + d*x])^2))/(2*b^3*d)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3271, 3042, 3500, 3042, 3214, 3042, 3138, 218}

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.

\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

\(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {2 x \left (a^2-b^2\right )^2}{b}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{b}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {2 x \left (a^2-b^2\right )^2}{b}-\frac {a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {2 x \left (a^2-b^2\right )^2}{b}-\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {a^2 \left (2 a^2-5 b^2\right ) \sin (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {2 x \left (a^2-b^2\right )^2}{b}-\frac {2 a \left (2 a^4-5 a^2 b^2+6 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}\)

Input:

Int[Cos[c + d*x]^3/(a + b*Cos[c + d*x])^3,x]
 

Output:

-1/2*(a^2*Cos[c + d*x]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]) 
^2) - (-(((2*(a^2 - b^2)^2*x)/b - (2*a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTan[ 
(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d) 
)/(b*(a^2 - b^2))) + (a^2*(2*a^2 - 5*b^2)*Sin[c + d*x])/(b*(a^2 - b^2)*d*( 
a + b*Cos[c + d*x])))/(2*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* 
(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin 
[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x] 
, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
IntegersQ[2*m, 2*n])
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {-\frac {2 a \left (\frac {\frac {\left (2 a^{2}-a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (2 a^{2}+a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (2 a^{4}-5 a^{2} b^{2}+6 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\)	\(233\)
default	\(\frac {-\frac {2 a \left (\frac {\frac {\left (2 a^{2}-a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (2 a^{2}+a b -6 b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (2 a^{4}-5 a^{2} b^{2}+6 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\)	\(233\)
risch	\(\frac {x}{b^{3}}-\frac {i a^{2} \left (4 a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-7 a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+8 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-17 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}+3 a^{2} b^{2}-6 b^{4}\right )}{b^{3} \left (a^{2}-b^{2}\right )^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{3}}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d b}-\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{3}}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d b}+\frac {3 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(683\)

Input:

int(cos(d*x+c)^3/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2*a/b^3*((1/2*(2*a^2-a*b-6*b^2)*a*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d* 
x+1/2*c)^3+1/2*(2*a^2+a*b-6*b^2)*a*b/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2 
*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2+1/2*(2*a^4-5*a^ 
2*b^2+6*b^4)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2* 
d*x+1/2*c)/((a-b)*(a+b))^(1/2)))+2/b^3*arctan(tan(1/2*d*x+1/2*c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (166) = 332\).

Time = 0.13 (sec) , antiderivative size = 913, normalized size of antiderivative = 5.10 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
 

Output:

[1/4*(4*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*x*cos(d*x + c)^2 + 8*(a^ 
7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*x*cos(d*x + c) + 4*(a^8 - 3*a^6*b^2 
 + 3*a^4*b^4 - a^2*b^6)*d*x - (2*a^7 - 5*a^5*b^2 + 6*a^3*b^4 + (2*a^5*b^2 
- 5*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5 
)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*c 
os(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 
 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 2*(2*a^7*b - 
7*a^5*b^3 + 5*a^3*b^5 + 3*(a^6*b^2 - 3*a^4*b^4 + 2*a^2*b^6)*cos(d*x + c))* 
sin(d*x + c))/((a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d*cos(d*x + c)^2 + 
 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 
3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d), 1/2*(2*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b 
^6 - b^8)*d*x*cos(d*x + c)^2 + 4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d 
*x*cos(d*x + c) + 2*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x - (2*a^7 - 
 5*a^5*b^2 + 6*a^3*b^4 + (2*a^5*b^2 - 5*a^3*b^4 + 6*a*b^6)*cos(d*x + c)^2 
+ 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan 
(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (2*a^7*b - 7*a^5* 
b^3 + 5*a^3*b^5 + 3*(a^6*b^2 - 3*a^4*b^4 + 2*a^2*b^6)*cos(d*x + c))*sin(d* 
x + c))/((a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d*cos(d*x + c)^2 + 2*(a^ 
7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 3*a^6* 
b^5 + 3*a^4*b^7 - a^2*b^9)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**3/(a+b*cos(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.78 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, a^{5} - 5 \, a^{3} b^{2} + 6 \, a b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}} + \frac {d x + c}{b^{3}}}{d} \] Input:

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^3,x, algorithm="giac")
 

Output:

((2*a^5 - 5*a^3*b^2 + 6*a*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a 
+ 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^ 
2 - b^2)))/((a^4*b^3 - 2*a^2*b^5 + b^7)*sqrt(a^2 - b^2)) - (2*a^5*tan(1/2* 
d*x + 1/2*c)^3 - 3*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 5*a^3*b^2*tan(1/2*d*x + 
1/2*c)^3 + 6*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*a^5*tan(1/2*d*x + 1/2*c) + 
 3*a^4*b*tan(1/2*d*x + 1/2*c) - 5*a^3*b^2*tan(1/2*d*x + 1/2*c) - 6*a^2*b^3 
*tan(1/2*d*x + 1/2*c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*(a*tan(1/2*d*x + 1/2*c 
)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) + (d*x + c)/b^3)/d
 

Mupad [B] (verification not implemented)

Time = 50.31 (sec) , antiderivative size = 5102, normalized size of antiderivative = 28.50 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int(cos(c + d*x)^3/(a + b*cos(c + d*x))^3,x)
 

Output:

(2*atan((((((8*(12*a*b^14 - 4*b^15 + 8*a^2*b^13 - 34*a^3*b^12 - 6*a^4*b^11 
 + 36*a^5*b^10 + 4*a^6*b^9 - 18*a^7*b^8 - 2*a^8*b^7 + 4*a^9*b^6))/(a*b^12 
+ b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b 
^6) - (tan(c/2 + (d*x)/2)*(8*a*b^15 - 8*a^2*b^14 - 32*a^3*b^13 + 32*a^4*b^ 
12 + 48*a^5*b^11 - 48*a^6*b^10 - 32*a^7*b^9 + 32*a^8*b^8 + 8*a^9*b^7 - 8*a 
^10*b^6)*8i)/(b^3*(a*b^10 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a 
^5*b^6 - a^6*b^5 - a^7*b^4)))*1i)/b^3 + (8*tan(c/2 + (d*x)/2)*(8*a^10 - 8* 
a^9*b - 8*a*b^9 + 4*b^10 + 24*a^2*b^8 + 32*a^3*b^7 - 52*a^4*b^6 - 48*a^5*b 
^5 + 57*a^6*b^4 + 32*a^7*b^3 - 32*a^8*b^2))/(a*b^10 + b^11 - 3*a^2*b^9 - 3 
*a^3*b^8 + 3*a^4*b^7 + 3*a^5*b^6 - a^6*b^5 - a^7*b^4))/b^3 - ((((8*(12*a*b 
^14 - 4*b^15 + 8*a^2*b^13 - 34*a^3*b^12 - 6*a^4*b^11 + 36*a^5*b^10 + 4*a^6 
*b^9 - 18*a^7*b^8 - 2*a^8*b^7 + 4*a^9*b^6))/(a*b^12 + b^13 - 3*a^2*b^11 - 
3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) + (tan(c/2 + (d*x) 
/2)*(8*a*b^15 - 8*a^2*b^14 - 32*a^3*b^13 + 32*a^4*b^12 + 48*a^5*b^11 - 48* 
a^6*b^10 - 32*a^7*b^9 + 32*a^8*b^8 + 8*a^9*b^7 - 8*a^10*b^6)*8i)/(b^3*(a*b 
^10 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a^5*b^6 - a^6*b^5 - a^7 
*b^4)))*1i)/b^3 - (8*tan(c/2 + (d*x)/2)*(8*a^10 - 8*a^9*b - 8*a*b^9 + 4*b^ 
10 + 24*a^2*b^8 + 32*a^3*b^7 - 52*a^4*b^6 - 48*a^5*b^5 + 57*a^6*b^4 + 32*a 
^7*b^3 - 32*a^8*b^2))/(a*b^10 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 
 3*a^5*b^6 - a^6*b^5 - a^7*b^4))/b^3)/((((((8*(12*a*b^14 - 4*b^15 + 8*a...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 1004, normalized size of antiderivative = 5.61 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^3/(a+b*cos(d*x+c))^3,x)
 

Output:

( - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*cos(c + d*x)*a**6*b + 20*sqrt(a**2 - b**2)*atan((tan((c + 
d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**4*b**3 
- 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt 
(a**2 - b**2))*cos(c + d*x)*a**2*b**5 + 4*sqrt(a**2 - b**2)*atan((tan((c + 
 d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**5*b 
**2 - 10*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ 
sqrt(a**2 - b**2))*sin(c + d*x)**2*a**3*b**4 + 12*sqrt(a**2 - b**2)*atan(( 
tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)** 
2*a*b**6 - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2) 
*b)/sqrt(a**2 - b**2))*a**7 + 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a 
 - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**5*b**2 - 2*sqrt(a**2 - b**2)* 
atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**3*b** 
4 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq 
rt(a**2 - b**2))*a*b**6 - 3*cos(c + d*x)*sin(c + d*x)*a**6*b**2 + 9*cos(c 
+ d*x)*sin(c + d*x)*a**4*b**4 - 6*cos(c + d*x)*sin(c + d*x)*a**2*b**6 + 4* 
cos(c + d*x)*a**7*b*d*x - 12*cos(c + d*x)*a**5*b**3*d*x + 12*cos(c + d*x)* 
a**3*b**5*d*x - 4*cos(c + d*x)*a*b**7*d*x - 2*sin(c + d*x)**2*a**6*b**2*d* 
x + 6*sin(c + d*x)**2*a**4*b**4*d*x - 6*sin(c + d*x)**2*a**2*b**6*d*x + 2* 
sin(c + d*x)**2*b**8*d*x - 2*sin(c + d*x)*a**7*b + 7*sin(c + d*x)*a**5*...