\(\int \cos ^6(c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [792]

Optimal result

Integrand size = 40, antiderivative size = 221 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) x+\frac {\left (4 a^3 B+14 a b^2 B+15 a^2 b C+5 b^3 C\right ) \sin (c+d x)}{5 d}+\frac {\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \sin ^3(c+d x)}{15 d} \]

[Out]

1/8*(9*B*a^2*b+4*B*b^3+3*C*a^3+12*C*a*b^2)*x+1/5*(4*B*a^3+14*B*a*b^2+15*C*a^2*b+5*C*b^3)*sin(d*x+c)/d+1/8*(9*B
*a^2*b+4*B*b^3+3*C*a^3+12*C*a*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/20*a^2*(7*B*b+5*C*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/
5*a*B*cos(d*x+c)^4*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-1/15*a*(4*B*a^2+12*B*b^2+15*C*a*b)*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4157, 4110, 4159, 4132, 2715, 8, 4129, 3092} \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {a \left (4 a^2 B+15 a b C+12 b^2 B\right ) \sin ^3(c+d x)}{15 d}+\frac {a^2 (5 a C+7 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {\left (4 a^3 B+15 a^2 b C+14 a b^2 B+5 b^3 C\right ) \sin (c+d x)}{5 d}+\frac {\left (3 a^3 C+9 a^2 b B+12 a b^2 C+4 b^3 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^3 C+9 a^2 b B+12 a b^2 C+4 b^3 B\right )+\frac {a B \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

[In]

Int[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((9*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*x)/8 + ((4*a^3*B + 14*a*b^2*B + 15*a^2*b*C + 5*b^3*C)*Sin[c + d*
x])/(5*d) + ((9*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*(7*b*B + 5*a
*C)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (a*B*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) - (a*
(4*a^2*B + 12*b^2*B + 15*a*b*C)*Sin[c + d*x]^3)/(15*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 4110

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

Rule 4129

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[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 4157

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 4159

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]
+ Dist[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]

Rubi steps \begin{align*} \text {integral}& = \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (-a (7 b B+5 a C)-\left (4 a^2 B+5 b^2 B+10 a b C\right ) \sec (c+d x)-b (2 a B+5 b C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^3(c+d x) \left (4 a \left (4 a^2 B+12 b^2 B+15 a b C\right )+5 \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \sec (c+d x)+4 b^2 (2 a B+5 b C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^3(c+d x) \left (4 a \left (4 a^2 B+12 b^2 B+15 a b C\right )+4 b^2 (2 a B+5 b C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{4} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \int \cos ^2(c+d x) \, dx \\ & = \frac {\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos (c+d x) \left (4 b^2 (2 a B+5 b C)+4 a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \cos ^2(c+d x)\right ) \, dx+\frac {1}{8} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \int 1 \, dx \\ & = \frac {1}{8} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) x+\frac {\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {\text {Subst}\left (\int \left (4 b^2 (2 a B+5 b C)+4 a \left (4 a^2 B+12 b^2 B+15 a b C\right )-4 a \left (4 a^2 B+12 b^2 B+15 a b C\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d} \\ & = \frac {1}{8} \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) x+\frac {\left (4 a^3 B+14 a b^2 B+15 a^2 b C+5 b^3 C\right ) \sin (c+d x)}{5 d}+\frac {\left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (7 b B+5 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a B \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {a \left (4 a^2 B+12 b^2 B+15 a b C\right ) \sin ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.48 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.80 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {60 \left (9 a^2 b B+4 b^3 B+3 a^3 C+12 a b^2 C\right ) (c+d x)+60 \left (5 a^3 B+18 a b^2 B+18 a^2 b C+8 b^3 C\right ) \sin (c+d x)+120 \left (3 a^2 b B+b^3 B+a^3 C+3 a b^2 C\right ) \sin (2 (c+d x))+10 a \left (5 a^2 B+12 b^2 B+12 a b C\right ) \sin (3 (c+d x))+15 a^2 (3 b B+a C) \sin (4 (c+d x))+6 a^3 B \sin (5 (c+d x))}{480 d} \]

[In]

Integrate[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(60*(9*a^2*b*B + 4*b^3*B + 3*a^3*C + 12*a*b^2*C)*(c + d*x) + 60*(5*a^3*B + 18*a*b^2*B + 18*a^2*b*C + 8*b^3*C)*
Sin[c + d*x] + 120*(3*a^2*b*B + b^3*B + a^3*C + 3*a*b^2*C)*Sin[2*(c + d*x)] + 10*a*(5*a^2*B + 12*b^2*B + 12*a*
b*C)*Sin[3*(c + d*x)] + 15*a^2*(3*b*B + a*C)*Sin[4*(c + d*x)] + 6*a^3*B*Sin[5*(c + d*x)])/(480*d)

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(\frac {120 \left (3 B \,a^{2} b +B \,b^{3}+a^{3} C +3 C a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+10 \left (5 B \,a^{3}+12 B a \,b^{2}+12 a^{2} b C \right ) \sin \left (3 d x +3 c \right )+15 \left (3 B \,a^{2} b +a^{3} C \right ) \sin \left (4 d x +4 c \right )+6 B \,a^{3} \sin \left (5 d x +5 c \right )+60 \left (5 B \,a^{3}+18 B a \,b^{2}+18 a^{2} b C +8 C \,b^{3}\right ) \sin \left (d x +c \right )+540 \left (B \,a^{2} b +\frac {4}{9} B \,b^{3}+\frac {1}{3} a^{3} C +\frac {4}{3} C a \,b^{2}\right ) x d}{480 d}\)	\(179\)
derivativedivides	\(\frac {\frac {B \,a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 B \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+a^{2} b C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+B \,b^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C a \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b^{3}}{d}\)	\(227\)
default	\(\frac {\frac {B \,a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 B \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B a \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+a^{2} b C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+B \,b^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C a \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b^{3}}{d}\)	\(227\)
risch	\(\frac {9 B \,a^{2} b x}{8}+\frac {x B \,b^{3}}{2}+\frac {3 a^{3} x C}{8}+\frac {3 x C a \,b^{2}}{2}+\frac {5 a^{3} B \sin \left (d x +c \right )}{8 d}+\frac {9 \sin \left (d x +c \right ) B a \,b^{2}}{4 d}+\frac {9 \sin \left (d x +c \right ) a^{2} b C}{4 d}+\frac {\sin \left (d x +c \right ) C \,b^{3}}{d}+\frac {B \,a^{3} \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 \sin \left (4 d x +4 c \right ) B \,a^{2} b}{32 d}+\frac {\sin \left (4 d x +4 c \right ) a^{3} C}{32 d}+\frac {5 B \,a^{3} \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) B a \,b^{2}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) a^{2} b C}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,b^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} C}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) C a \,b^{2}}{4 d}\)	\(278\)

[In]

int(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/480*(120*(3*B*a^2*b+B*b^3+C*a^3+3*C*a*b^2)*sin(2*d*x+2*c)+10*(5*B*a^3+12*B*a*b^2+12*C*a^2*b)*sin(3*d*x+3*c)+
15*(3*B*a^2*b+C*a^3)*sin(4*d*x+4*c)+6*B*a^3*sin(5*d*x+5*c)+60*(5*B*a^3+18*B*a*b^2+18*C*a^2*b+8*C*b^3)*sin(d*x+
c)+540*(B*a^2*b+4/9*B*b^3+1/3*a^3*C+4/3*C*a*b^2)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.79 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, C a^{3} + 9 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} d x + {\left (24 \, B a^{3} \cos \left (d x + c\right )^{4} + 64 \, B a^{3} + 240 \, C a^{2} b + 240 \, B a b^{2} + 120 \, C b^{3} + 30 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, B a^{3} + 15 \, C a^{2} b + 15 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, C a^{3} + 9 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(3*C*a^3 + 9*B*a^2*b + 12*C*a*b^2 + 4*B*b^3)*d*x + (24*B*a^3*cos(d*x + c)^4 + 64*B*a^3 + 240*C*a^2*b
+ 240*B*a*b^2 + 120*C*b^3 + 30*(C*a^3 + 3*B*a^2*b)*cos(d*x + c)^3 + 8*(4*B*a^3 + 15*C*a^2*b + 15*B*a*b^2)*cos
(d*x + c)^2 + 15*(3*C*a^3 + 9*B*a^2*b + 12*C*a*b^2 + 4*B*b^3)*cos(d*x + c))*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*(a+b*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.98 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b^{2} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 480 \, C b^{3} \sin \left (d x + c\right )}{480 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 672 vs. \(2 (209) = 418\).

Time = 0.33 (sec) , antiderivative size = 672, normalized size of antiderivative = 3.04 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, C a^{3} + 9 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 90 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 480 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 720 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C 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} \]

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(3*C*a^3 + 9*B*a^2*b + 12*C*a*b^2 + 4*B*b^3)*(d*x + c) + 2*(120*B*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*C*
a^3*tan(1/2*d*x + 1/2*c)^9 - 225*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a
*b^2*tan(1/2*d*x + 1/2*c)^9 - 180*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*B*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*C*b^3
*tan(1/2*d*x + 1/2*c)^9 + 160*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 30*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 90*B*a^2*b*tan(
1/2*d*x + 1/2*c)^7 + 960*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 960*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 360*C*a*b^2*tan
(1/2*d*x + 1/2*c)^7 - 120*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 480*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*B*a^3*tan(1/2*
d*x + 1/2*c)^5 + 1200*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1200*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*C*b^3*tan(1/2
*d*x + 1/2*c)^5 + 160*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 90*B*a^2*b*tan(1/2*d*x
+ 1/2*c)^3 + 960*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 960*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 360*C*a*b^2*tan(1/2*d*x
+ 1/2*c)^3 + 120*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*B*a^3*tan(1/2*d*x + 1/
2*c) + 75*C*a^3*tan(1/2*d*x + 1/2*c) + 225*B*a^2*b*tan(1/2*d*x + 1/2*c) + 360*C*a^2*b*tan(1/2*d*x + 1/2*c) + 3
60*B*a*b^2*tan(1/2*d*x + 1/2*c) + 180*C*a*b^2*tan(1/2*d*x + 1/2*c) + 60*B*b^3*tan(1/2*d*x + 1/2*c) + 120*C*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 = 17.28 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.25 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B\,b^3\,x}{2}+\frac {3\,C\,a^3\,x}{8}+\frac {9\,B\,a^2\,b\,x}{8}+\frac {3\,C\,a\,b^2\,x}{2}+\frac {5\,B\,a^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {C\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,a^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a^2\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {9\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,C\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d} \]

[In]

int(cos(c + d*x)^6*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^3,x)

[Out]

(B*b^3*x)/2 + (3*C*a^3*x)/8 + (9*B*a^2*b*x)/8 + (3*C*a*b^2*x)/2 + (5*B*a^3*sin(c + d*x))/(8*d) + (C*b^3*sin(c
+ d*x))/d + (5*B*a^3*sin(3*c + 3*d*x))/(48*d) + (B*a^3*sin(5*c + 5*d*x))/(80*d) + (B*b^3*sin(2*c + 2*d*x))/(4*
d) + (C*a^3*sin(2*c + 2*d*x))/(4*d) + (C*a^3*sin(4*c + 4*d*x))/(32*d) + (3*B*a^2*b*sin(2*c + 2*d*x))/(4*d) + (
B*a*b^2*sin(3*c + 3*d*x))/(4*d) + (3*B*a^2*b*sin(4*c + 4*d*x))/(32*d) + (3*C*a*b^2*sin(2*c + 2*d*x))/(4*d) + (
C*a^2*b*sin(3*c + 3*d*x))/(4*d) + (9*B*a*b^2*sin(c + d*x))/(4*d) + (9*C*a^2*b*sin(c + d*x))/(4*d)