\(\int \frac {\cos (c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\) [979]

Optimal result

Integrand size = 39, antiderivative size = 144 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) x}{2 b^3}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}+\frac {(b B-a C) \sin (c+d x)}{b^2 d}+\frac {C \cos (c+d x) \sin (c+d x)}{2 b d} \] Output:

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

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {2 \left (2 A b^2-2 a b B+2 a^2 C+b^2 C\right ) (c+d x)+\frac {8 a \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+4 b (b B-a C) \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^3 d} \] Input:

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

Output:

(2*(2*A*b^2 - 2*a*b*B + 2*a^2*C + b^2*C)*(c + d*x) + (8*a*(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] + 4*b*(b*B - a*C)*Sin[c + d*x] + b^2*C*Sin[2*(c + d*x)])/(4*b^3*d)
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3528, 3042, 3502, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

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 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 = 0.66 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {-\frac {2 a \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (B \,b^{2}-a b C -\frac {1}{2} C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (B \,b^{2}-a b C +\frac {1}{2} C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 A \,b^{2}-2 B a b +2 a^{2} C +C \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\)	\(180\)
default	\(\frac {-\frac {2 a \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (B \,b^{2}-a b C -\frac {1}{2} C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (B \,b^{2}-a b C +\frac {1}{2} C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 A \,b^{2}-2 B a b +2 a^{2} C +C \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\)	\(180\)
risch	\(\frac {x A}{b}-\frac {x B a}{b^{2}}+\frac {x C \,a^{2}}{b^{3}}+\frac {C x}{2 b}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 b d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} C a}{2 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} C a}{2 b^{2} d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d b}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {\sin \left (2 d x +2 c \right ) C}{4 b d}\)	\(577\)

Input:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 457, normalized size of antiderivative = 3.17 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [\frac {{\left (2 \, C a^{4} - 2 \, B a^{3} b + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} b^{4}\right )} d x - {\left (C a^{3} - B a^{2} b + A a b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (2 \, C a^{3} b - 2 \, B a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}, \frac {{\left (2 \, C a^{4} - 2 \, B a^{3} b + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} b^{4}\right )} d x - 2 \, {\left (C a^{3} - B a^{2} b + A a b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, C a^{3} b - 2 \, B a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, a 
lgorithm="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 = 0.14 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.66 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2} + C b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {4 \, {\left (C a^{3} - B a^{2} b + A a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.07 (sec) , antiderivative size = 5594, normalized size of antiderivative = 38.85 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)*(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)*(2*B*b - 2*C*a + C*b))/b^2 - (tan(c/2 + (d*x)/2)^3*(2 
*C*a - 2*B*b + C*b))/b^2)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^ 
4 + 1)) - (atan(((((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 
- 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b 
^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C 
^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 
- 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 
 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32* 
A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^ 
4*b^3 - 32*B*C*a^5*b^2))/b^4 + (((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8 
*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a 
*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (8*tan(c/2 + (d*x)/2)*(8*a*b^8 - 16*a 
^2*b^7 + 8*a^3*b^6)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2) - B*a*b*1i))/b^7)*(C 
*a^2*1i + b^2*(A*1i + (C*1i)/2) - B*a*b*1i))/b^3)*(C*a^2*1i + b^2*(A*1i + 
(C*1i)/2) - B*a*b*1i)*1i)/b^3 + (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2 
*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^ 
5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^ 
2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b 
^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 
 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28...
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.74 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} c +\cos \left (d x +c \right )^{2} a^{2} b^{2} c d x -\cos \left (d x +c \right )^{2} b^{4} c d x +\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2} c -\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4} c +\sin \left (d x +c \right )^{2} a^{2} b^{2} c d x -\sin \left (d x +c \right )^{2} b^{4} c d x -2 \sin \left (d x +c \right ) a^{3} b c +2 \sin \left (d x +c \right ) a^{2} b^{3}+2 \sin \left (d x +c \right ) a \,b^{3} c -2 \sin \left (d x +c \right ) b^{5}+2 a^{4} c d x -2 a^{2} b^{2} c d x}{2 b^{3} d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*a**3*c + cos(c + d*x)**2*a**2*b**2*c*d*x - cos(c + d*x)**2 
*b**4*c*d*x + cos(c + d*x)*sin(c + d*x)*a**2*b**2*c - cos(c + d*x)*sin(c + 
 d*x)*b**4*c + sin(c + d*x)**2*a**2*b**2*c*d*x - sin(c + d*x)**2*b**4*c*d* 
x - 2*sin(c + d*x)*a**3*b*c + 2*sin(c + d*x)*a**2*b**3 + 2*sin(c + d*x)*a* 
b**3*c - 2*sin(c + d*x)*b**5 + 2*a**4*c*d*x - 2*a**2*b**2*c*d*x)/(2*b**3*d 
*(a**2 - b**2))