\(\int \frac {\cos ^2(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\) [794]

Optimal result

Integrand size = 40, antiderivative size = 178 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\left (2 a^2+b^2\right ) (b B-a C) x}{2 b^4}-\frac {2 a^3 (b B-a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}-\frac {\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac {(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d} \] Output:

1/2*(2*a^2+b^2)*(B*b-C*a)*x/b^4-2*a^3*(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*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 = 1.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {6 \left (2 a^2+b^2\right ) (b B-a C) (c+d x)-\frac {24 a^3 (-b B+a C) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+3 b \left (-4 a b B+4 a^2 C+3 b^2 C\right ) \sin (c+d x)+3 b^2 (b B-a C) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^4 d} \] Input:

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

Output:

(6*(2*a^2 + b^2)*(b*B - a*C)*(c + d*x) - (24*a^3*(-(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 
*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 = 1.17 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3469, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3469

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

Int[(Cos[c + d*x]^2*(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^2)*(b*B - a*C)*x)/b - (4*a^3*(b*B - 
a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sq 
rt[a + b]*d)))/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 3469

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( 
n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
 f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( 
m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m 
+ n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !(IGt 
Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 3508

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[1/b^2   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*C, 0]
 

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 = 0.69 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {-\frac {2 a^{3} \left (B b -C a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-B a \,b^{2}-\frac {1}{2} B \,b^{3}+a^{2} b C +\frac {1}{2} C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 B a \,b^{2}+2 a^{2} b C +\frac {2}{3} C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-B a \,b^{2}+a^{2} b C +C \,b^{3}+\frac {1}{2} B \,b^{3}-\frac {1}{2} C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\left (2 B \,a^{2} b +B \,b^{3}-2 a^{3} C -C a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\)	\(240\)
default	\(\frac {-\frac {2 a^{3} \left (B b -C a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 \left (\left (-B a \,b^{2}-\frac {1}{2} B \,b^{3}+a^{2} b C +\frac {1}{2} C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-2 B a \,b^{2}+2 a^{2} b C +\frac {2}{3} C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-B a \,b^{2}+a^{2} b C +C \,b^{3}+\frac {1}{2} B \,b^{3}-\frac {1}{2} C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\left (2 B \,a^{2} b +B \,b^{3}-2 a^{3} C -C a \,b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\)	\(240\)
risch	\(\frac {x B \,a^{2}}{b^{3}}+\frac {x B}{2 b}-\frac {x \,a^{3} C}{b^{4}}-\frac {a C x}{2 b^{2}}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B a}{2 b^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B a}{2 b^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{2 b^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{2 b^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 b d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C}{8 b d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{4}}+\frac {C \sin \left (3 d x +3 c \right )}{12 b d}+\frac {\sin \left (2 d x +2 c \right ) B}{4 b d}-\frac {\sin \left (2 d x +2 c \right ) C a}{4 b^{2} d}\)	\(515\)

Input:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 541, normalized size of antiderivative = 3.04 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\left [-\frac {3 \, {\left (2 \, C a^{5} - 2 \, B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} d x - 3 \, {\left (C a^{4} - B a^{3} b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (6 \, C a^{4} b - 6 \, B a^{3} b^{2} - 2 \, C a^{2} b^{3} + 6 \, B a b^{4} - 4 \, C b^{5} + 2 \, {\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}, -\frac {3 \, {\left (2 \, C a^{5} - 2 \, B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} d x - 6 \, {\left (C a^{4} - B a^{3} b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (6 \, C a^{4} b - 6 \, B a^{3} b^{2} - 2 \, C a^{2} b^{3} + 6 \, B a b^{4} - 4 \, C b^{5} + 2 \, {\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \] Input:

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

Output:

[-1/6*(3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4 + B*b^5)*d 
*x - 3*(C*a^4 - B*a^3*b)*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*C*a^2*b^3 + 6*B*a*b^4 - 4*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 - 
C*a^3*b^2 + B*a^2*b^3 - C*a*b^4 + B*b^5)*d*x - 6*(C*a^4 - B*a^3*b)*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*C*a^2*b^3 + 6*B*a*b^4 - 4*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. \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

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

integrate(cos(d*x+c)^2*(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (161) = 322\).

Time = 0.14 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=-\frac {\frac {3 \, {\left (2 \, C a^{3} - 2 \, B a^{2} b + C a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {12 \, {\left (C a^{4} - B a^{3} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {2 \, {\left (6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \] Input:

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

Output:

-1/6*(3*(2*C*a^3 - 2*B*a^2*b + C*a*b^2 - B*b^3)*(d*x + c)/b^4 + 12*(C*a^4 
- B*a^3*b)*(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^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 - 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 + 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) + 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 = 3.40 (sec) , antiderivative size = 4568, normalized size of antiderivative = 25.66 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

int((cos(c + d*x)^2*(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)*(B*b^2 + 2*C*a^2 + 2*C*b^2 - 2*B*a*b - C*a*b))/b^3 + 
(tan(c/2 + (d*x)/2)^5*(2*C*a^2 - B*b^2 + 2*C*b^2 - 2*B*a*b + C*a*b))/b^3 + 
 (4*tan(c/2 + (d*x)/2)^3*(3*C*a^2 + C*b^2 - 3*B*a*b))/(3*b^3))/(d*(3*tan(c 
/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (a 
tan((((2*a^2 + b^2)*(B*b - C*a)*((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 + 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 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 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 
*(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 - 2*B*a*b^12 - 2*C*a*b^12))/b^9 - ( 
tan(c/2 + (d*x)/2)*(2*a^2 + b^2)*(B*b - C*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^ 
3*b^8)*4i)/b^10)*(2*a^2 + b^2)*(B*b - C*a)*1i)/(2*b^4)))/(2*b^4) + ((2*a^2 
 + b^2)*(B*b - C*a)*((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 + 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 - 2*B* 
C*a*b^8 + 16*B*C*a^8*b + 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*(2*B*b^1...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.22 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{4} c -12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} b^{2}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2} c +3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{4}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{4} c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{6}-2 \sin \left (d x +c \right )^{3} a^{2} b^{3} c +2 \sin \left (d x +c \right )^{3} b^{5} c +6 \sin \left (d x +c \right ) a^{4} b c -6 \sin \left (d x +c \right ) a^{3} b^{3}+6 \sin \left (d x +c \right ) a \,b^{5}-6 \sin \left (d x +c \right ) b^{5} c -6 a^{5} c^{2}-6 a^{5} c d x +6 a^{4} b^{2} c +6 a^{4} b^{2} d x +3 a^{3} b^{2} c^{2}+3 a^{3} b^{2} c d x -3 a^{2} b^{4} c -3 a^{2} b^{4} d x +3 a \,b^{4} c^{2}+3 a \,b^{4} c d x -3 b^{6} c -3 b^{6} d x}{6 b^{4} d \left (a^{2}-b^{2}\right )} \] Input:

int(cos(d*x+c)^2*(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 - 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**3*b**2 - 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*cos(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)*a**3*b**3 + 6*sin(c + d*x)*a*b**5 - 6*sin(c + d*x)*b**5*c - 6*a**5 
*c**2 - 6*a**5*c*d*x + 6*a**4*b**2*c + 6*a**4*b**2*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))