\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx\) [268]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F(-2)]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 29, antiderivative size = 180 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}+\frac {a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.77 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\frac {-\frac {2 \left (-3 a A b+a^2 B+2 b^2 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {a (A b-a B) \sin (c+d x)}{(a-b) b (a+b) (a+b \cos (c+d x))^2}+\frac {\left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \sin (c+d x)}{(a-b)^2 b (a+b)^2 (a+b \cos (c+d x))}}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {3042, 3447, 3042, 3500, 3042, 3233, 25, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

\(\displaystyle \int \frac {A \cos (c+d x)+B \cos ^2(c+d x)}{(a+b \cos (c+d x))^3}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {-\frac {\int -\frac {b \left (-B a^2+3 A b a-2 b^2 B\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {\left (a^3 B+a^2 A b-4 a b^2 B+2 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {\int \frac {b \left (-B a^2+3 A b a-2 b^2 B\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {\left (a^3 B+a^2 A b-4 a b^2 B+2 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\frac {2 b \left (a^2 (-B)+3 a A b-2 b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\left (a^3 B+a^2 A b-4 a b^2 B+2 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 b \left (a^2-b^2\right )}\)

Input:

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

Output:

(a*(A*b - a*B)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - 
((2*b*(3*a*A*b - a^2*B - 2*b^2*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sq 
rt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - ((a^2*A*b + 2*A*b^3 
+ a^3*B - 4*a*b^2*B)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/( 
2*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 3233
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

rule 3447
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

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 = 1.53 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.30

method result size
derivativedivides \(\frac {-\frac {2 \left (-\frac {\left (2 a^{2} A +A a b +2 A \,b^{2}-a^{2} B -4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{2} A -A a b +2 A \,b^{2}+a^{2} B -4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}-\frac {\left (3 A a b -a^{2} B -2 B \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) \(234\)
default \(\frac {-\frac {2 \left (-\frac {\left (2 a^{2} A +A a b +2 A \,b^{2}-a^{2} B -4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a^{2} A -A a b +2 A \,b^{2}+a^{2} B -4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}-\frac {\left (3 A a b -a^{2} B -2 B \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) \(234\)
risch \(\frac {i \left (3 A a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+2 B \,a^{4} b \,{\mathrm e}^{3 i \left (d x +c \right )}-5 B \,a^{2} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 A \,a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+5 A \,a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 A \,b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+2 B \,a^{5} {\mathrm e}^{2 i \left (d x +c \right )}-7 B \,a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 B a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+4 A \,a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}+5 A a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}+2 B \,a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}-11 B \,a^{2} b^{3} {\mathrm e}^{i \left (d x +c \right )}+A \,a^{2} b^{3}+2 A \,b^{5}+B \,a^{3} b^{2}-4 B a \,b^{4}\right )}{b^{2} \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 {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 ) A}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {a^{2} \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 ) B}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\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 ) B \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\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 ) A}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {a^{2} \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 ) B}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\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 ) B \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) \(802\)

Input:

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

Output:

1/d*(-2*(-1/2*(2*A*a^2+A*a*b+2*A*b^2-B*a^2-4*B*a*b)/(a-b)/(a^2+2*a*b+b^2)* 
tan(1/2*d*x+1/2*c)^3-1/2*(2*A*a^2-A*a*b+2*A*b^2+B*a^2-4*B*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-(3*A*a*b-B*a^2-2*B*b^2)/(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)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (165) = 330\).

Time = 0.13 (sec) , antiderivative size = 740, normalized size of antiderivative = 4.11 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorithm="fri 
cas")
 

Output:

[-1/4*((B*a^4 - 3*A*a^3*b + 2*B*a^2*b^2 + (B*a^2*b^2 - 3*A*a*b^3 + 2*B*b^4 
)*cos(d*x + c)^2 + 2*(B*a^3*b - 3*A*a^2*b^2 + 2*B*a*b^3)*cos(d*x + c))*sqr 
t(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*s 
qrt(-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*a^5 - 3*B*a^4*b - A*a^3*b 
^2 + 3*B*a^2*b^3 - A*a*b^4 + (B*a^5 + A*a^4*b - 5*B*a^3*b^2 + A*a^2*b^3 + 
4*B*a*b^4 - 2*A*b^5)*cos(d*x + c))*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3 
*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^ 
7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d), 1/2*((B*a^ 
4 - 3*A*a^3*b + 2*B*a^2*b^2 + (B*a^2*b^2 - 3*A*a*b^3 + 2*B*b^4)*cos(d*x + 
c)^2 + 2*(B*a^3*b - 3*A*a^2*b^2 + 2*B*a*b^3)*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*a^5 - 
 3*B*a^4*b - A*a^3*b^2 + 3*B*a^2*b^3 - A*a*b^4 + (B*a^5 + A*a^4*b - 5*B*a^ 
3*b^2 + A*a^2*b^3 + 4*B*a*b^4 - 2*A*b^5)*cos(d*x + c))*sin(d*x + c))/((a^6 
*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^ 
3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2 
*b^6)*d)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorithm="max 
ima")
 

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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (165) = 330\).

Time = 0.23 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.17 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\left (B a^{2} - 3 \, A a b + 2 \, B b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \] Input:

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

Output:

((B*a^2 - 3*A*a*b + 2*B*b^2)*(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 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (2*A*a^3*tan(1/2*d*x + 
1/2*c)^3 - B*a^3*tan(1/2*d*x + 1/2*c)^3 - A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 
 3*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*B*a 
*b^2*tan(1/2*d*x + 1/2*c)^3 - 2*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*A*a^3*tan 
(1/2*d*x + 1/2*c) + B*a^3*tan(1/2*d*x + 1/2*c) + A*a^2*b*tan(1/2*d*x + 1/2 
*c) - 3*B*a^2*b*tan(1/2*d*x + 1/2*c) + A*a*b^2*tan(1/2*d*x + 1/2*c) - 4*B* 
a*b^2*tan(1/2*d*x + 1/2*c) + 2*A*b^3*tan(1/2*d*x + 1/2*c))/((a^4 - 2*a^2*b 
^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) 
)/d
 

Mupad [B] (verification not implemented)

Time = 26.83 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.38 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,a^2+2\,A\,b^2-B\,a^2+A\,a\,b-4\,B\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^2+2\,A\,b^2+B\,a^2-A\,a\,b-4\,B\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (B\,a^2-3\,A\,a\,b+2\,B\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \] Input:

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

Output:

((tan(c/2 + (d*x)/2)^3*(2*A*a^2 + 2*A*b^2 - B*a^2 + A*a*b - 4*B*a*b))/((a 
+ b)^2*(a - b)) + (tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + B*a^2 - A*a*b - 
 4*B*a*b))/((a + b)*(a^2 - 2*a*b + b^2)))/(d*(2*a*b + tan(c/2 + (d*x)/2)^2 
*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2 + b^2)) 
+ (atan((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(a^2 - 2*a*b + b^2))/(2*(a + b)^(1 
/2)*(a - b)^(5/2)))*(B*a^2 + 2*B*b^2 - 3*A*a*b))/(d*(a + b)^(5/2)*(a - b)^ 
(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.07 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) b^{2}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a b +\sin \left (d x +c \right ) a^{3}-\sin \left (d x +c \right ) a \,b^{2}}{d \left (\cos \left (d x +c \right ) a^{4} b -2 \cos \left (d x +c \right ) a^{2} b^{3}+\cos \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

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

Output:

( - 2*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)*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*b + sin(c + d*x)*a**3 - 
sin(c + d*x)*a*b**2)/(d*(cos(c + d*x)*a**4*b - 2*cos(c + d*x)*a**2*b**3 + 
cos(c + d*x)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))