Integrand size = 31, antiderivative size = 158 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {7}{8} a^4 (4 A+5 B) x+\frac {8 a^4 (4 A+5 B) \sin (c+d x)}{5 d}+\frac {27 a^4 (4 A+5 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a^4 (4 A+5 B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{5 d}-\frac {4 a^4 (4 A+5 B) \sin ^3(c+d x)}{15 d} \]
7/8*a^4*(4*A+5*B)*x+8/5*a^4*(4*A+5*B)*sin(d*x+c)/d+27/40*a^4*(4*A+5*B)*cos (d*x+c)*sin(d*x+c)/d+1/20*a^4*(4*A+5*B)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*A*co s(d*x+c)^4*(a+a*sec(d*x+c))^4*sin(d*x+c)/d-4/15*a^4*(4*A+5*B)*sin(d*x+c)^3 /d
Time = 0.67 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.75 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 \sin (c+d x) \left (664 A+800 B+15 (28 A+27 B) \cos (c+d x)+16 (17 A+10 B) \cos ^2(c+d x)+30 (4 A+B) \cos ^3(c+d x)+24 A \cos ^4(c+d x)+\frac {210 (4 A+5 B) \arcsin \left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{\sqrt {\sin ^2(c+d x)}}\right )}{120 d} \]
(a^4*Sin[c + d*x]*(664*A + 800*B + 15*(28*A + 27*B)*Cos[c + d*x] + 16*(17* A + 10*B)*Cos[c + d*x]^2 + 30*(4*A + B)*Cos[c + d*x]^3 + 24*A*Cos[c + d*x] ^4 + (210*(4*A + 5*B)*ArcSin[Sqrt[Sin[(c + d*x)/2]^2]])/Sqrt[Sin[c + d*x]^ 2]))/(120*d)
Time = 0.49 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 4501, 3042, 4278, 2009}
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.
| \(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a)^4 (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \int \cos ^4(c+d x) (\sec (c+d x) a+a)^4dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 4278 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \int \left (\cos ^4(c+d x) a^4+4 \cos ^3(c+d x) a^4+6 \cos ^2(c+d x) a^4+4 \cos (c+d x) a^4+a^4\right )dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} (4 A+5 B) \left (-\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {8 a^4 \sin (c+d x)}{d}+\frac {a^4 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {27 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {35 a^4 x}{8}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
(A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(5*d) + ((4*A + 5*B )*((35*a^4*x)/8 + (8*a^4*Sin[c + d*x])/d + (27*a^4*Cos[c + d*x]*Sin[c + d* x])/(8*d) + (a^4*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (4*a^4*Sin[c + d*x]^ 3)/(3*d)))/5
3.1.79.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f *x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I GtQ[m, 0] && RationalQ[n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), 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[(a*A*m - b*B*n)/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] , x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a ^2 - b^2, 0] && EqQ[m + n + 1, 0] && !LeQ[m, -1]
Time = 3.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.59
| method | result | size |
| parallelrisch | \(\frac {\left (\left (16 A +14 B \right ) \sin \left (2 d x +2 c \right )+\left (\frac {29 A}{6}+\frac {8 B}{3}\right ) \sin \left (3 d x +3 c \right )+\left (A +\frac {B}{4}\right ) \sin \left (4 d x +4 c \right )+\frac {A \sin \left (5 d x +5 c \right )}{10}+\left (49 A +56 B \right ) \sin \left (d x +c \right )+28 d \left (A +\frac {5 B}{4}\right ) x \right ) a^{4}}{8 d}\) | \(94\) |
| risch | \(\frac {7 a^{4} A x}{2}+\frac {35 a^{4} x B}{8}+\frac {49 \sin \left (d x +c \right ) a^{4} A}{8 d}+\frac {7 \sin \left (d x +c \right ) B \,a^{4}}{d}+\frac {a^{4} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{4} A \sin \left (4 d x +4 c \right )}{8 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{4}}{32 d}+\frac {29 a^{4} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{4}}{3 d}+\frac {2 \sin \left (2 d x +2 c \right ) a^{4} A}{d}+\frac {7 \sin \left (2 d x +2 c \right ) B \,a^{4}}{4 d}\) | \(172\) |
| derivativedivides | \(\frac {a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{4} \sin \left (d x +c \right )+2 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+6 B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {a^{4} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(248\) |
| default | \(\frac {a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{4} \sin \left (d x +c \right )+2 a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+6 B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 B \,a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {a^{4} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(248\) |
1/8*((16*A+14*B)*sin(2*d*x+2*c)+(29/6*A+8/3*B)*sin(3*d*x+3*c)+(A+1/4*B)*si n(4*d*x+4*c)+1/10*A*sin(5*d*x+5*c)+(49*A+56*B)*sin(d*x+c)+28*d*(A+5/4*B)*x )*a^4/d
Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.70 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {105 \, {\left (4 \, A + 5 \, B\right )} a^{4} d x + {\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 30 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 16 \, {\left (17 \, A + 10 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \, {\left (28 \, A + 27 \, B\right )} a^{4} \cos \left (d x + c\right ) + 8 \, {\left (83 \, A + 100 \, B\right )} a^{4}\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(105*(4*A + 5*B)*a^4*d*x + (24*A*a^4*cos(d*x + c)^4 + 30*(4*A + B)*a ^4*cos(d*x + c)^3 + 16*(17*A + 10*B)*a^4*cos(d*x + c)^2 + 15*(28*A + 27*B) *a^4*cos(d*x + c) + 8*(83*A + 100*B)*a^4)*sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.49 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} + 60 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 480 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 480 \, {\left (d x + c\right )} B a^{4} + 480 \, A a^{4} \sin \left (d x + c\right ) + 1920 \, B a^{4} \sin \left (d x + c\right )}{480 \, d} \]
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 - 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 + 60*(12*d*x + 12*c + sin(4*d *x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 + 480*(2*d*x + 2*c + sin(2*d*x + 2*c ))*A*a^4 - 640*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^4 + 720*(2*d*x + 2*c + sin(2 *d*x + 2*c))*B*a^4 + 480*(d*x + c)*B*a^4 + 480*A*a^4*sin(d*x + c) + 1920*B *a^4*sin(d*x + c))/d
Time = 0.34 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.33 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {105 \, {\left (4 \, A a^{4} + 5 \, B a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (420 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1960 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2450 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3584 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3160 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3950 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1500 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]
1/120*(105*(4*A*a^4 + 5*B*a^4)*(d*x + c) + 2*(420*A*a^4*tan(1/2*d*x + 1/2* c)^9 + 525*B*a^4*tan(1/2*d*x + 1/2*c)^9 + 1960*A*a^4*tan(1/2*d*x + 1/2*c)^ 7 + 2450*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 3584*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 4480*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 3160*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 3950*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 1500*A*a^4*tan(1/2*d*x + 1/2*c) + 1395 *B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 16.23 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.57 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {\left (7\,A\,a^4+\frac {35\,B\,a^4}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {98\,A\,a^4}{3}+\frac {245\,B\,a^4}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {896\,A\,a^4}{15}+\frac {224\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {158\,A\,a^4}{3}+\frac {395\,B\,a^4}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (25\,A\,a^4+\frac {93\,B\,a^4}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {7\,a^4\,\mathrm {atan}\left (\frac {7\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+5\,B\right )}{4\,\left (7\,A\,a^4+\frac {35\,B\,a^4}{4}\right )}\right )\,\left (4\,A+5\,B\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*(25*A*a^4 + (93*B*a^4)/4) + tan(c/2 + (d*x)/2)^9*(7*A* a^4 + (35*B*a^4)/4) + tan(c/2 + (d*x)/2)^7*((98*A*a^4)/3 + (245*B*a^4)/6) + tan(c/2 + (d*x)/2)^3*((158*A*a^4)/3 + (395*B*a^4)/6) + tan(c/2 + (d*x)/2 )^5*((896*A*a^4)/15 + (224*B*a^4)/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)) + (7*a^4*atan((7*a^4*tan(c/2 + (d*x)/2)*(4*A + 5*B ))/(4*(7*A*a^4 + (35*B*a^4)/4)))*(4*A + 5*B))/(4*d)