Integrand size = 21, antiderivative size = 160 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {1}{8} b \left (9 a^2+4 b^2\right ) x+\frac {a \left (4 a^2+15 b^2\right ) \sin (c+d x)}{5 d}+\frac {b \left (9 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {11 a^2 b \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a^2 \cos ^4(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{5 d}-\frac {a \left (4 a^2+15 b^2\right ) \sin ^3(c+d x)}{15 d} \]
1/8*b*(9*a^2+4*b^2)*x+1/5*a*(4*a^2+15*b^2)*sin(d*x+c)/d+1/8*b*(9*a^2+4*b^2 )*cos(d*x+c)*sin(d*x+c)/d+11/20*a^2*b*cos(d*x+c)^3*sin(d*x+c)/d+1/5*a^2*co s(d*x+c)^4*(a+b*sec(d*x+c))*sin(d*x+c)/d-1/15*a*(4*a^2+15*b^2)*sin(d*x+c)^ 3/d
Time = 0.48 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.81 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {540 a^2 b c+240 b^3 c+540 a^2 b d x+240 b^3 d x+60 a \left (5 a^2+18 b^2\right ) \sin (c+d x)+120 \left (3 a^2 b+b^3\right ) \sin (2 (c+d x))+50 a^3 \sin (3 (c+d x))+120 a b^2 \sin (3 (c+d x))+45 a^2 b \sin (4 (c+d x))+6 a^3 \sin (5 (c+d x))}{480 d} \]
(540*a^2*b*c + 240*b^3*c + 540*a^2*b*d*x + 240*b^3*d*x + 60*a*(5*a^2 + 18* b^2)*Sin[c + d*x] + 120*(3*a^2*b + b^3)*Sin[2*(c + d*x)] + 50*a^3*Sin[3*(c + d*x)] + 120*a*b^2*Sin[3*(c + d*x)] + 45*a^2*b*Sin[4*(c + d*x)] + 6*a^3* Sin[5*(c + d*x)])/(480*d)
Time = 0.75 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.90, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4328, 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.
| \(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4328 |
| \(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) \left (11 b a^2+\left (4 a^2+15 b^2\right ) \sec (c+d x) a+b \left (3 a^2+5 b^2\right ) \sec ^2(c+d x)\right )dx+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {11 b a^2+\left (4 a^2+15 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+b \left (3 a^2+5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (a \left (4 a^2+15 b^2\right ) \int \cos ^3(c+d x)dx+\int \cos ^4(c+d x) \left (11 b a^2+b \left (3 a^2+5 b^2\right ) \sec ^2(c+d x)\right )dx\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (a \left (4 a^2+15 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+\int \frac {11 b a^2+b \left (3 a^2+5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {11 b a^2+b \left (3 a^2+5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {a \left (4 a^2+15 b^2\right ) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {11 b a^2+b \left (3 a^2+5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {a \left (4 a^2+15 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} b \left (9 a^2+4 b^2\right ) \int \cos ^2(c+d x)dx-\frac {a \left (4 a^2+15 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {11 a^2 b \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} b \left (9 a^2+4 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a \left (4 a^2+15 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {11 a^2 b \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} b \left (9 a^2+4 b^2\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a \left (4 a^2+15 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {11 a^2 b \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (-\frac {a \left (4 a^2+15 b^2\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5}{4} b \left (9 a^2+4 b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )+\frac {11 a^2 b \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a^2 \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))}{5 d}\) |
(a^2*Cos[c + d*x]^4*(a + b*Sec[c + d*x])*Sin[c + d*x])/(5*d) + ((11*a^2*b* Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (5*b*(9*a^2 + 4*b^2)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4 - (a*(4*a^2 + 15*b^2)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d)/5
3.5.74.3.1 Defintions of rubi rules used
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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[a^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* ((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2* (n + 1))*Csc[e + f*x] - b*(b^2*n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && ((Int egerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))
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]
Time = 1.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.71
| method | result | size |
| parallelrisch | \(\frac {\left (360 a^{2} b +120 b^{3}\right ) \sin \left (2 d x +2 c \right )+\left (50 a^{3}+120 a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+45 a^{2} b \sin \left (4 d x +4 c \right )+6 a^{3} \sin \left (5 d x +5 c \right )+\left (300 a^{3}+1080 a \,b^{2}\right ) \sin \left (d x +c \right )+540 d x b \left (a^{2}+\frac {4 b^{2}}{9}\right )}{480 d}\) | \(113\) |
| derivativedivides | \(\frac {\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(123\) |
| default | \(\frac {\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(123\) |
| risch | \(\frac {9 a^{2} b x}{8}+\frac {b^{3} x}{2}+\frac {5 a^{3} \sin \left (d x +c \right )}{8 d}+\frac {9 \sin \left (d x +c \right ) a \,b^{2}}{4 d}+\frac {a^{3} \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 a^{2} b \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 a^{3} \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) a \,b^{2}}{4 d}+\frac {3 a^{2} b \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) b^{3}}{4 d}\) | \(149\) |
| norman | \(\frac {\left (\frac {9}{8} a^{2} b +\frac {1}{2} b^{3}\right ) x +\left (-\frac {45}{8} a^{2} b -\frac {5}{2} b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {45}{8} a^{2} b -\frac {5}{2} b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {9}{8} a^{2} b +\frac {1}{2} b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {9}{8} a^{2} b +\frac {1}{2} b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {9}{8} a^{2} b +\frac {1}{2} b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {27}{8} a^{2} b +\frac {3}{2} b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {27}{8} a^{2} b +\frac {3}{2} b^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\frac {\left (8 a^{3}-15 a^{2} b +24 a \,b^{2}-4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {\left (8 a^{3}+15 a^{2} b +24 a \,b^{2}+4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (88 a^{3}-15 a^{2} b -120 a \,b^{2}+60 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 d}+\frac {\left (88 a^{3}+15 a^{2} b -120 a \,b^{2}-60 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}-\frac {8 a \left (19 a^{2}+15 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}-\frac {2 a \left (2 a^{2}-9 a b -6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {2 a \left (2 a^{2}+9 a b -6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(468\) |
1/480*((360*a^2*b+120*b^3)*sin(2*d*x+2*c)+(50*a^3+120*a*b^2)*sin(3*d*x+3*c )+45*a^2*b*sin(4*d*x+4*c)+6*a^3*sin(5*d*x+5*c)+(300*a^3+1080*a*b^2)*sin(d* x+c)+540*d*x*b*(a^2+4/9*b^2))/d
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.69 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} d x + {\left (24 \, a^{3} \cos \left (d x + c\right )^{4} + 90 \, a^{2} b \cos \left (d x + c\right )^{3} + 64 \, a^{3} + 240 \, a b^{2} + 8 \, {\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(15*(9*a^2*b + 4*b^3)*d*x + (24*a^3*cos(d*x + c)^4 + 90*a^2*b*cos(d* x + c)^3 + 64*a^3 + 240*a*b^2 + 8*(4*a^3 + 15*a*b^2)*cos(d*x + c)^2 + 15*( 9*a^2*b + 4*b^3)*cos(d*x + c))*sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3}}{480 \, d} \]
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3 + 4 5*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2*b - 480*(sin (d*x + c)^3 - 3*sin(d*x + c))*a*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c)) *b^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (148) = 296\).
Time = 0.32 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.08 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {15 \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]
1/120*(15*(9*a^2*b + 4*b^3)*(d*x + c) + 2*(120*a^3*tan(1/2*d*x + 1/2*c)^9 - 225*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60 *b^3*tan(1/2*d*x + 1/2*c)^9 + 160*a^3*tan(1/2*d*x + 1/2*c)^7 - 90*a^2*b*ta n(1/2*d*x + 1/2*c)^7 + 960*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*b^3*tan(1/2* d*x + 1/2*c)^7 + 464*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 160*a^3*tan(1/2*d*x + 1/2*c)^3 + 90*a^2*b*tan(1/2*d*x + 1/2*c) ^3 + 960*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*b^3*tan(1/2*d*x + 1/2*c)^3 + 1 20*a^3*tan(1/2*d*x + 1/2*c) + 225*a^2*b*tan(1/2*d*x + 1/2*c) + 360*a*b^2*t an(1/2*d*x + 1/2*c) + 60*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 16.28 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.79 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {\left (2\,a^3-\frac {15\,a^2\,b}{4}+6\,a\,b^2-b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,a^3}{3}-\frac {3\,a^2\,b}{2}+16\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^3}{15}+20\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,a^3}{3}+\frac {3\,a^2\,b}{2}+16\,a\,b^2+2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^3+\frac {15\,a^2\,b}{4}+6\,a\,b^2+b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^2+4\,b^2\right )}{4\,\left (\frac {9\,a^2\,b}{4}+b^3\right )}\right )\,\left (9\,a^2+4\,b^2\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)^3*(16*a*b^2 + (3*a^2*b)/2 + (8*a^3)/3 + 2*b^3) + tan(c /2 + (d*x)/2)^9*(6*a*b^2 - (15*a^2*b)/4 + 2*a^3 - b^3) + tan(c/2 + (d*x)/2 )^7*(16*a*b^2 - (3*a^2*b)/2 + (8*a^3)/3 - 2*b^3) + tan(c/2 + (d*x)/2)*(6*a *b^2 + (15*a^2*b)/4 + 2*a^3 + b^3) + tan(c/2 + (d*x)/2)^5*(20*a*b^2 + (116 *a^3)/15))/(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^2 + 4*b^2))/(4*((9*a^2*b)/4 + b^3)))*(9* a^2 + 4*b^2))/(4*d)