\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [363]

Optimal result

Integrand size = 41, antiderivative size = 246 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {(23 A-13 B+6 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}+\frac {4 (34 A-19 B+9 C) \tan (c+d x)}{5 a^3 d}-\frac {(23 A-13 B+6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(13 A-8 B+3 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(23 A-13 B+6 C) \sec ^2(c+d x) \tan (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {4 (34 A-19 B+9 C) \tan ^3(c+d x)}{15 a^3 d} \] Output:

-1/2*(23*A-13*B+6*C)*arctanh(sin(d*x+c))/a^3/d+4/5*(34*A-19*B+9*C)*tan(d*x 
+c)/a^3/d-1/2*(23*A-13*B+6*C)*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*(A-B+C)*sec( 
d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(13*A-8*B+3*C)*sec(d*x+c)^2* 
tan(d*x+c)/a/d/(a+a*cos(d*x+c))^2-1/3*(23*A-13*B+6*C)*sec(d*x+c)^2*tan(d*x 
+c)/d/(a^3+a^3*cos(d*x+c))+4/15*(34*A-19*B+9*C)*tan(d*x+c)^3/a^3/d
 

Mathematica [A] (verified)

Time = 3.70 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.10 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {960 (23 A-13 B+6 C) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+2 \cos \left (\frac {1}{2} (c+d x)\right ) (4321 A-2331 B+1146 C+(7814 A-4274 B+2124 C) \cos (c+d x)+8 (691 A-381 B+186 C) \cos (2 (c+d x))+3098 A \cos (3 (c+d x))-1718 B \cos (3 (c+d x))+828 C \cos (3 (c+d x))+1287 A \cos (4 (c+d x))-717 B \cos (4 (c+d x))+342 C \cos (4 (c+d x))+272 A \cos (5 (c+d x))-152 B \cos (5 (c+d x))+72 C \cos (5 (c+d x))) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )}{240 a^3 d (1+\cos (c+d x))^3} \] Input:

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a* 
Cos[c + d*x])^3,x]
 

Output:

(960*(23*A - 13*B + 6*C)*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[(c 
 + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 2*Cos[(c + d*x)/ 
2]*(4321*A - 2331*B + 1146*C + (7814*A - 4274*B + 2124*C)*Cos[c + d*x] + 8 
*(691*A - 381*B + 186*C)*Cos[2*(c + d*x)] + 3098*A*Cos[3*(c + d*x)] - 1718 
*B*Cos[3*(c + d*x)] + 828*C*Cos[3*(c + d*x)] + 1287*A*Cos[4*(c + d*x)] - 7 
17*B*Cos[4*(c + d*x)] + 342*C*Cos[4*(c + d*x)] + 272*A*Cos[5*(c + d*x)] - 
152*B*Cos[5*(c + d*x)] + 72*C*Cos[5*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d* 
x)/2])/(240*a^3*d*(1 + Cos[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.50 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 3520, 3042, 3457, 3042, 3457, 27, 3042, 3227, 3042, 4254, 2009, 4255, 3042, 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.

\(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

\(\displaystyle \frac {\int \frac {(a (8 A-3 B+3 C)-5 a (A-B) \cos (c+d x)) \sec ^4(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {\left (3 a^2 (21 A-11 B+6 C)-4 a^2 (13 A-8 B+3 C) \cos (c+d x)\right ) \sec ^4(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {3 a^2 (21 A-11 B+6 C)-4 a^2 (13 A-8 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\frac {\int 3 \left (4 a^3 (34 A-19 B+9 C)-5 a^3 (23 A-13 B+6 C) \cos (c+d x)\right ) \sec ^4(c+d x)dx}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \int \left (4 a^3 (34 A-19 B+9 C)-5 a^3 (23 A-13 B+6 C) \cos (c+d x)\right ) \sec ^4(c+d x)dx}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \int \frac {4 a^3 (34 A-19 B+9 C)-5 a^3 (23 A-13 B+6 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {\frac {3 \left (4 a^3 (34 A-19 B+9 C) \int \sec ^4(c+d x)dx-5 a^3 (23 A-13 B+6 C) \int \sec ^3(c+d x)dx\right )}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (4 a^3 (34 A-19 B+9 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx-5 a^3 (23 A-13 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {\frac {3 \left (-\frac {4 a^3 (34 A-19 B+9 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}-5 a^3 (23 A-13 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 (23 A-13 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a^3 (34 A-19 B+9 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 (23 A-13 B+6 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^3 (34 A-19 B+9 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 (23 A-13 B+6 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^3 (34 A-19 B+9 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 (23 A-13 B+6 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^3 (34 A-19 B+9 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {5 a^2 (23 A-13 B+6 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (13 A-8 B+3 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

Input:

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c 
+ d*x])^3,x]
 

Output:

-1/5*((A - B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^3) 
+ (-1/3*(a*(13*A - 8*B + 3*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Cos[c 
 + d*x])^2) + ((-5*a^2*(23*A - 13*B + 6*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d 
*(a + a*Cos[c + d*x])) + (3*(-5*a^3*(23*A - 13*B + 6*C)*(ArcTanh[Sin[c + d 
*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (4*a^3*(34*A - 19*B + 9* 
C)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d))/a^2)/(3*a^2))/(5*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 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 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 3520

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[(a*A - b*B + 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))   Int[(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)) + B*(b*c*m + a 
*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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 4255

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

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 = 0.95 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.98

method	result	size
parallelrisch	\(\frac {1380 \left (A -\frac {13 B}{23}+\frac {6 C}{23}\right ) \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-1380 \left (A -\frac {13 B}{23}+\frac {6 C}{23}\right ) \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+1549 \left (\left (A -\frac {859 B}{1549}+\frac {414 C}{1549}\right ) \cos \left (3 d x +3 c \right )+\frac {4 \left (691 A -381 B +186 C \right ) \cos \left (2 d x +2 c \right )}{1549}+\frac {3 \left (\frac {429 A}{2}-\frac {239 B}{2}+57 C \right ) \cos \left (4 d x +4 c \right )}{1549}+\frac {4 \left (34 A -19 B +9 C \right ) \cos \left (5 d x +5 c \right )}{1549}+\frac {\left (3907 A -2137 B +1062 C \right ) \cos \left (d x +c \right )}{1549}+\frac {4321 A}{3098}-\frac {2331 B}{3098}+\frac {573 C}{1549}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{120 d \,a^{3} \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(241\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-16 A +4 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-46 A +26 B -12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {34 A -14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {4 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {16 A -4 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {34 A -14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (46 A -26 B +12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}}{4 d \,a^{3}}\)	\(302\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-16 A +4 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-46 A +26 B -12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {34 A -14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {4 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {16 A -4 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {34 A -14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (46 A -26 B +12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}}{4 d \,a^{3}}\)	\(302\)
norman	\(\frac {\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}+\frac {\left (47 A -37 B +27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{60 d a}-\frac {\left (93 A -51 B +25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {\left (177 A -109 B +45 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d a}+\frac {7 \left (411 A -241 B +111 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d a}+\frac {\left (679 A -419 B +219 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{60 d a}-\frac {\left (1849 A -959 B +429 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d a}-\frac {\left (1887 A -967 B +447 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{60 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a^{2}}+\frac {\left (23 A -13 B +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}-\frac {\left (23 A -13 B +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\)	\(323\)
risch	\(\frac {i \left (144 C +544 A -304 B -4550 B \,{\mathrm e}^{7 i \left (d x +c \right )}+11684 A \,{\mathrm e}^{6 i \left (d x +c \right )}+3144 C \,{\mathrm e}^{6 i \left (d x +c \right )}+12622 A \,{\mathrm e}^{4 i \left (d x +c \right )}+3372 C \,{\mathrm e}^{4 i \left (d x +c \right )}+5347 A \,{\mathrm e}^{2 i \left (d x +c \right )}+1422 C \,{\mathrm e}^{2 i \left (d x +c \right )}+13340 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3480 C \,{\mathrm e}^{5 i \left (d x +c \right )}+9230 A \,{\mathrm e}^{3 i \left (d x +c \right )}+2460 C \,{\mathrm e}^{3 i \left (d x +c \right )}+2375 A \,{\mathrm e}^{i \left (d x +c \right )}-2977 B \,{\mathrm e}^{2 i \left (d x +c \right )}+630 C \,{\mathrm e}^{i \left (d x +c \right )}-7380 B \,{\mathrm e}^{5 i \left (d x +c \right )}-6604 B \,{\mathrm e}^{6 i \left (d x +c \right )}-5130 B \,{\mathrm e}^{3 i \left (d x +c \right )}-1325 B \,{\mathrm e}^{i \left (d x +c \right )}+4370 A \,{\mathrm e}^{8 i \left (d x +c \right )}+8050 A \,{\mathrm e}^{7 i \left (d x +c \right )}+2100 C \,{\mathrm e}^{7 i \left (d x +c \right )}-2470 \,{\mathrm e}^{8 i \left (d x +c \right )} B +1140 C \,{\mathrm e}^{8 i \left (d x +c \right )}+90 C \,{\mathrm e}^{10 i \left (d x +c \right )}+450 C \,{\mathrm e}^{9 i \left (d x +c \right )}+1725 A \,{\mathrm e}^{9 i \left (d x +c \right )}+345 A \,{\mathrm e}^{10 i \left (d x +c \right )}-195 B \,{\mathrm e}^{10 i \left (d x +c \right )}-7002 B \,{\mathrm e}^{4 i \left (d x +c \right )}-975 B \,{\mathrm e}^{9 i \left (d x +c \right )}\right )}{15 a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {23 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}-\frac {13 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}-\frac {23 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}+\frac {13 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}\)	\(539\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x,meth 
od=_RETURNVERBOSE)
 

Output:

1/120*(1380*(A-13/23*B+6/23*C)*(cos(3*d*x+3*c)+3*cos(d*x+c))*ln(tan(1/2*d* 
x+1/2*c)-1)-1380*(A-13/23*B+6/23*C)*(cos(3*d*x+3*c)+3*cos(d*x+c))*ln(tan(1 
/2*d*x+1/2*c)+1)+1549*((A-859/1549*B+414/1549*C)*cos(3*d*x+3*c)+4/1549*(69 
1*A-381*B+186*C)*cos(2*d*x+2*c)+3/1549*(429/2*A-239/2*B+57*C)*cos(4*d*x+4* 
c)+4/1549*(34*A-19*B+9*C)*cos(5*d*x+5*c)+1/1549*(3907*A-2137*B+1062*C)*cos 
(d*x+c)+4321/3098*A-2331/3098*B+573/1549*C)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x 
+1/2*c)^4)/d/a^3/(cos(3*d*x+3*c)+3*cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.41 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {15 \, {\left ({\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (23 \, A - 13 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (34 \, A - 19 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (429 \, A - 239 \, B + 114 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (869 \, A - 479 \, B + 234 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (19 \, A - 9 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 10 \, A\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{6} + 3 \, a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + a^{3} d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3, 
x, algorithm="fricas")
 

Output:

-1/60*(15*((23*A - 13*B + 6*C)*cos(d*x + c)^6 + 3*(23*A - 13*B + 6*C)*cos( 
d*x + c)^5 + 3*(23*A - 13*B + 6*C)*cos(d*x + c)^4 + (23*A - 13*B + 6*C)*co 
s(d*x + c)^3)*log(sin(d*x + c) + 1) - 15*((23*A - 13*B + 6*C)*cos(d*x + c) 
^6 + 3*(23*A - 13*B + 6*C)*cos(d*x + c)^5 + 3*(23*A - 13*B + 6*C)*cos(d*x 
+ c)^4 + (23*A - 13*B + 6*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(1 
6*(34*A - 19*B + 9*C)*cos(d*x + c)^5 + 3*(429*A - 239*B + 114*C)*cos(d*x + 
 c)^4 + (869*A - 479*B + 234*C)*cos(d*x + c)^3 + 5*(19*A - 9*B + 6*C)*cos( 
d*x + c)^2 - 15*(A - B)*cos(d*x + c) + 10*A)*sin(d*x + c))/(a^3*d*cos(d*x 
+ c)^6 + 3*a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + a^3*d*cos(d*x + 
 c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))* 
*3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 630 vs. \(2 (232) = 464\).

Time = 0.08 (sec) , antiderivative size = 630, normalized size of antiderivative = 2.56 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3, 
x, algorithm="maxima")
 

Output:

1/60*(A*(20*(33*sin(d*x + c)/(cos(d*x + c) + 1) - 76*sin(d*x + c)^3/(cos(d 
*x + c) + 1)^3 + 51*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^3 - 3*a^3*sin( 
d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^ 
4 - a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (735*sin(d*x + c)/(cos(d*x 
+ c) + 1) + 50*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos 
(d*x + c) + 1)^5)/a^3 - 690*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 
 690*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - B*(60*(5*sin(d*x + c) 
/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3* 
sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1 
)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^3/(cos(d*x + 
 c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x 
+ c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) 
 - 1)/a^3) + 3*C*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) 
 + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*si 
n(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a 
^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c) 
/(cos(d*x + c) + 1) - 1)/a^3))/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.45 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {30 \, {\left (23 \, A - 13 \, B + 6 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (23 \, A - 13 \, B + 6 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {20 \, {\left (51 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 76 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 50 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 735 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3, 
x, algorithm="giac")
 

Output:

-1/60*(30*(23*A - 13*B + 6*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30* 
(23*A - 13*B + 6*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 20*(51*A*tan( 
1/2*d*x + 1/2*c)^5 - 21*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c 
)^5 - 76*A*tan(1/2*d*x + 1/2*c)^3 + 36*B*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan 
(1/2*d*x + 1/2*c)^3 + 33*A*tan(1/2*d*x + 1/2*c) - 15*B*tan(1/2*d*x + 1/2*c 
) + 6*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^3) - (3*A* 
a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*t 
an(1/2*d*x + 1/2*c)^5 + 50*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^12*tan(1 
/2*d*x + 1/2*c)^3 + 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 735*A*a^12*tan(1/2* 
d*x + 1/2*c) - 465*B*a^12*tan(1/2*d*x + 1/2*c) + 255*C*a^12*tan(1/2*d*x + 
1/2*c))/a^15)/d
 

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.11 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {6\,A-4\,B+2\,C}{a^3}-\frac {5\,B-15\,A+C}{4\,a^3}+\frac {5\,\left (A-B+C\right )}{2\,a^3}\right )}{d}-\frac {\left (17\,A-7\,B+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (12\,B-\frac {76\,A}{3}-4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (11\,A-5\,B+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {6\,A-4\,B+2\,C}{12\,a^3}+\frac {A-B+C}{3\,a^3}\right )}{d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (23\,A-13\,B+6\,C\right )}{a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B+C\right )}{20\,a^3\,d} \] Input:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + 
 d*x))^3),x)
 

Output:

(tan(c/2 + (d*x)/2)*((6*A - 4*B + 2*C)/a^3 - (5*B - 15*A + C)/(4*a^3) + (5 
*(A - B + C))/(2*a^3)))/d - (tan(c/2 + (d*x)/2)*(11*A - 5*B + 2*C) + tan(c 
/2 + (d*x)/2)^5*(17*A - 7*B + 2*C) - tan(c/2 + (d*x)/2)^3*((76*A)/3 - 12*B 
 + 4*C))/(d*(3*a^3*tan(c/2 + (d*x)/2)^2 - 3*a^3*tan(c/2 + (d*x)/2)^4 + a^3 
*tan(c/2 + (d*x)/2)^6 - a^3)) + (tan(c/2 + (d*x)/2)^3*((6*A - 4*B + 2*C)/( 
12*a^3) + (A - B + C)/(3*a^3)))/d - (atanh(tan(c/2 + (d*x)/2))*(23*A - 13* 
B + 6*C))/(a^3*d) + (tan(c/2 + (d*x)/2)^5*(A - B + C))/(20*a^3*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 854, normalized size of antiderivative = 3.47 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x)
 

Output:

(690*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**6*a - 390*log(tan((c + d* 
x)/2) - 1)*tan((c + d*x)/2)**6*b + 180*log(tan((c + d*x)/2) - 1)*tan((c + 
d*x)/2)**6*c - 2070*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a + 1170 
*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*b - 540*log(tan((c + d*x)/2 
) - 1)*tan((c + d*x)/2)**4*c + 2070*log(tan((c + d*x)/2) - 1)*tan((c + d*x 
)/2)**2*a - 1170*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 540*log 
(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*c - 690*log(tan((c + d*x)/2) - 
1)*a + 390*log(tan((c + d*x)/2) - 1)*b - 180*log(tan((c + d*x)/2) - 1)*c - 
 690*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**6*a + 390*log(tan((c + d* 
x)/2) + 1)*tan((c + d*x)/2)**6*b - 180*log(tan((c + d*x)/2) + 1)*tan((c + 
d*x)/2)**6*c + 2070*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a - 1170 
*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*b + 540*log(tan((c + d*x)/2 
) + 1)*tan((c + d*x)/2)**4*c - 2070*log(tan((c + d*x)/2) + 1)*tan((c + d*x 
)/2)**2*a + 1170*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b - 540*log 
(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 690*log(tan((c + d*x)/2) + 
1)*a - 390*log(tan((c + d*x)/2) + 1)*b + 180*log(tan((c + d*x)/2) + 1)*c + 
 3*tan((c + d*x)/2)**11*a - 3*tan((c + d*x)/2)**11*b + 3*tan((c + d*x)/2)* 
*11*c + 41*tan((c + d*x)/2)**9*a - 31*tan((c + d*x)/2)**9*b + 21*tan((c + 
d*x)/2)**9*c + 594*tan((c + d*x)/2)**7*a - 354*tan((c + d*x)/2)**7*b + 174 
*tan((c + d*x)/2)**7*c - 3078*tan((c + d*x)/2)**5*a + 1698*tan((c + d*x...