\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{a+a \cos (c+d x)} \, dx\) [40]

Optimal result

Integrand size = 33, antiderivative size = 124 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {(2 A+3 C) x}{2 a}+\frac {(3 A+4 C) \sin (c+d x)}{a d}-\frac {(2 A+3 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(3 A+4 C) \sin ^3(c+d x)}{3 a d} \] Output:

-1/2*(2*A+3*C)*x/a+(3*A+4*C)*sin(d*x+c)/a/d-1/2*(2*A+3*C)*cos(d*x+c)*sin(d 
*x+c)/a/d-(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))-1/3*(3*A+4*C)*s 
in(d*x+c)^3/a/d
 

Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.81 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-12 (2 A+3 C) d x \cos \left (\frac {d x}{2}\right )-12 (2 A+3 C) d x \cos \left (c+\frac {d x}{2}\right )+60 A \sin \left (\frac {d x}{2}\right )+69 C \sin \left (\frac {d x}{2}\right )+12 A \sin \left (c+\frac {d x}{2}\right )+21 C \sin \left (c+\frac {d x}{2}\right )+12 A \sin \left (c+\frac {3 d x}{2}\right )+18 C \sin \left (c+\frac {3 d x}{2}\right )+12 A \sin \left (2 c+\frac {3 d x}{2}\right )+18 C \sin \left (2 c+\frac {3 d x}{2}\right )-2 C \sin \left (2 c+\frac {5 d x}{2}\right )-2 C \sin \left (3 c+\frac {5 d x}{2}\right )+C \sin \left (3 c+\frac {7 d x}{2}\right )+C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{24 a d (1+\cos (c+d x))} \] Input:

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

Output:

(Cos[(c + d*x)/2]*Sec[c/2]*(-12*(2*A + 3*C)*d*x*Cos[(d*x)/2] - 12*(2*A + 3 
*C)*d*x*Cos[c + (d*x)/2] + 60*A*Sin[(d*x)/2] + 69*C*Sin[(d*x)/2] + 12*A*Si 
n[c + (d*x)/2] + 21*C*Sin[c + (d*x)/2] + 12*A*Sin[c + (3*d*x)/2] + 18*C*Si 
n[c + (3*d*x)/2] + 12*A*Sin[2*c + (3*d*x)/2] + 18*C*Sin[2*c + (3*d*x)/2] - 
 2*C*Sin[2*c + (5*d*x)/2] - 2*C*Sin[3*c + (5*d*x)/2] + C*Sin[3*c + (7*d*x) 
/2] + C*Sin[4*c + (7*d*x)/2]))/(24*a*d*(1 + Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.87, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 3521, 25, 3042, 3227, 3042, 3113, 2009, 3115, 24}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3113

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3113

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] 
 && IGtQ[(n - 1)/2, 0]
 

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3521

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.59

method	result	size
parallelrisch	\(\frac {\left (-C \cos \left (2 d x +2 c \right )+\cos \left (3 d x +3 c \right ) C +\left (12 A +17 C \right ) \cos \left (d x +c \right )+24 A +31 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \left (A +\frac {3 C}{2}\right ) x d}{12 d a}\)	\(73\)
derivativedivides	\(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \left (\left (-\frac {A}{2}-\frac {5 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-A -\frac {4 C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {A}{2}-\frac {3 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\left (2 A +3 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(123\)
default	\(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \left (\left (-\frac {A}{2}-\frac {5 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-A -\frac {4 C}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {A}{2}-\frac {3 C}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\left (2 A +3 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(123\)
risch	\(-\frac {x A}{a}-\frac {3 C x}{2 a}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 a d}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )} C}{8 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 a d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 a d}+\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {C \sin \left (3 d x +3 c \right )}{12 a d}-\frac {\sin \left (2 d x +2 c \right ) C}{4 a d}\)	\(174\)
norman	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}+\frac {\left (3 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\left (2 A +3 C \right ) x}{2 a}+\frac {3 \left (2 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {2 \left (2 A +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {3 \left (2 A +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {2 \left (2 A +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {\left (2 A +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}+\frac {\left (30 A +37 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\left (36 A +49 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(249\)

Input:

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

1/12*((-C*cos(2*d*x+2*c)+cos(3*d*x+3*c)*C+(12*A+17*C)*cos(d*x+c)+24*A+31*C 
)*tan(1/2*d*x+1/2*c)-12*(A+3/2*C)*x*d)/d/a
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {3 \, {\left (2 \, A + 3 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (2 \, A + 3 \, C\right )} d x - {\left (2 \, C \cos \left (d x + c\right )^{3} - C \cos \left (d x + c\right )^{2} + {\left (6 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 12 \, A + 16 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c)),x, algorithm="f 
ricas")
 

Output:

-1/6*(3*(2*A + 3*C)*d*x*cos(d*x + c) + 3*(2*A + 3*C)*d*x - (2*C*cos(d*x + 
c)^3 - C*cos(d*x + c)^2 + (6*A + 7*C)*cos(d*x + c) + 12*A + 16*C)*sin(d*x 
+ c))/(a*d*cos(d*x + c) + a*d)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1163 vs. \(2 (100) = 200\).

Time = 1.25 (sec) , antiderivative size = 1163, normalized size of antiderivative = 9.38 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((-6*A*d*x*tan(c/2 + d*x/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a* 
d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 18*A*d*x*tan 
(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 
 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 18*A*d*x*tan(c/2 + d*x/2)**2/(6*a*d 
*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2 
)**2 + 6*a*d) - 6*A*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/ 
2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 6*A*tan(c/2 + d*x/2)**7/(6*a 
*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x 
/2)**2 + 6*a*d) + 30*A*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**6 + 18 
*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 42*A*tan( 
c/2 + d*x/2)**3/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 
18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 18*A*tan(c/2 + d*x/2)/(6*a*d*tan(c/2 
 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6 
*a*d) - 9*C*d*x*tan(c/2 + d*x/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*ta 
n(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 27*C*d*x*tan(c/2 
 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18* 
a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 27*C*d*x*tan(c/2 + d*x/2)**2/(6*a*d*tan 
(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 
 + 6*a*d) - 9*C*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)** 
4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 6*C*tan(c/2 + d*x/2)**7/(6*a*...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (118) = 236\).

Time = 0.12 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.17 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{3 \, d} \] Input:

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c)),x, algorithm="m 
axima")
 

Output:

1/3*(C*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + 
c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a + 3*a*sin(d*x + c)^ 
2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(d 
*x + c)^6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1) 
)/a + 3*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*A*(2*arctan(sin(d*x + c)/ 
(cos(d*x + c) + 1))/a - 2*sin(d*x + c)/((a + a*sin(d*x + c)^2/(cos(d*x + c 
) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} {\left (2 \, A + 3 \, C\right )}}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \] Input:

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c)),x, algorithm="g 
iac")
 

Output:

-1/6*(3*(d*x + c)*(2*A + 3*C)/a - 6*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d* 
x + 1/2*c))/a - 2*(6*A*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1/2*c)^ 
5 + 12*A*tan(1/2*d*x + 1/2*c)^3 + 16*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*tan(1/ 
2*d*x + 1/2*c) + 9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3 
*a))/d
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {A\,\sin \left (c+d\,x\right )}{a\,d}-\frac {3\,C\,x}{2\,a}-\frac {A\,x}{a}+\frac {7\,C\,\sin \left (c+d\,x\right )}{4\,a\,d}+\frac {A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d}-\frac {C\,\sin \left (2\,c+2\,d\,x\right )}{4\,a\,d}+\frac {C\,\sin \left (3\,c+3\,d\,x\right )}{12\,a\,d}+\frac {C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d} \] Input:

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

Output:

(A*sin(c + d*x))/(a*d) - (3*C*x)/(2*a) - (A*x)/a + (7*C*sin(c + d*x))/(4*a 
*d) + (A*tan(c/2 + (d*x)/2))/(a*d) - (C*sin(2*c + 2*d*x))/(4*a*d) + (C*sin 
(3*c + 3*d*x))/(12*a*d) + (C*tan(c/2 + (d*x)/2))/(a*d)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c -6 \cos \left (d x +c \right ) a -6 \cos \left (d x +c \right ) c -2 \sin \left (d x +c \right )^{4} c +6 \sin \left (d x +c \right )^{2} a +12 \sin \left (d x +c \right )^{2} c -6 \sin \left (d x +c \right ) a d x -9 \sin \left (d x +c \right ) c d x +6 a +6 c}{6 \sin \left (d x +c \right ) a d} \] Input:

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

Output:

( - 3*cos(c + d*x)*sin(c + d*x)**2*c - 6*cos(c + d*x)*a - 6*cos(c + d*x)*c 
 - 2*sin(c + d*x)**4*c + 6*sin(c + d*x)**2*a + 12*sin(c + d*x)**2*c - 6*si 
n(c + d*x)*a*d*x - 9*sin(c + d*x)*c*d*x + 6*a + 6*c)/(6*sin(c + d*x)*a*d)