\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [53]

Optimal result

Integrand size = 33, antiderivative size = 91 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 A \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {(10 A+C) \tan (c+d x)}{3 a^2 d}-\frac {2 A \tan (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A+C) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:

-2*A*arctanh(sin(d*x+c))/a^2/d+1/3*(10*A+C)*tan(d*x+c)/a^2/d-2*A*tan(d*x+c 
)/a^2/d/(1+cos(d*x+c))-1/3*(A+C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^2
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(288\) vs. \(2(91)=182\).

Time = 2.55 (sec) , antiderivative size = 288, normalized size of antiderivative = 3.16 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left ((A+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (7 A+C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+6 A \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+(A+C) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2 (2 A+C+C \cos (2 (c+d x)))} \] Input:

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

Output:

(4*Cos[(c + d*x)/2]*Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*((A + C)*Sec[c/2 
]*Sin[(d*x)/2] + 2*(7*A + C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 6* 
A*Cos[(c + d*x)/2]^3*(2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*Log[C 
os[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos 
[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] 
+ Sin[(c + d*x)/2]))) + (A + C)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + 
Cos[c + d*x])^2*(2*A + C + C*Cos[2*(c + d*x)]))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 3521, 3042, 3457, 3042, 3227, 3042, 4254, 24, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \left (a^2 (10 A+C)-6 a^2 A \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {6 A \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 (10 A+C)-6 a^2 A \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {6 A \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {a^2 (10 A+C) \int \sec ^2(c+d x)dx-6 a^2 A \int \sec (c+d x)dx}{a^2}-\frac {6 A \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (10 A+C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-6 a^2 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {6 A \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {-\frac {a^2 (10 A+C) \int 1d(-\tan (c+d x))}{d}-6 a^2 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {6 A \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {\frac {a^2 (10 A+C) \tan (c+d x)}{d}-6 a^2 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {6 A \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {\frac {a^2 (10 A+C) \tan (c+d x)}{d}-\frac {6 a^2 A \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {6 A \tan (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A+C) \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

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

rule 3042

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

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 3457

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

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)]
 

rule 4254

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

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.30

method	result	size
parallelrisch	\(\frac {24 \cos \left (d x +c \right ) A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-24 \cos \left (d x +c \right ) A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+28 \left (\left (\frac {5 A}{14}+\frac {C}{28}\right ) \cos \left (2 d x +2 c \right )+\left (A +\frac {C}{7}\right ) \cos \left (d x +c \right )+\frac {4 A}{7}+\frac {C}{28}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{12 d \,a^{2} \cos \left (d x +c \right )}\)	\(118\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}\)	\(123\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}\)	\(123\)
risch	\(\frac {2 i \left (6 A \,{\mathrm e}^{4 i \left (d x +c \right )}+18 A \,{\mathrm e}^{3 i \left (d x +c \right )}+3 C \,{\mathrm e}^{3 i \left (d x +c \right )}+22 A \,{\mathrm e}^{2 i \left (d x +c \right )}+C \,{\mathrm e}^{2 i \left (d x +c \right )}+24 A \,{\mathrm e}^{i \left (d x +c \right )}+3 C \,{\mathrm e}^{i \left (d x +c \right )}+10 A +C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {2 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}\)	\(169\)
norman	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {2 \left (4 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}-\frac {\left (9 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {2 \left (10 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a}+\frac {2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}-\frac {2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}\)	\(193\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+a*cos(d*x+c))^2,x,method=_RETURNVER 
BOSE)
 

Output:

1/12*(24*cos(d*x+c)*A*ln(tan(1/2*d*x+1/2*c)-1)-24*cos(d*x+c)*A*ln(tan(1/2* 
d*x+1/2*c)+1)+28*((5/14*A+1/28*C)*cos(2*d*x+2*c)+(A+1/7*C)*cos(d*x+c)+4/7* 
A+1/28*C)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2)/d/a^2/cos(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.84 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {3 \, {\left (A \cos \left (d x + c\right )^{3} + 2 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A \cos \left (d x + c\right )^{3} + 2 \, A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (10 \, A + C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A + C\right )} \cos \left (d x + c\right ) + 3 \, A\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

-1/3*(3*(A*cos(d*x + c)^3 + 2*A*cos(d*x + c)^2 + A*cos(d*x + c))*log(sin(d 
*x + c) + 1) - 3*(A*cos(d*x + c)^3 + 2*A*cos(d*x + c)^2 + A*cos(d*x + c))* 
log(-sin(d*x + c) + 1) - ((10*A + C)*cos(d*x + c)^2 + 2*(7*A + C)*cos(d*x 
+ c) + 3*A)*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + 
 a^2*d*cos(d*x + c))
 

Sympy [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:

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

Output:

(Integral(A*sec(c + d*x)**2/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + I 
ntegral(C*cos(c + d*x)**2*sec(c + d*x)**2/(cos(c + d*x)**2 + 2*cos(c + d*x 
) + 1), x))/a**2
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (87) = 174\).

Time = 0.05 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.10 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {A {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {C {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \] Input:

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

Output:

1/6*(A*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) 
 + 1)^3)/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(si 
n(d*x + c)/(cos(d*x + c) + 1) - 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))) + C*(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 = 0.33 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.56 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {12 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {12 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.24 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{a^2}+\frac {3\,A-C}{2\,a^2}\right )}{d}-\frac {4\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*((A + C)/a^2 + (3*A - C)/(2*a^2)))/d - (4*A*atanh(tan( 
c/2 + (d*x)/2)))/(a^2*d) - (2*A*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 = 184, normalized size of antiderivative = 2.02 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c}{6 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )} \] Input:

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

Output:

(12*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 12*log(tan((c + d*x) 
/2) - 1)*a - 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a + 12*log(t 
an((c + d*x)/2) + 1)*a + tan((c + d*x)/2)**5*a + tan((c + d*x)/2)**5*c + 1 
4*tan((c + d*x)/2)**3*a + 2*tan((c + d*x)/2)**3*c - 27*tan((c + d*x)/2)*a 
- 3*tan((c + d*x)/2)*c)/(6*a**2*d*(tan((c + d*x)/2)**2 - 1))