\(\int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [661]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 33, antiderivative size = 218 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) x+\frac {a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {3 A b \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d} \] Output:

1/8*b*(4*b^2*(A+2*C)+3*a^2*(3*A+4*C))*x+1/15*a*(15*b^2*(2*A+3*C)+2*a^2*(4* 
A+5*C))*sin(d*x+c)/d+3/40*b*(2*A*b^2+5*a^2*(3*A+4*C))*cos(d*x+c)*sin(d*x+c 
)/d+1/30*a*(3*A*b^2+2*a^2*(4*A+5*C))*cos(d*x+c)^2*sin(d*x+c)/d+3/20*A*b*co 
s(d*x+c)^3*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+b*sec(d*x 
+c))^3*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.71 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {60 b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) (c+d x)+60 a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sin (c+d x)+120 b \left (A b^2+3 a^2 (A+C)\right ) \sin (2 (c+d x))+10 a \left (12 A b^2+a^2 (5 A+4 C)\right ) \sin (3 (c+d x))+45 a^2 A b \sin (4 (c+d x))+6 a^3 A \sin (5 (c+d x))}{480 d} \] Input:

Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]
 

Output:

(60*b*(4*b^2*(A + 2*C) + 3*a^2*(3*A + 4*C))*(c + d*x) + 60*a*(6*b^2*(3*A + 
 4*C) + a^2*(5*A + 6*C))*Sin[c + d*x] + 120*b*(A*b^2 + 3*a^2*(A + C))*Sin[ 
2*(c + d*x)] + 10*a*(12*A*b^2 + a^2*(5*A + 4*C))*Sin[3*(c + d*x)] + 45*a^2 
*A*b*Sin[4*(c + d*x)] + 6*a^3*A*Sin[5*(c + d*x)])/(480*d)
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4583, 3042, 4582, 3042, 4562, 25, 3042, 4535, 3042, 3117, 4533, 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.

\(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 4583

\(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (b (A+5 C) \sec ^2(c+d x)+a (4 A+5 C) \sec (c+d x)+3 A b\right )dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (b (A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )+3 A b\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4582

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (b^2 (7 A+20 C) \sec ^2(c+d x)+a b (29 A+40 C) \sec (c+d x)+2 \left (2 (4 A+5 C) a^2+3 A b^2\right )\right )dx+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b^2 (7 A+20 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a b (29 A+40 C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 \left (2 (4 A+5 C) a^2+3 A b^2\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4562

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) \left (3 (7 A+20 C) \sec ^2(c+d x) b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b+4 a \left (2 (4 A+5 C) a^2+15 b^2 (2 A+3 C)\right ) \sec (c+d x)\right )dx\right )+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \cos ^2(c+d x) \left (3 (7 A+20 C) \sec ^2(c+d x) b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b+4 a \left (2 (4 A+5 C) a^2+15 b^2 (2 A+3 C)\right ) \sec (c+d x)\right )dx+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \frac {3 (7 A+20 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b+4 a \left (2 (4 A+5 C) a^2+15 b^2 (2 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (4 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \int \cos (c+d x)dx+\int \cos ^2(c+d x) \left (3 (7 A+20 C) \sec ^2(c+d x) b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b\right )dx\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (4 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+\int \frac {3 (7 A+20 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 3117

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\int \frac {3 (7 A+20 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^3+9 \left (5 (3 A+4 C) a^2+2 A b^2\right ) b}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {4 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \sin (c+d x)}{d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {15}{2} b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int 1dx+\frac {4 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \sin (c+d x)}{d}+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {1}{3} \left (\frac {4 a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \sin (c+d x)}{d}+\frac {9 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {15}{2} b x \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )\right )\right )+\frac {3 A b \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\)

Input:

Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]
 

Output:

(A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(5*d) + ((3*A*b*Cos 
[c + d*x]^3*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(4*d) + ((2*a*(3*A*b^2 + 
2*a^2*(4*A + 5*C))*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((15*b*(4*b^2*(A + 
 2*C) + 3*a^2*(3*A + 4*C))*x)/2 + (4*a*(15*b^2*(2*A + 3*C) + 2*a^2*(4*A + 
5*C))*Sin[c + d*x])/d + (9*b*(2*A*b^2 + 5*a^2*(3*A + 4*C))*Cos[c + d*x]*Si 
n[c + d*x])/(2*d))/3)/4)/5
 

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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3117
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; 
 FreeQ[{c, d}, x]
 

rule 4533
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

rule 4535
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

rule 4562
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si 
mp[1/(d*n)   Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* 
b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ 
{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
 

rule 4582
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e 
 + f*x])^n/(f*n)), x] - Simp[1/(d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d* 
Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs 
c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
 

rule 4583
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co 
t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( 
d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m 
 - a*(C*n + A*(n + 1))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^ 
2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && Gt 
Q[m, 0] && LeQ[n, -1]
 
Maple [A] (verified)

Time = 3.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.67

method result size
parallelrisch \(\frac {360 \left (a^{2} \left (A +C \right )+\frac {A \,b^{2}}{3}\right ) b \sin \left (2 d x +2 c \right )+\left (\left (50 A +40 C \right ) a^{3}+120 a A \,b^{2}\right ) \sin \left (3 d x +3 c \right )+45 A \,a^{2} b \sin \left (4 d x +4 c \right )+6 a^{3} A \sin \left (5 d x +5 c \right )+300 \left (a^{2} \left (A +\frac {6 C}{5}\right )+\frac {18 b^{2} \left (A +\frac {4 C}{3}\right )}{5}\right ) a \sin \left (d x +c \right )+540 \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{9}\right ) b x d}{480 d}\) \(147\)
derivativedivides \(\frac {\frac {a^{3} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {a^{3} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C a \,b^{2} \sin \left (d x +c \right )+C \,b^{3} \left (d x +c \right )}{d}\) \(201\)
default \(\frac {\frac {a^{3} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {a^{3} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C a \,b^{2} \sin \left (d x +c \right )+C \,b^{3} \left (d x +c \right )}{d}\) \(201\)
risch \(\frac {9 a^{2} A b x}{8}+\frac {A \,b^{3} x}{2}+\frac {3 C \,a^{2} b x}{2}+C \,b^{3} x +\frac {5 a^{3} A \sin \left (d x +c \right )}{8 d}+\frac {9 \sin \left (d x +c \right ) a A \,b^{2}}{4 d}+\frac {3 a^{3} C \sin \left (d x +c \right )}{4 d}+\frac {3 \sin \left (d x +c \right ) C a \,b^{2}}{d}+\frac {a^{3} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 A \,a^{2} b \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a^{3} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) a A \,b^{2}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) a^{3} C}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) A \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{2} b C}{4 d}\) \(241\)

Input:

int(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVER 
BOSE)
 

Output:

1/480*(360*(a^2*(A+C)+1/3*A*b^2)*b*sin(2*d*x+2*c)+((50*A+40*C)*a^3+120*a*A 
*b^2)*sin(3*d*x+3*c)+45*A*a^2*b*sin(4*d*x+4*c)+6*a^3*A*sin(5*d*x+5*c)+300* 
(a^2*(A+6/5*C)+18/5*b^2*(A+4/3*C))*a*sin(d*x+c)+540*((A+4/3*C)*a^2+4/9*b^2 
*(A+2*C))*b*x*d)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.70 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} d x + {\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 90 \, A a^{2} b \cos \left (d x + c\right )^{3} + 16 \, {\left (4 \, A + 5 \, C\right )} a^{3} + 120 \, {\left (2 \, A + 3 \, C\right )} a b^{2} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm= 
"fricas")
 

Output:

1/120*(15*(3*(3*A + 4*C)*a^2*b + 4*(A + 2*C)*b^3)*d*x + (24*A*a^3*cos(d*x 
+ c)^4 + 90*A*a^2*b*cos(d*x + c)^3 + 16*(4*A + 5*C)*a^3 + 120*(2*A + 3*C)* 
a*b^2 + 8*((4*A + 5*C)*a^3 + 15*A*a*b^2)*cos(d*x + c)^2 + 15*(3*(3*A + 4*C 
)*a^2*b + 4*A*b^3)*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*(a+b*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 480 \, {\left (d x + c\right )} C b^{3} + 1440 \, C a b^{2} \sin \left (d x + c\right )}{480 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm= 
"maxima")
 

Output:

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^3 - 
 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 45*(12*d*x + 12*c + sin(4*d 
*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2*b + 360*(2*d*x + 2*c + sin(2*d*x + 2 
*c))*C*a^2*b - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a*b^2 + 120*(2*d*x 
+ 2*c + sin(2*d*x + 2*c))*A*b^3 + 480*(d*x + c)*C*b^3 + 1440*C*a*b^2*sin(d 
*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 606 vs. \(2 (206) = 412\).

Time = 0.28 (sec) , antiderivative size = 606, normalized size of antiderivative = 2.78 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm= 
"giac")
 

Output:

1/120*(15*(9*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 8*C*b^3)*(d*x + c) + 2*(120* 
A*a^3*tan(1/2*d*x + 1/2*c)^9 + 120*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 225*A*a^ 
2*b*tan(1/2*d*x + 1/2*c)^9 - 180*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*A*a* 
b^2*tan(1/2*d*x + 1/2*c)^9 + 360*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*b^3 
*tan(1/2*d*x + 1/2*c)^9 + 160*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 320*C*a^3*tan 
(1/2*d*x + 1/2*c)^7 - 90*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 360*C*a^2*b*tan( 
1/2*d*x + 1/2*c)^7 + 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 1440*C*a*b^2*tan 
(1/2*d*x + 1/2*c)^7 - 120*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^3*tan(1/2 
*d*x + 1/2*c)^5 + 400*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2* 
d*x + 1/2*c)^5 + 2160*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 160*A*a^3*tan(1/2*d 
*x + 1/2*c)^3 + 320*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 90*A*a^2*b*tan(1/2*d*x 
+ 1/2*c)^3 + 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 960*A*a*b^2*tan(1/2*d*x 
+ 1/2*c)^3 + 1440*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*A*b^3*tan(1/2*d*x + 
 1/2*c)^3 + 120*A*a^3*tan(1/2*d*x + 1/2*c) + 120*C*a^3*tan(1/2*d*x + 1/2*c 
) + 225*A*a^2*b*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b*tan(1/2*d*x + 1/2*c) + 
360*A*a*b^2*tan(1/2*d*x + 1/2*c) + 360*C*a*b^2*tan(1/2*d*x + 1/2*c) + 60*A 
*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
 

Mupad [B] (verification not implemented)

Time = 14.81 (sec) , antiderivative size = 442, normalized size of antiderivative = 2.03 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (2\,A\,a^3-A\,b^3+2\,C\,a^3+6\,A\,a\,b^2-\frac {15\,A\,a^2\,b}{4}+6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,A\,a^3}{3}-2\,A\,b^3+\frac {16\,C\,a^3}{3}+16\,A\,a\,b^2-\frac {3\,A\,a^2\,b}{2}+24\,C\,a\,b^2-6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^3}{15}+\frac {20\,C\,a^3}{3}+20\,A\,a\,b^2+36\,C\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,A\,a^3}{3}+2\,A\,b^3+\frac {16\,C\,a^3}{3}+16\,A\,a\,b^2+\frac {3\,A\,a^2\,b}{2}+24\,C\,a\,b^2+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+A\,b^3+2\,C\,a^3+6\,A\,a\,b^2+\frac {15\,A\,a^2\,b}{4}+6\,C\,a\,b^2+3\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,A\,a^2+4\,A\,b^2+12\,C\,a^2+8\,C\,b^2\right )}{4\,\left (A\,b^3+2\,C\,b^3+\frac {9\,A\,a^2\,b}{4}+3\,C\,a^2\,b\right )}\right )\,\left (9\,A\,a^2+4\,A\,b^2+12\,C\,a^2+8\,C\,b^2\right )}{4\,d} \] Input:

int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^3,x)
 

Output:

(tan(c/2 + (d*x)/2)^9*(2*A*a^3 - A*b^3 + 2*C*a^3 + 6*A*a*b^2 - (15*A*a^2*b 
)/4 + 6*C*a*b^2 - 3*C*a^2*b) + tan(c/2 + (d*x)/2)^3*((8*A*a^3)/3 + 2*A*b^3 
 + (16*C*a^3)/3 + 16*A*a*b^2 + (3*A*a^2*b)/2 + 24*C*a*b^2 + 6*C*a^2*b) + t 
an(c/2 + (d*x)/2)^7*((8*A*a^3)/3 - 2*A*b^3 + (16*C*a^3)/3 + 16*A*a*b^2 - ( 
3*A*a^2*b)/2 + 24*C*a*b^2 - 6*C*a^2*b) + tan(c/2 + (d*x)/2)^5*((116*A*a^3) 
/15 + (20*C*a^3)/3 + 20*A*a*b^2 + 36*C*a*b^2) + tan(c/2 + (d*x)/2)*(2*A*a^ 
3 + A*b^3 + 2*C*a^3 + 6*A*a*b^2 + (15*A*a^2*b)/4 + 6*C*a*b^2 + 3*C*a^2*b)) 
/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x) 
/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (b*atan((b* 
tan(c/2 + (d*x)/2)*(9*A*a^2 + 4*A*b^2 + 12*C*a^2 + 8*C*b^2))/(4*(A*b^3 + 2 
*C*b^3 + (9*A*a^2*b)/4 + 3*C*a^2*b)))*(9*A*a^2 + 4*A*b^2 + 12*C*a^2 + 8*C* 
b^2))/(4*d)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.01 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-90 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3} b +225 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b +180 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b c +60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}+24 \sin \left (d x +c \right )^{5} a^{4}-80 \sin \left (d x +c \right )^{3} a^{4}-40 \sin \left (d x +c \right )^{3} a^{3} c -120 \sin \left (d x +c \right )^{3} a^{2} b^{2}+120 \sin \left (d x +c \right ) a^{4}+120 \sin \left (d x +c \right ) a^{3} c +360 \sin \left (d x +c \right ) a^{2} b^{2}+360 \sin \left (d x +c \right ) a \,b^{2} c +135 a^{3} b d x +180 a^{2} b c d x +60 a \,b^{3} d x +120 b^{3} c d x}{120 d} \] Input:

int(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x)
 

Output:

( - 90*cos(c + d*x)*sin(c + d*x)**3*a**3*b + 225*cos(c + d*x)*sin(c + d*x) 
*a**3*b + 180*cos(c + d*x)*sin(c + d*x)*a**2*b*c + 60*cos(c + d*x)*sin(c + 
 d*x)*a*b**3 + 24*sin(c + d*x)**5*a**4 - 80*sin(c + d*x)**3*a**4 - 40*sin( 
c + d*x)**3*a**3*c - 120*sin(c + d*x)**3*a**2*b**2 + 120*sin(c + d*x)*a**4 
 + 120*sin(c + d*x)*a**3*c + 360*sin(c + d*x)*a**2*b**2 + 360*sin(c + d*x) 
*a*b**2*c + 135*a**3*b*d*x + 180*a**2*b*c*d*x + 60*a*b**3*d*x + 120*b**3*c 
*d*x)/(120*d)