\(\int \frac {\cos (c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^2} \, dx\) [50]

Optimal result

Integrand size = 31, antiderivative size = 90 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 C x}{a^2}+\frac {(A+4 C) \sin (c+d x)}{3 a^2 d}+\frac {2 C \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:

-2*C*x/a^2+1/3*(A+4*C)*sin(d*x+c)/a^2/d+2*C*sin(d*x+c)/a^2/d/(1+cos(d*x+c) 
)-1/3*(A+C)*cos(d*x+c)^2*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. \(195\) vs. \(2(90)=180\).

Time = 1.75 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.17 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-36 C d x \cos \left (\frac {d x}{2}\right )-36 C d x \cos \left (c+\frac {d x}{2}\right )-12 C d x \cos \left (c+\frac {3 d x}{2}\right )-12 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+12 A \sin \left (\frac {d x}{2}\right )+66 C \sin \left (\frac {d x}{2}\right )-12 A \sin \left (c+\frac {d x}{2}\right )-30 C \sin \left (c+\frac {d x}{2}\right )+8 A \sin \left (c+\frac {3 d x}{2}\right )+41 C \sin \left (c+\frac {3 d x}{2}\right )+9 C \sin \left (2 c+\frac {3 d x}{2}\right )+3 C \sin \left (2 c+\frac {5 d x}{2}\right )+3 C \sin \left (3 c+\frac {5 d x}{2}\right )\right )}{48 a^2 d} \] Input:

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

Output:

(Sec[c/2]*Sec[(c + d*x)/2]^3*(-36*C*d*x*Cos[(d*x)/2] - 36*C*d*x*Cos[c + (d 
*x)/2] - 12*C*d*x*Cos[c + (3*d*x)/2] - 12*C*d*x*Cos[2*c + (3*d*x)/2] + 12* 
A*Sin[(d*x)/2] + 66*C*Sin[(d*x)/2] - 12*A*Sin[c + (d*x)/2] - 30*C*Sin[c + 
(d*x)/2] + 8*A*Sin[c + (3*d*x)/2] + 41*C*Sin[c + (3*d*x)/2] + 9*C*Sin[2*c 
+ (3*d*x)/2] + 3*C*Sin[2*c + (5*d*x)/2] + 3*C*Sin[3*c + (5*d*x)/2]))/(48*a 
^2*d)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3521, 3042, 3447, 3042, 3502, 27, 3042, 3214, 3042, 3127}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\int -\frac {6 a^2 C \cos (c+d x)}{\cos (c+d x) a+a}dx}{a}+\frac {(A+4 C) \sin (c+d x)}{d}}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {(A+4 C) \sin (c+d x)}{d}-6 a C \int \frac {\cos (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(A+4 C) \sin (c+d x)}{d}-6 a C \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {(A+4 C) \sin (c+d x)}{d}-6 a C \left (\frac {x}{a}-\int \frac {1}{\cos (c+d x) a+a}dx\right )}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {(A+4 C) \sin (c+d x)}{d}-6 a C \left (\frac {x}{a}-\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx\right )}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3127

\(\displaystyle \frac {\frac {(A+4 C) \sin (c+d x)}{d}-6 a C \left (\frac {x}{a}-\frac {\sin (c+d x)}{d (a \cos (c+d x)+a)}\right )}{3 a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((A + C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (( 
(A + 4*C)*Sin[c + d*x])/d - 6*a*C*(x/a - Sin[c + d*x]/(d*(a + a*Cos[c + d* 
x]))))/(3*a^2)
 

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 3042

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

rule 3127

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + 
 d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[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 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, 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 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 = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.77

method	result	size
parallelrisch	\(\frac {\left (8 A +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (3 C \cos \left (2 d x +2 c \right )+28 \cos \left (d x +c \right ) C -2 A +23 C \right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-24 d x C}{12 a^{2} d}\)	\(69\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 C \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{2 d \,a^{2}}\)	\(103\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 C \left (-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{2 d \,a^{2}}\)	\(103\)
risch	\(-\frac {2 C x}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a^{2} d}+\frac {2 i \left (3 A \,{\mathrm e}^{2 i \left (d x +c \right )}+9 C \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A \,{\mathrm e}^{i \left (d x +c \right )}+15 C \,{\mathrm e}^{i \left (d x +c \right )}+2 A +8 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(124\)
norman	\(\frac {\frac {\left (A +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {2 C x}{a}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {6 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {6 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {2 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (A +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {2 \left (2 A +17 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a}\)	\(195\)

Input:

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

Output:

1/12*((8*A+sec(1/2*d*x+1/2*c)^2*(3*C*cos(2*d*x+2*c)+28*cos(d*x+c)*C-2*A+23 
*C))*tan(1/2*d*x+1/2*c)-24*d*x*C)/a^2/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.13 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {6 \, C d x \cos \left (d x + c\right )^{2} + 12 \, C d x \cos \left (d x + c\right ) + 6 \, C d x - {\left (3 \, C \cos \left (d x + c\right )^{2} + 2 \, {\left (A + 7 \, C\right )} \cos \left (d x + c\right ) + A + 10 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

-1/3*(6*C*d*x*cos(d*x + c)^2 + 12*C*d*x*cos(d*x + c) + 6*C*d*x - (3*C*cos( 
d*x + c)^2 + 2*(A + 7*C)*cos(d*x + c) + A + 10*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. 335 vs. \(2 (87) = 174\).

Time = 1.14 (sec) , antiderivative size = 335, normalized size of antiderivative = 3.72 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\begin {cases} - \frac {A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {2 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {3 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {12 C d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {12 C d x}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {C \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {14 C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {27 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-A*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d 
) + 2*A*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 3* 
A*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 12*C*d*x*ta 
n(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 12*C*d*x/(6* 
a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - C*tan(c/2 + d*x/2)**5/(6*a**2*d*t 
an(c/2 + d*x/2)**2 + 6*a**2*d) + 14*C*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/ 
2 + d*x/2)**2 + 6*a**2*d) + 27*C*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/ 
2)**2 + 6*a**2*d), Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(c)/(a*cos(c) + a)** 
2, True))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.83 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.27 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {12 \, {\left (d x + c\right )} C}{a^{2}} - \frac {12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:

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

Output:

-1/6*(12*(d*x + c)*C/a^2 - 12*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c 
)^2 + 1)*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c) 
^3 - 3*A*a^4*tan(1/2*d*x + 1/2*c) - 15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
 

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{a^2}-\frac {A-3\,C}{2\,a^2}\right )}{d}-\frac {2\,C\,x}{a^2}+\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*((A + C)/a^2 - (A - 3*C)/(2*a^2)))/d - (2*C*x)/a^2 + ( 
2*C*tan(c/2 + (d*x)/2))/(d*(a^2*tan(c/2 + (d*x)/2)^2 + a^2)) - (tan(c/2 + 
(d*x)/2)^3*(A + C))/(6*a^2*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.39 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a +14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c d x +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c -12 c d x}{6 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:

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

Output:

( - tan((c + d*x)/2)**5*a - tan((c + d*x)/2)**5*c + 2*tan((c + d*x)/2)**3* 
a + 14*tan((c + d*x)/2)**3*c - 12*tan((c + d*x)/2)**2*c*d*x + 3*tan((c + d 
*x)/2)*a + 27*tan((c + d*x)/2)*c - 12*c*d*x)/(6*a**2*d*(tan((c + d*x)/2)** 
2 + 1))