\(\int \cos ^4(c+d x) (a+b \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [652]

Optimal result

Integrand size = 33, antiderivative size = 145 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac {2 a b (2 A+3 C) \sin (c+d x)}{3 d}+\frac {\left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d} \]

[Out]

1/8*(4*b^2*(A+2*C)+a^2*(3*A+4*C))*x+2/3*a*b*(2*A+3*C)*sin(d*x+c)/d+1/8*(2*A*b^2+a^2*(3*A+4*C))*cos(d*x+c)*sin(
d*x+c)/d+1/6*a*A*b*cos(d*x+c)^2*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+b*sec(d*x+c))^2*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4180, 4159, 4132, 2717, 4130, 8} \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right )+\frac {2 a b (2 A+3 C) \sin (c+d x)}{3 d}+\frac {a A b \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

((4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*x)/8 + (2*a*b*(2*A + 3*C)*Sin[c + d*x])/(3*d) + ((2*A*b^2 + a^2*(3*A + 4*
C))*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*A*b*Cos[c + d*x]^2*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]^3*(a + b*Se
c[c + d*x])^2*Sin[c + d*x])/(4*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 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 4180

Int[((A_.) + 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] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - 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, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (2 A b+a (3 A+4 C) \sec (c+d x)+b (A+4 C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{12} \int \cos ^2(c+d x) \left (-3 \left (2 A b^2+a^2 (3 A+4 C)\right )-8 a b (2 A+3 C) \sec (c+d x)-3 b^2 (A+4 C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{12} \int \cos ^2(c+d x) \left (-3 \left (2 A b^2+a^2 (3 A+4 C)\right )-3 b^2 (A+4 C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{3} (2 a b (2 A+3 C)) \int \cos (c+d x) \, dx \\ & = \frac {2 a b (2 A+3 C) \sin (c+d x)}{3 d}+\frac {\left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}-\frac {1}{8} \left (-4 b^2 (A+2 C)-a^2 (3 A+4 C)\right ) \int 1 \, dx \\ & = \frac {1}{8} \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac {2 a b (2 A+3 C) \sin (c+d x)}{3 d}+\frac {\left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a A b \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.72 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {12 \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) (c+d x)+48 a b (3 A+4 C) \sin (c+d x)+24 \left (A b^2+a^2 (A+C)\right ) \sin (2 (c+d x))+16 a A b \sin (3 (c+d x))+3 a^2 A \sin (4 (c+d x))}{96 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(12*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*(c + d*x) + 48*a*b*(3*A + 4*C)*Sin[c + d*x] + 24*(A*b^2 + a^2*(A + C))
*Sin[2*(c + d*x)] + 16*a*A*b*Sin[3*(c + d*x)] + 3*a^2*A*Sin[4*(c + d*x)])/(96*d)

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(\frac {24 \left (\left (A +C \right ) a^{2}+A \,b^{2}\right ) \sin \left (2 d x +2 c \right )+16 a A b \sin \left (3 d x +3 c \right )+3 a^{2} A \sin \left (4 d x +4 c \right )+144 a b \left (A +\frac {4 C}{3}\right ) \sin \left (d x +c \right )+36 x d \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{3}\right )}{96 d}\)	\(99\)
derivativedivides	\(\frac {a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a A b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \sin \left (d x +c \right )+C \,b^{2} \left (d x +c \right )}{d}\)	\(140\)
default	\(\frac {a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a A b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \sin \left (d x +c \right )+C \,b^{2} \left (d x +c \right )}{d}\)	\(140\)
risch	\(\frac {3 a^{2} A x}{8}+\frac {x A \,b^{2}}{2}+\frac {a^{2} x C}{2}+x C \,b^{2}+\frac {3 a A b \sin \left (d x +c \right )}{2 d}+\frac {2 \sin \left (d x +c \right ) C a b}{d}+\frac {a^{2} A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a A b \sin \left (3 d x +3 c \right )}{6 d}+\frac {a^{2} A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{4 d}\)	\(146\)
norman	\(\frac {\left (-\frac {3}{8} a^{2} A -\frac {1}{2} A \,b^{2}-\frac {1}{2} C \,a^{2}-C \,b^{2}\right ) x +\left (-\frac {9}{8} a^{2} A -\frac {3}{2} A \,b^{2}-\frac {3}{2} C \,a^{2}-3 C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-\frac {9}{8} a^{2} A -\frac {3}{2} A \,b^{2}-\frac {3}{2} C \,a^{2}-3 C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-\frac {3}{8} a^{2} A -\frac {1}{2} A \,b^{2}-\frac {1}{2} C \,a^{2}-C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{8} a^{2} A +\frac {1}{2} A \,b^{2}+\frac {1}{2} C \,a^{2}+C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {3}{8} a^{2} A +\frac {1}{2} A \,b^{2}+\frac {1}{2} C \,a^{2}+C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {9}{8} a^{2} A +\frac {3}{2} A \,b^{2}+\frac {3}{2} C \,a^{2}+3 C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {9}{8} a^{2} A +\frac {3}{2} A \,b^{2}+\frac {3}{2} C \,a^{2}+3 C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (7 a^{2} A -4 A \,b^{2}-4 C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (27 a^{2} A -32 a A b +12 A \,b^{2}+12 C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {\left (27 a^{2} A +32 a A b +12 A \,b^{2}+12 C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {\left (5 a^{2} A -16 a A b +4 A \,b^{2}+4 C \,a^{2}-16 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}-\frac {\left (5 a^{2} A +16 a A b +4 A \,b^{2}+4 C \,a^{2}+16 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (81 a^{2} A -16 a A b -12 A \,b^{2}-12 C \,a^{2}-144 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {\left (81 a^{2} A +16 a A b -12 A \,b^{2}-12 C \,a^{2}+144 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}\)	\(616\)

[In]

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/96*(24*((A+C)*a^2+A*b^2)*sin(2*d*x+2*c)+16*a*A*b*sin(3*d*x+3*c)+3*a^2*A*sin(4*d*x+4*c)+144*a*b*(A+4/3*C)*sin
(d*x+c)+36*x*d*((A+4/3*C)*a^2+4/3*b^2*(A+2*C)))/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.72 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \, {\left (A + 2 \, C\right )} b^{2}\right )} d x + {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 16 \, A a b \cos \left (d x + c\right )^{2} + 16 \, {\left (2 \, A + 3 \, C\right )} a b + 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \, A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*((3*A + 4*C)*a^2 + 4*(A + 2*C)*b^2)*d*x + (6*A*a^2*cos(d*x + c)^3 + 16*A*a*b*cos(d*x + c)^2 + 16*(2*A
+ 3*C)*a*b + 3*((3*A + 4*C)*a^2 + 4*A*b^2)*cos(d*x + c))*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.90 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} + 96 \, {\left (d x + c\right )} C b^{2} + 192 \, C a b \sin \left (d x + c\right )}{96 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2 + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*
a^2 - 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a*b + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^2 + 96*(d*x + c)*C*
b^2 + 192*C*a*b*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (135) = 270\).

Time = 0.31 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.61 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 8 \, C b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 144 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 144 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(3*A*a^2 + 4*C*a^2 + 4*A*b^2 + 8*C*b^2)*(d*x + c) - 2*(15*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^2*tan(
1/2*d*x + 1/2*c)^7 - 48*A*a*b*tan(1/2*d*x + 1/2*c)^7 - 48*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^2*tan(1/2*d*x
+ 1/2*c)^7 - 9*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 80*A*a*b*tan(1/2*d*x + 1/2*c)^
5 - 144*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*C
*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 144*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*ta
n(1/2*d*x + 1/2*c)^3 - 15*A*a^2*tan(1/2*d*x + 1/2*c) - 12*C*a^2*tan(1/2*d*x + 1/2*c) - 48*A*a*b*tan(1/2*d*x +
1/2*c) - 48*C*a*b*tan(1/2*d*x + 1/2*c) - 12*A*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

Mupad [B] (verification not implemented)

Time = 15.62 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,A\,a^2\,x}{8}+\frac {A\,b^2\,x}{2}+\frac {C\,a^2\,x}{2}+C\,b^2\,x+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {A\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,A\,a\,b\,\sin \left (c+d\,x\right )}{2\,d}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{6\,d} \]

[In]

int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^2,x)

[Out]

(3*A*a^2*x)/8 + (A*b^2*x)/2 + (C*a^2*x)/2 + C*b^2*x + (A*a^2*sin(2*c + 2*d*x))/(4*d) + (A*a^2*sin(4*c + 4*d*x)
)/(32*d) + (A*b^2*sin(2*c + 2*d*x))/(4*d) + (C*a^2*sin(2*c + 2*d*x))/(4*d) + (3*A*a*b*sin(c + d*x))/(2*d) + (2
*C*a*b*sin(c + d*x))/d + (A*a*b*sin(3*c + 3*d*x))/(6*d)