\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\) [978]

Optimal result

Integrand size = 41, antiderivative size = 206 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\left (2 a^2 b B+b^3 B-2 a^3 C-a b^2 (2 A+C)\right ) x}{2 b^4}+\frac {2 a^2 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}+\frac {\left (3 A b^2-3 a b B+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac {(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d} \] Output:

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

Mathematica [A] (verified)

Time = 2.06 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {-6 \left (-2 a^2 b B-b^3 B+2 a^3 C+a b^2 (2 A+C)\right ) (c+d x)-\frac {24 a^2 \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+3 b \left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right ) \sin (c+d x)+3 b^2 (b B-a C) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^4 d} \] Input:

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b* 
Cos[c + d*x]),x]
 

Output:

(-6*(-2*a^2*b*B - b^3*B + 2*a^3*C + a*b^2*(2*A + C))*(c + d*x) - (24*a^2*( 
A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b 
^2]])/Sqrt[-a^2 + b^2] + 3*b*(4*A*b^2 - 4*a*b*B + 4*a^2*C + 3*b^2*C)*Sin[c 
 + d*x] + 3*b^2*(b*B - a*C)*Sin[2*(c + d*x)] + b^3*C*Sin[3*(c + d*x)])/(12 
*b^4*d)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 3528, 3042, 3528, 3042, 3502, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + b*Cos[c 
+ d*x]),x]
 

Output:

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

Defintions of rubi rules used

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 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 3502

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

rule 3528

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {\frac {2 a^{2} \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-A \,b^{3}+B a \,b^{2}+\frac {1}{2} B \,b^{3}-a^{2} b C -\frac {1}{2} C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 A \,b^{3}+2 B a \,b^{2}-2 a^{2} b C -\frac {2}{3} C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A \,b^{3}+B a \,b^{2}-a^{2} b C -C \,b^{3}-\frac {1}{2} B \,b^{3}+\frac {1}{2} C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {\left (2 a A \,b^{2}-2 B \,a^{2} b -B \,b^{3}+2 a^{3} C +C a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\)	\(275\)
default	\(\frac {\frac {2 a^{2} \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\frac {\left (-A \,b^{3}+B a \,b^{2}+\frac {1}{2} B \,b^{3}-a^{2} b C -\frac {1}{2} C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 A \,b^{3}+2 B a \,b^{2}-2 a^{2} b C -\frac {2}{3} C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-A \,b^{3}+B a \,b^{2}-a^{2} b C -C \,b^{3}-\frac {1}{2} B \,b^{3}+\frac {1}{2} C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {\left (2 a A \,b^{2}-2 B \,a^{2} b -B \,b^{3}+2 a^{3} C +C a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}\right )}{b^{4}}}{d}\)	\(275\)
risch	\(-\frac {x a A}{b^{2}}+\frac {x B \,a^{2}}{b^{3}}+\frac {x B}{2 b}-\frac {x \,a^{3} C}{b^{4}}-\frac {a C x}{2 b^{2}}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 b d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B a}{2 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 b d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{2 b^{3} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C}{8 b d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{2 b^{3} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B a}{2 b^{2} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {C \sin \left (3 d x +3 c \right )}{12 b d}+\frac {\sin \left (2 d x +2 c \right ) B}{4 b d}-\frac {\sin \left (2 d x +2 c \right ) C a}{4 b^{2} d}\)	\(706\)

Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x,method 
=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 599, normalized size of antiderivative = 2.91 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [-\frac {3 \, {\left (2 \, C a^{5} - 2 \, B a^{4} b + {\left (2 \, A - C\right )} a^{3} b^{2} + B a^{2} b^{3} - {\left (2 \, A + C\right )} a b^{4} + B b^{5}\right )} d x + 3 \, {\left (C a^{4} - B a^{3} b + A a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (6 \, C a^{4} b - 6 \, B a^{3} b^{2} + 2 \, {\left (3 \, A - C\right )} a^{2} b^{3} + 6 \, B a b^{4} - 2 \, {\left (3 \, A + 2 \, C\right )} b^{5} + 2 \, {\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}, -\frac {3 \, {\left (2 \, C a^{5} - 2 \, B a^{4} b + {\left (2 \, A - C\right )} a^{3} b^{2} + B a^{2} b^{3} - {\left (2 \, A + C\right )} a b^{4} + B b^{5}\right )} d x - 6 \, {\left (C a^{4} - B a^{3} b + A a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (6 \, C a^{4} b - 6 \, B a^{3} b^{2} + 2 \, {\left (3 \, A - C\right )} a^{2} b^{3} + 6 \, B a b^{4} - 2 \, {\left (3 \, A + 2 \, C\right )} b^{5} + 2 \, {\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, 
 algorithm="fricas")
 

Output:

[-1/6*(3*(2*C*a^5 - 2*B*a^4*b + (2*A - C)*a^3*b^2 + B*a^2*b^3 - (2*A + C)* 
a*b^4 + B*b^5)*d*x + 3*(C*a^4 - B*a^3*b + A*a^2*b^2)*sqrt(-a^2 + b^2)*log( 
(2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(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)) - (6*C*a^4*b - 6*B*a^3*b^2 + 2*(3*A - C)*a^2*b^3 + 6 
*B*a*b^4 - 2*(3*A + 2*C)*b^5 + 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C 
*a^3*b^2 - B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^2* 
b^4 - b^6)*d), -1/6*(3*(2*C*a^5 - 2*B*a^4*b + (2*A - C)*a^3*b^2 + B*a^2*b^ 
3 - (2*A + C)*a*b^4 + B*b^5)*d*x - 6*(C*a^4 - B*a^3*b + A*a^2*b^2)*sqrt(a^ 
2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (6 
*C*a^4*b - 6*B*a^3*b^2 + 2*(3*A - C)*a^2*b^3 + 6*B*a*b^4 - 2*(3*A + 2*C)*b 
^5 + 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - B*a^2*b^3 - C*a 
*b^4 + B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)), 
x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (188) = 376\).

Time = 0.15 (sec) , antiderivative size = 424, normalized size of antiderivative = 2.06 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {\frac {3 \, {\left (2 \, C a^{3} - 2 \, B a^{2} b + 2 \, A a b^{2} + C a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {12 \, {\left (C a^{4} - B a^{3} b + A a^{2} 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 )}}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {2 \, {\left (6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, 
 algorithm="giac")
 

Output:

-1/6*(3*(2*C*a^3 - 2*B*a^2*b + 2*A*a*b^2 + C*a*b^2 - B*b^3)*(d*x + c)/b^4 
+ 12*(C*a^4 - B*a^3*b + A*a^2*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))/sqr 
t(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^4) - 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 
 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2 
*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2 
*d*x + 1/2*c)^5 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*B*a*b*tan(1/2*d*x + 
 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^2*tan(1/2*d*x + 1/2*c) 
^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 6*B*a*b*tan(1/2*d*x + 1/2*c) - 3*C*a*b 
*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x 
 + 1/2*c) + 6*C*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3* 
b^3))/d
 

Mupad [B] (verification not implemented)

Time = 5.66 (sec) , antiderivative size = 7119, normalized size of antiderivative = 34.56 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c 
+ d*x)),x)
                                                                                    
                                                                                    
 

Output:

((tan(c/2 + (d*x)/2)^5*(2*A*b^2 - B*b^2 + 2*C*a^2 + 2*C*b^2 - 2*B*a*b + C* 
a*b))/b^3 + (4*tan(c/2 + (d*x)/2)^3*(3*A*b^2 + 3*C*a^2 + C*b^2 - 3*B*a*b)) 
/(3*b^3) + (tan(c/2 + (d*x)/2)*(2*A*b^2 + B*b^2 + 2*C*a^2 + 2*C*b^2 - 2*B* 
a*b - C*a*b))/b^3)/(d*(3*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + t 
an(c/2 + (d*x)/2)^6 + 1)) + (atan(((((8*tan(c/2 + (d*x)/2)*(B^2*b^9 - 8*C^ 
2*a^9 - 3*B^2*a*b^8 + 16*C^2*a^8*b + 4*A^2*a^2*b^7 - 12*A^2*a^3*b^6 + 16*A 
^2*a^4*b^5 - 8*A^2*a^5*b^4 + 7*B^2*a^2*b^7 - 13*B^2*a^3*b^6 + 16*B^2*a^4*b 
^5 - 16*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^7 - 3*C^2 
*a^3*b^6 + 7*C^2*a^4*b^5 - 13*C^2*a^5*b^4 + 16*C^2*a^6*b^3 - 16*C^2*a^7*b^ 
2 - 4*A*B*a*b^8 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 12*A*B*a^2*b^7 - 20*A*B*a^3 
*b^6 + 28*A*B*a^4*b^5 - 32*A*B*a^5*b^4 + 16*A*B*a^6*b^3 + 4*A*C*a^2*b^7 - 
12*A*C*a^3*b^6 + 20*A*C*a^4*b^5 - 28*A*C*a^5*b^4 + 32*A*C*a^6*b^3 - 16*A*C 
*a^7*b^2 + 6*B*C*a^2*b^7 - 14*B*C*a^3*b^6 + 26*B*C*a^4*b^5 - 32*B*C*a^5*b^ 
4 + 32*B*C*a^6*b^3 - 32*B*C*a^7*b^2))/b^6 + (((8*(4*A*a^3*b^10 - 8*A*a^2*b 
^11 - 2*B*b^13 - 2*B*a^2*b^11 + 6*B*a^3*b^10 - 4*B*a^4*b^9 - 2*C*a^2*b^11 
+ 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*A*a*b^12 + 2*B*a*b^12 + 2*C 
*a*b^12))/b^9 - (8*tan(c/2 + (d*x)/2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)* 
((B*b^3*1i)/2 - C*a^3*1i - b^2*(A*a*1i + (C*a*1i)/2) + B*a^2*b*1i))/b^10)* 
((B*b^3*1i)/2 - C*a^3*1i - b^2*(A*a*1i + (C*a*1i)/2) + B*a^2*b*1i))/b^4)*( 
(B*b^3*1i)/2 - C*a^3*1i - b^2*(A*a*1i + (C*a*1i)/2) + B*a^2*b*1i)*1i)/b...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4} c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2} c +3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{4}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{4} c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{6}-2 \sin \left (d x +c \right )^{3} a^{2} b^{3} c +2 \sin \left (d x +c \right )^{3} b^{5} c +6 \sin \left (d x +c \right ) a^{4} b c -6 \sin \left (d x +c \right ) b^{5} c -6 a^{5} c^{2}-6 a^{5} c d x +3 a^{3} b^{2} c^{2}+3 a^{3} b^{2} c d x +3 a^{2} b^{4} c +3 a^{2} b^{4} d x +3 a \,b^{4} c^{2}+3 a \,b^{4} c d x -3 b^{6} c -3 b^{6} d x}{6 b^{4} d \left (a^{2}-b^{2}\right )} \] Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)
 

Output:

(12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( 
a**2 - b**2))*a**4*c - 3*cos(c + d*x)*sin(c + d*x)*a**3*b**2*c + 3*cos(c + 
 d*x)*sin(c + d*x)*a**2*b**4 + 3*cos(c + d*x)*sin(c + d*x)*a*b**4*c - 3*co 
s(c + d*x)*sin(c + d*x)*b**6 - 2*sin(c + d*x)**3*a**2*b**3*c + 2*sin(c + d 
*x)**3*b**5*c + 6*sin(c + d*x)*a**4*b*c - 6*sin(c + d*x)*b**5*c - 6*a**5*c 
**2 - 6*a**5*c*d*x + 3*a**3*b**2*c**2 + 3*a**3*b**2*c*d*x + 3*a**2*b**4*c 
+ 3*a**2*b**4*d*x + 3*a*b**4*c**2 + 3*a*b**4*c*d*x - 3*b**6*c - 3*b**6*d*x 
)/(6*b**4*d*(a**2 - b**2))