Integrand size = 31, antiderivative size = 131 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a (3 A+4 C) x+\frac {a (4 A+5 C) \sin (c+d x)}{5 d}+\frac {a (3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 C) \sin ^3(c+d x)}{15 d} \] Output:
1/8*a*(3*A+4*C)*x+1/5*a*(4*A+5*C)*sin(d*x+c)/d+1/8*a*(3*A+4*C)*cos(d*x+c)* sin(d*x+c)/d+1/4*a*A*cos(d*x+c)^3*sin(d*x+c)/d+1/5*a*A*cos(d*x+c)^4*sin(d* x+c)/d-1/15*a*(4*A+5*C)*sin(d*x+c)^3/d
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.66 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (480 (A+C) \sin (c+d x)-160 (2 A+C) \sin ^3(c+d x)+96 A \sin ^5(c+d x)+15 (4 (3 A+4 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+A \sin (4 (c+d x)))\right )}{480 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]
Output:
(a*(480*(A + C)*Sin[c + d*x] - 160*(2*A + C)*Sin[c + d*x]^3 + 96*A*Sin[c + d*x]^5 + 15*(4*(3*A + 4*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + A*Sin [4*(c + d*x)])))/(480*d)
Time = 0.67 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4563, 25, 3042, 4535, 3042, 3113, 2009, 4533, 3042, 3115, 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.
| \(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4563 |
| \(\displaystyle \frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}-\frac {1}{5} \int -\cos ^4(c+d x) \left (5 a C \sec ^2(c+d x)+a (4 A+5 C) \sec (c+d x)+5 a A\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) \left (5 a C \sec ^2(c+d x)+a (4 A+5 C) \sec (c+d x)+5 a A\right )dx+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )+5 a A}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (a (4 A+5 C) \int \cos ^3(c+d x)dx+\int \cos ^4(c+d x) \left (5 a C \sec ^2(c+d x)+5 a A\right )dx\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (a (4 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+\int \frac {5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {a (4 A+5 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (3 A+4 C) \int \cos ^2(c+d x)dx-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (3 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (3 A+4 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (-\frac {a (4 A+5 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5}{4} a (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )+\frac {5 a A \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]
Output:
(a*A*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) + ((5*a*A*Cos[c + d*x]^3*Sin[c + d *x])/(4*d) + (5*a*(3*A + 4*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4 - (a*(4*A + 5*C)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d)/5
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
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]
Int[((A_.) + 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] + Simp[1/(d*n) Int[(d*Csc[e + f*x] )^(n + 1)*Simp[A*b*n + a*(C*n + 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, C}, x] && LtQ[n, -1]
Time = 0.51 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.66
| method | result | size |
| parallelrisch | \(\frac {\left (8 \left (A +C \right ) \sin \left (2 d x +2 c \right )+\frac {2 \left (5 A +4 C \right ) \sin \left (3 d x +3 c \right )}{3}+A \sin \left (4 d x +4 c \right )+\frac {2 A \sin \left (5 d x +5 c \right )}{5}+4 \left (5 A +6 C \right ) \sin \left (d x +c \right )+12 x d \left (A +\frac {4 C}{3}\right )\right ) a}{32 d}\) | \(87\) |
| derivativedivides | \(\frac {\frac {a A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(117\) |
| default | \(\frac {\frac {a A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(117\) |
| risch | \(\frac {3 a A x}{8}+\frac {a x C}{2}+\frac {5 a A \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) C a}{4 d}+\frac {a A \sin \left (5 d x +5 c \right )}{80 d}+\frac {a A \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) C a}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\) | \(134\) |
| norman | \(\frac {\frac {a \left (3 A +4 C \right ) x}{8}+\frac {2 a \left (A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {a \left (3 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {3 a \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {a \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}-\frac {5 a \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}-\frac {5 a \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {a \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {3 a \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {a \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}-\frac {2 a \left (5 A -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {a \left (13 A +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {8 a \left (19 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}+\frac {a \left (83 A +20 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 d}+\frac {a \left (93 A -100 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\) | \(362\) |
Input:
int(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBO SE)
Output:
1/32*(8*(A+C)*sin(2*d*x+2*c)+2/3*(5*A+4*C)*sin(3*d*x+3*c)+A*sin(4*d*x+4*c) +2/5*A*sin(5*d*x+5*c)+4*(5*A+6*C)*sin(d*x+c)+12*x*d*(A+4/3*C))*a/d
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.72 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, A + 4 \, C\right )} a d x + {\left (24 \, A a \cos \left (d x + c\right )^{4} + 30 \, A a \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A + 4 \, C\right )} a \cos \left (d x + c\right ) + 16 \, {\left (4 \, A + 5 \, C\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="f ricas")
Output:
1/120*(15*(3*A + 4*C)*a*d*x + (24*A*a*cos(d*x + c)^4 + 30*A*a*cos(d*x + c) ^3 + 8*(4*A + 5*C)*a*cos(d*x + c)^2 + 15*(3*A + 4*C)*a*cos(d*x + c) + 16*( 4*A + 5*C)*a)*sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a}{480 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="m axima")
Output:
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a + 1 5*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a - 160*(sin(d *x + c)^3 - 3*sin(d*x + c))*C*a + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a )/d
Time = 0.33 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.42 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, A a + 4 \, C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 130 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 200 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 190 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 440 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="g iac")
Output:
1/120*(15*(3*A*a + 4*C*a)*(d*x + c) + 2*(45*A*a*tan(1/2*d*x + 1/2*c)^9 + 6 0*C*a*tan(1/2*d*x + 1/2*c)^9 + 130*A*a*tan(1/2*d*x + 1/2*c)^7 + 200*C*a*ta n(1/2*d*x + 1/2*c)^7 + 464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*C*a*tan(1/2*d* x + 1/2*c)^5 + 190*A*a*tan(1/2*d*x + 1/2*c)^3 + 440*C*a*tan(1/2*d*x + 1/2* c)^3 + 195*A*a*tan(1/2*d*x + 1/2*c) + 180*C*a*tan(1/2*d*x + 1/2*c))/(tan(1 /2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 14.42 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.66 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {3\,A\,a}{4}+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {13\,A\,a}{6}+\frac {10\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {19\,A\,a}{6}+\frac {22\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,C\right )}{4\,\left (\frac {3\,A\,a}{4}+C\,a\right )}\right )\,\left (3\,A+4\,C\right )}{4\,d} \] Input:
int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x)),x)
Output:
(tan(c/2 + (d*x)/2)*((13*A*a)/4 + 3*C*a) + tan(c/2 + (d*x)/2)^9*((3*A*a)/4 + C*a) + tan(c/2 + (d*x)/2)^7*((13*A*a)/6 + (10*C*a)/3) + tan(c/2 + (d*x) /2)^3*((19*A*a)/6 + (22*C*a)/3) + tan(c/2 + (d*x)/2)^5*((116*A*a)/15 + (20 *C*a)/3))/(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)) + (a *atan((a*tan(c/2 + (d*x)/2)*(3*A + 4*C))/(4*((3*A*a)/4 + C*a)))*(3*A + 4*C ))/(4*d)
Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +24 \sin \left (d x +c \right )^{5} a -80 \sin \left (d x +c \right )^{3} a -40 \sin \left (d x +c \right )^{3} c +120 \sin \left (d x +c \right ) a +120 \sin \left (d x +c \right ) c +45 a d x +60 c d x \right )}{120 d} \] Input:
int(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x)
Output:
(a*( - 30*cos(c + d*x)*sin(c + d*x)**3*a + 75*cos(c + d*x)*sin(c + d*x)*a + 60*cos(c + d*x)*sin(c + d*x)*c + 24*sin(c + d*x)**5*a - 80*sin(c + d*x)* *3*a - 40*sin(c + d*x)**3*c + 120*sin(c + d*x)*a + 120*sin(c + d*x)*c + 45 *a*d*x + 60*c*d*x))/(120*d)