\(\int \frac {\cos ^2(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^2} \, dx\) [262]

Optimal result

Integrand size = 40, antiderivative size = 147 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {(4 B-7 C) x}{2 a^2}+\frac {2 (5 B-8 C) \sin (c+d x)}{3 a^2 d}-\frac {(4 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {(5 B-8 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac {(B-C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:

-1/2*(4*B-7*C)*x/a^2+2/3*(5*B-8*C)*sin(d*x+c)/a^2/d-1/2*(4*B-7*C)*cos(d*x+ 
c)*sin(d*x+c)/a^2/d+1/3*(5*B-8*C)*cos(d*x+c)^2*sin(d*x+c)/a^2/d/(1+cos(d*x 
+c))+1/3*(B-C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^2
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(315\) vs. \(2(147)=294\).

Time = 1.93 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.14 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-36 (4 B-7 C) d x \cos \left (\frac {d x}{2}\right )-36 (4 B-7 C) d x \cos \left (c+\frac {d x}{2}\right )-48 B d x \cos \left (c+\frac {3 d x}{2}\right )+84 C d x \cos \left (c+\frac {3 d x}{2}\right )-48 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+264 B \sin \left (\frac {d x}{2}\right )-381 C \sin \left (\frac {d x}{2}\right )-120 B \sin \left (c+\frac {d x}{2}\right )+147 C \sin \left (c+\frac {d x}{2}\right )+164 B \sin \left (c+\frac {3 d x}{2}\right )-239 C \sin \left (c+\frac {3 d x}{2}\right )+36 B \sin \left (2 c+\frac {3 d x}{2}\right )-63 C \sin \left (2 c+\frac {3 d x}{2}\right )+12 B \sin \left (2 c+\frac {5 d x}{2}\right )-15 C \sin \left (2 c+\frac {5 d x}{2}\right )+12 B \sin \left (3 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {5 d x}{2}\right )+3 C \sin \left (3 c+\frac {7 d x}{2}\right )+3 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \] Input:

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

Output:

(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(4*B - 7*C)*d*x*Cos[(d*x)/2] - 36*(4*B - 7 
*C)*d*x*Cos[c + (d*x)/2] - 48*B*d*x*Cos[c + (3*d*x)/2] + 84*C*d*x*Cos[c + 
(3*d*x)/2] - 48*B*d*x*Cos[2*c + (3*d*x)/2] + 84*C*d*x*Cos[2*c + (3*d*x)/2] 
 + 264*B*Sin[(d*x)/2] - 381*C*Sin[(d*x)/2] - 120*B*Sin[c + (d*x)/2] + 147* 
C*Sin[c + (d*x)/2] + 164*B*Sin[c + (3*d*x)/2] - 239*C*Sin[c + (3*d*x)/2] + 
 36*B*Sin[2*c + (3*d*x)/2] - 63*C*Sin[2*c + (3*d*x)/2] + 12*B*Sin[2*c + (5 
*d*x)/2] - 15*C*Sin[2*c + (5*d*x)/2] + 12*B*Sin[3*c + (5*d*x)/2] - 15*C*Si 
n[3*c + (5*d*x)/2] + 3*C*Sin[3*c + (7*d*x)/2] + 3*C*Sin[4*c + (7*d*x)/2])) 
/(48*a^2*d*(1 + Cos[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3508, 3042, 3456, 3042, 3456, 3042, 3213}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\int \frac {\cos ^2(c+d x) (3 a (B-C)-a (2 B-5 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (3 a (B-C)-a (2 B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int \cos (c+d x) \left (2 a^2 (5 B-8 C)-3 a^2 (4 B-7 C) \cos (c+d x)\right )dx}{a^2}+\frac {(5 B-8 C) \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 a^2 (5 B-8 C)-3 a^2 (4 B-7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {(5 B-8 C) \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3213

\(\displaystyle \frac {\frac {\frac {2 a^2 (5 B-8 C) \sin (c+d x)}{d}-\frac {3 a^2 (4 B-7 C) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {3}{2} a^2 x (4 B-7 C)}{a^2}+\frac {(5 B-8 C) \sin (c+d x) \cos ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}+\frac {(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

Input:

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

Output:

((B - C)*Cos[c + d*x]^3*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (((5* 
B - 8*C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + ((-3*a^2*(4 
*B - 7*C)*x)/2 + (2*a^2*(5*B - 8*C)*Sin[c + d*x])/d - (3*a^2*(4*B - 7*C)*C 
os[c + d*x]*Sin[c + d*x])/(2*d))/a^2)/(3*a^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3213

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co 
s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free 
Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3456

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

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]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60

method	result	size
parallelrisch	\(\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {3 \left (B -C \right ) \cos \left (2 d x +2 c \right )}{28}+\frac {3 \cos \left (3 d x +3 c \right ) C}{112}+\left (B -\frac {163 C}{112}\right ) \cos \left (d x +c \right )+\frac {23 B}{28}-\frac {5 C}{4}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-6 x d \left (B -\frac {7 C}{4}\right )}{3 a^{2} d}\)	\(88\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \left (\left (\frac {5 C}{2}-B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {3 C}{2}-B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-2 \left (4 B -7 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(135\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 \left (\left (\frac {5 C}{2}-B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {3 C}{2}-B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-2 \left (4 B -7 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(135\)
risch	\(-\frac {2 x B}{a^{2}}+\frac {7 C x}{2 a^{2}}-\frac {i C \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{a^{2} d}+\frac {i C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}+\frac {2 i \left (9 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}+15 B \,{\mathrm e}^{i \left (d x +c \right )}-21 C \,{\mathrm e}^{i \left (d x +c \right )}+8 B -11 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(207\)
norman	\(\frac {\frac {\left (11 B -18 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (4 B -7 C \right ) x}{2 a}-\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}-\frac {2 \left (4 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {3 \left (4 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {2 \left (4 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {\left (4 B -7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}+\frac {\left (9 B -13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (11 B -17 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (61 B -100 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}+\frac {\left (95 B -149 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a}\)	\(281\)

Input:

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

Output:

1/3*(7*tan(1/2*d*x+1/2*c)*(3/28*(B-C)*cos(2*d*x+2*c)+3/112*cos(3*d*x+3*c)* 
C+(B-163/112*C)*cos(d*x+c)+23/28*B-5/4*C)*sec(1/2*d*x+1/2*c)^2-6*x*d*(B-7/ 
4*C))/a^2/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {3 \, {\left (4 \, B - 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (4 \, B - 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (4 \, B - 7 \, C\right )} d x - {\left (3 \, C \cos \left (d x + c\right )^{3} + 6 \, {\left (B - C\right )} \cos \left (d x + c\right )^{2} + {\left (28 \, B - 43 \, C\right )} \cos \left (d x + c\right ) + 20 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

-1/6*(3*(4*B - 7*C)*d*x*cos(d*x + c)^2 + 6*(4*B - 7*C)*d*x*cos(d*x + c) + 
3*(4*B - 7*C)*d*x - (3*C*cos(d*x + c)^3 + 6*(B - C)*cos(d*x + c)^2 + (28*B 
 - 43*C)*cos(d*x + c) + 20*B - 32*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 
 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 848 vs. \(2 (136) = 272\).

Time = 2.37 (sec) , antiderivative size = 848, normalized size of antiderivative = 5.77 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((-12*B*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 1 
2*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 24*B*d*x*tan(c/2 + d*x/2)**2/(6 
*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 
12*B*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6 
*a**2*d) - B*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d 
*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 13*B*tan(c/2 + d*x/2)**5/(6*a**2*d*tan( 
c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 41*B*tan(c/2 
 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 
 + 6*a**2*d) + 27*B*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a* 
*2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x*tan(c/2 + d*x/2)**4/(6*a** 
2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 42*C 
*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 
 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a** 
2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + C*tan(c/2 + d*x/2)**7/(6*a**2*d*tan( 
c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 19*C*tan(c/2 
 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 
 + 6*a**2*d) - 71*C*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12 
*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 39*C*tan(c/2 + d*x/2)/(6*a**2*d* 
tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)) 
, (x*(B*cos(c) + C*cos(c)**2)*cos(c)**2/(a*cos(c) + a)**2, True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (137) = 274\).

Time = 0.15 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.93 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \] Input:

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

Output:

-1/6*(C*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x 
+ c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d* 
x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin 
(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + 
c) + 1))/a^2) - B*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(c 
os(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 
 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x 
 + c) + 1))))/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} {\left (4 \, B - 7 \, C\right )}}{a^{2}} - \frac {6 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:

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

Output:

-1/6*(3*(d*x + c)*(4*B - 7*C)/a^2 - 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5*C*ta 
n(1/2*d*x + 1/2*c)^3 + 2*B*tan(1/2*d*x + 1/2*c) - 3*C*tan(1/2*d*x + 1/2*c) 
)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C 
*a^4*tan(1/2*d*x + 1/2*c)^3 - 15*B*a^4*tan(1/2*d*x + 1/2*c) + 21*C*a^4*tan 
(1/2*d*x + 1/2*c))/a^6)/d
 

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^2}+\frac {2\,B-4\,C}{2\,a^2}\right )}{d}-\frac {x\,\left (4\,B-7\,C\right )}{2\,a^2}+\frac {\left (2\,B-5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B-3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*((3*(B - C))/(2*a^2) + (2*B - 4*C)/(2*a^2)))/d - (x*(4 
*B - 7*C))/(2*a^2) + (tan(c/2 + (d*x)/2)^3*(2*B - 5*C) + tan(c/2 + (d*x)/2 
)*(2*B - 3*C))/(d*(2*a^2*tan(c/2 + (d*x)/2)^2 + a^2*tan(c/2 + (d*x)/2)^4 + 
 a^2)) - (tan(c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^2(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b -9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c -12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b d x +21 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c d x +2 \cos \left (d x +c \right ) b -2 \cos \left (d x +c \right ) c -3 \sin \left (d x +c \right )^{4} c +22 \sin \left (d x +c \right )^{2} b -31 \sin \left (d x +c \right )^{2} c -12 \sin \left (d x +c \right ) b d x +21 \sin \left (d x +c \right ) c d x -2 b +2 c}{6 \sin \left (d x +c \right ) a^{2} d \left (\cos \left (d x +c \right )+1\right )} \] Input:

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

Output:

(6*cos(c + d*x)*sin(c + d*x)**2*b - 9*cos(c + d*x)*sin(c + d*x)**2*c - 12* 
cos(c + d*x)*sin(c + d*x)*b*d*x + 21*cos(c + d*x)*sin(c + d*x)*c*d*x + 2*c 
os(c + d*x)*b - 2*cos(c + d*x)*c - 3*sin(c + d*x)**4*c + 22*sin(c + d*x)** 
2*b - 31*sin(c + d*x)**2*c - 12*sin(c + d*x)*b*d*x + 21*sin(c + d*x)*c*d*x 
 - 2*b + 2*c)/(6*sin(c + d*x)*a**2*d*(cos(c + d*x) + 1))