\(\int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\) [309]

Optimal result

Integrand size = 31, antiderivative size = 258 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {1}{8} \left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) x+\frac {\left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \sin (c+d x)}{15 d}+\frac {a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {a (8 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d} \]

[Out]

1/8*(12*A*a^3*b+16*A*a*b^3+3*B*a^4+24*B*a^2*b^2+8*B*b^4)*x+1/15*(8*A*a^4+60*A*a^2*b^2+15*A*b^4+40*B*a^3*b+60*B
*a*b^3)*sin(d*x+c)/d+1/40*a*(60*A*a^2*b+56*A*b^3+15*B*a^3+110*B*a*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/30*a^2*(8*A*a
^2+18*A*b^2+25*B*a*b)*cos(d*x+c)^2*sin(d*x+c)/d+1/20*a*(8*A*b+5*B*a)*cos(d*x+c)^3*(a+b*sec(d*x+c))^2*sin(d*x+c
)/d+1/5*a*A*cos(d*x+c)^4*(a+b*sec(d*x+c))^3*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4110, 4179, 4159, 4132, 2717, 4130, 8} \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{30 d}+\frac {a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {\left (8 a^4 A+40 a^3 b B+60 a^2 A b^2+60 a b^3 B+15 A b^4\right ) \sin (c+d x)}{15 d}+\frac {1}{8} x \left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right )+\frac {a (5 a B+8 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{20 d}+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d} \]

[In]

Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

((12*a^3*A*b + 16*a*A*b^3 + 3*a^4*B + 24*a^2*b^2*B + 8*b^4*B)*x)/8 + ((8*a^4*A + 60*a^2*A*b^2 + 15*A*b^4 + 40*
a^3*b*B + 60*a*b^3*B)*Sin[c + d*x])/(15*d) + (a*(60*a^2*A*b + 56*A*b^3 + 15*a^3*B + 110*a*b^2*B)*Cos[c + d*x]*
Sin[c + d*x])/(40*d) + (a^2*(8*a^2*A + 18*A*b^2 + 25*a*b*B)*Cos[c + d*x]^2*Sin[c + d*x])/(30*d) + (a*(8*A*b +
5*a*B)*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(20*d) + (a*A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3
*Sin[c + d*x])/(5*d)

Rule 8

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

Rule 2717

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

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 4130

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] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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 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]

Rule 4179

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] - Dist[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)))*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, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (-a (8 A b+5 a B)-\left (4 a^2 A+5 A b^2+10 a b B\right ) \sec (c+d x)-b (a A+5 b B) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a (8 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac {1}{20} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (-2 a \left (8 a^2 A+18 A b^2+25 a b B\right )-\left (44 a^2 A b+20 A b^3+15 a^3 B+60 a b^2 B\right ) \sec (c+d x)-b \left (12 a A b+5 a^2 B+20 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {a (8 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{60} \int \cos ^2(c+d x) \left (3 a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right )+4 \left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \sec (c+d x)+3 b^2 \left (12 a A b+5 a^2 B+20 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {a (8 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{60} \int \cos ^2(c+d x) \left (3 a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right )+3 b^2 \left (12 a A b+5 a^2 B+20 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{15} \left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \int \cos (c+d x) \, dx \\ & = \frac {\left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \sin (c+d x)}{15 d}+\frac {a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {a (8 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{8} \left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) \int 1 \, dx \\ & = \frac {1}{8} \left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) x+\frac {\left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \sin (c+d x)}{15 d}+\frac {a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {a (8 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.79 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.02 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {720 a^3 A b c+960 a A b^3 c+180 a^4 B c+1440 a^2 b^2 B c+480 b^4 B c+720 a^3 A b d x+960 a A b^3 d x+180 a^4 B d x+1440 a^2 b^2 B d x+480 b^4 B d x+60 \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \sin (c+d x)+120 a \left (4 a^2 A b+4 A b^3+a^3 B+6 a b^2 B\right ) \sin (2 (c+d x))+50 a^4 A \sin (3 (c+d x))+240 a^2 A b^2 \sin (3 (c+d x))+160 a^3 b B \sin (3 (c+d x))+60 a^3 A b \sin (4 (c+d x))+15 a^4 B \sin (4 (c+d x))+6 a^4 A \sin (5 (c+d x))}{480 d} \]

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(720*a^3*A*b*c + 960*a*A*b^3*c + 180*a^4*B*c + 1440*a^2*b^2*B*c + 480*b^4*B*c + 720*a^3*A*b*d*x + 960*a*A*b^3*
d*x + 180*a^4*B*d*x + 1440*a^2*b^2*B*d*x + 480*b^4*B*d*x + 60*(5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*a^3*b*B +
32*a*b^3*B)*Sin[c + d*x] + 120*a*(4*a^2*A*b + 4*A*b^3 + a^3*B + 6*a*b^2*B)*Sin[2*(c + d*x)] + 50*a^4*A*Sin[3*
(c + d*x)] + 240*a^2*A*b^2*Sin[3*(c + d*x)] + 160*a^3*b*B*Sin[3*(c + d*x)] + 60*a^3*A*b*Sin[4*(c + d*x)] + 15*
a^4*B*Sin[4*(c + d*x)] + 6*a^4*A*Sin[5*(c + d*x)])/(480*d)

Maple [A] (verified)

Time = 3.64 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(\frac {\left (480 A \,a^{3} b +480 A a \,b^{3}+120 B \,a^{4}+720 B \,a^{2} b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (50 a^{4} A +240 A \,a^{2} b^{2}+160 B \,a^{3} b \right ) \sin \left (3 d x +3 c \right )+\left (60 A \,a^{3} b +15 B \,a^{4}\right ) \sin \left (4 d x +4 c \right )+6 a^{4} A \sin \left (5 d x +5 c \right )+\left (300 a^{4} A +2160 A \,a^{2} b^{2}+480 A \,b^{4}+1440 B \,a^{3} b +1920 B a \,b^{3}\right ) \sin \left (d x +c \right )+720 \left (A \,a^{3} b +\frac {4}{3} A a \,b^{3}+\frac {1}{4} B \,a^{4}+2 B \,a^{2} b^{2}+\frac {2}{3} B \,b^{4}\right ) d x}{480 d}\)	\(201\)
derivativedivides	\(\frac {\frac {a^{4} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+4 A \,a^{3} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{2} b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {4 B \,a^{3} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 A a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 B \,a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{4} \sin \left (d x +c \right )+4 B a \,b^{3} \sin \left (d x +c \right )+B \,b^{4} \left (d x +c \right )}{d}\)	\(258\)
default	\(\frac {\frac {a^{4} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+4 A \,a^{3} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{2} b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {4 B \,a^{3} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 A a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 B \,a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{4} \sin \left (d x +c \right )+4 B a \,b^{3} \sin \left (d x +c \right )+B \,b^{4} \left (d x +c \right )}{d}\)	\(258\)
risch	\(\frac {3 A \,a^{3} b x}{2}+2 A a \,b^{3} x +\frac {3 a^{4} x B}{8}+3 B \,a^{2} b^{2} x +x B \,b^{4}+\frac {5 \sin \left (d x +c \right ) a^{4} A}{8 d}+\frac {9 \sin \left (d x +c \right ) A \,a^{2} b^{2}}{2 d}+\frac {\sin \left (d x +c \right ) A \,b^{4}}{d}+\frac {3 \sin \left (d x +c \right ) B \,a^{3} b}{d}+\frac {4 \sin \left (d x +c \right ) B a \,b^{3}}{d}+\frac {a^{4} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) A \,a^{3} b}{8 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{4}}{32 d}+\frac {5 a^{4} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{2} b^{2}}{2 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{3} b}{3 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{3} b}{d}+\frac {\sin \left (2 d x +2 c \right ) A a \,b^{3}}{d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{4}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B \,a^{2} b^{2}}{2 d}\)	\(308\)

[In]

int(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/480*((480*A*a^3*b+480*A*a*b^3+120*B*a^4+720*B*a^2*b^2)*sin(2*d*x+2*c)+(50*A*a^4+240*A*a^2*b^2+160*B*a^3*b)*s
in(3*d*x+3*c)+(60*A*a^3*b+15*B*a^4)*sin(4*d*x+4*c)+6*a^4*A*sin(5*d*x+5*c)+(300*A*a^4+2160*A*a^2*b^2+480*A*b^4+
1440*B*a^3*b+1920*B*a*b^3)*sin(d*x+c)+720*(A*a^3*b+4/3*A*a*b^3+1/4*B*a^4+2*B*a^2*b^2+2/3*B*b^4)*d*x)/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.76 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 8 \, B b^{4}\right )} d x + {\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 64 \, A a^{4} + 320 \, B a^{3} b + 480 \, A a^{2} b^{2} + 480 \, B a b^{3} + 120 \, A b^{4} + 30 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (2 \, A a^{4} + 10 \, B a^{3} b + 15 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*d*x + (24*A*a^4*cos(d*x + c)^4 + 64*A*a
^4 + 320*B*a^3*b + 480*A*a^2*b^2 + 480*B*a*b^3 + 120*A*b^4 + 30*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^3 + 16*(2*A*a
^4 + 10*B*a^3*b + 15*A*a^2*b^2)*cos(d*x + c)^2 + 15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3)*cos(d*x
+ c))*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.95 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 60 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} b - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b^{2} + 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b^{2} + 480 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{3} + 480 \, {\left (d x + c\right )} B b^{4} + 1920 \, B a b^{3} \sin \left (d x + c\right ) + 480 \, A b^{4} \sin \left (d x + c\right )}{480 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c
) + 8*sin(2*d*x + 2*c))*B*a^4 + 60*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3*b - 640*(sin(
d*x + c)^3 - 3*sin(d*x + c))*B*a^3*b - 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2*b^2 + 720*(2*d*x + 2*c + si
n(2*d*x + 2*c))*B*a^2*b^2 + 480*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b^3 + 480*(d*x + c)*B*b^4 + 1920*B*a*b^3*
sin(d*x + c) + 480*A*b^4*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 791 vs. \(2 (246) = 492\).

Time = 0.35 (sec) , antiderivative size = 791, normalized size of antiderivative = 3.07 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*(d*x + c) + 2*(120*A*a^4*tan(1/2*d*x +
1/2*c)^9 - 75*B*a^4*tan(1/2*d*x + 1/2*c)^9 - 300*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 480*B*a^3*b*tan(1/2*d*x + 1/
2*c)^9 + 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 360*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 240*A*a*b^3*tan(1/2*d*x
+ 1/2*c)^9 + 480*B*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*A*b^4*tan(1/2*d*x + 1/2*c)^9 + 160*A*a^4*tan(1/2*d*x +
1/2*c)^7 - 30*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 120*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 1280*B*a^3*b*tan(1/2*d*x + 1
/2*c)^7 + 1920*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 720*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 480*A*a*b^3*tan(1/2*d
*x + 1/2*c)^7 + 1920*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 480*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^4*tan(1/2*d*x
+ 1/2*c)^5 + 1600*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 2400*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 2880*B*a*b^3*tan(1
/2*d*x + 1/2*c)^5 + 720*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 160*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^4*tan(1/2*d*x
+ 1/2*c)^3 + 120*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 1280*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 1920*A*a^2*b^2*tan(1/
2*d*x + 1/2*c)^3 + 720*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 480*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 1920*B*a*b^3*ta
n(1/2*d*x + 1/2*c)^3 + 480*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^4*tan(1/2*d*x + 1/2*c) + 75*B*a^4*tan(1/2*d*
x + 1/2*c) + 300*A*a^3*b*tan(1/2*d*x + 1/2*c) + 480*B*a^3*b*tan(1/2*d*x + 1/2*c) + 720*A*a^2*b^2*tan(1/2*d*x +
1/2*c) + 360*B*a^2*b^2*tan(1/2*d*x + 1/2*c) + 240*A*a*b^3*tan(1/2*d*x + 1/2*c) + 480*B*a*b^3*tan(1/2*d*x + 1/
2*c) + 120*A*b^4*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 = 15.22 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.19 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {3\,B\,a^4\,x}{8}+B\,b^4\,x+2\,A\,a\,b^3\,x+\frac {3\,A\,a^3\,b\,x}{2}+\frac {5\,A\,a^4\,\sin \left (c+d\,x\right )}{8\,d}+\frac {A\,b^4\,\sin \left (c+d\,x\right )}{d}+3\,B\,a^2\,b^2\,x+\frac {5\,A\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {A\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {B\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {A\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {A\,a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {A\,a^3\,b\,\sin \left (4\,c+4\,d\,x\right )}{8\,d}+\frac {9\,A\,a^2\,b^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {B\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {A\,a^2\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{2\,d}+\frac {3\,B\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {4\,B\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,B\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \]

[In]

int(cos(c + d*x)^5*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^4,x)

[Out]

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