3.300 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=179 \[ \frac {a \left (3 a^2 A+12 a b B+10 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a^2 (2 a B+3 A b) \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {\left (2 a^3 B+6 a^2 A b+9 a b^2 B+3 A b^3\right ) \sin (c+d x)}{3 d}+\frac {1}{8} x \left (3 a^3 A+12 a^2 b B+12 a A b^2+8 b^3 B\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]

[Out]

1/8*(3*A*a^3+12*A*a*b^2+12*B*a^2*b+8*B*b^3)*x+1/3*(6*A*a^2*b+3*A*b^3+2*B*a^3+9*B*a*b^2)*sin(d*x+c)/d+1/8*a*(3*
A*a^2+10*A*b^2+12*B*a*b)*cos(d*x+c)*sin(d*x+c)/d+1/6*a^2*(3*A*b+2*B*a)*cos(d*x+c)^2*sin(d*x+c)/d+1/4*a*A*cos(d
*x+c)^3*(a+b*sec(d*x+c))^2*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.42, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4025, 4074, 4047, 2637, 4045, 8} \[ \frac {\left (6 a^2 A b+2 a^3 B+9 a b^2 B+3 A b^3\right ) \sin (c+d x)}{3 d}+\frac {a \left (3 a^2 A+12 a b B+10 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^3 A+12 a^2 b B+12 a A b^2+8 b^3 B\right )+\frac {a^2 (2 a B+3 A b) \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {a A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

((3*a^3*A + 12*a*A*b^2 + 12*a^2*b*B + 8*b^3*B)*x)/8 + ((6*a^2*A*b + 3*A*b^3 + 2*a^3*B + 9*a*b^2*B)*Sin[c + d*x
])/(3*d) + (a*(3*a^2*A + 10*A*b^2 + 12*a*b*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*(3*A*b + 2*a*B)*Cos[c +
d*x]^2*Sin[c + d*x])/(6*d) + (a*A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(4*d)

Rule 8

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

Rule 2637

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

Rule 4025

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 4045

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 4047

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 4074

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*} \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {a 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 (3 A b+2 a B)-\left (3 a^2 A+4 A b^2+8 a b B\right ) \sec (c+d x)-b (a A+4 b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 (3 A b+2 a B) \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {a 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 a \left (3 a^2 A+10 A b^2+12 a b B\right )+4 \left (6 a^2 A b+3 A b^3+2 a^3 B+9 a b^2 B\right ) \sec (c+d x)+3 b^2 (a A+4 b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 (3 A b+2 a B) \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {a 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 a \left (3 a^2 A+10 A b^2+12 a b B\right )+3 b^2 (a A+4 b B) \sec ^2(c+d x)\right ) \, dx+\frac {1}{3} \left (6 a^2 A b+3 A b^3+2 a^3 B+9 a b^2 B\right ) \int \cos (c+d x) \, dx\\ &=\frac {\left (6 a^2 A b+3 A b^3+2 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 d}+\frac {a \left (3 a^2 A+10 A b^2+12 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (3 A b+2 a B) \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac {1}{8} \left (3 a^3 A+12 a A b^2+12 a^2 b B+8 b^3 B\right ) \int 1 \, dx\\ &=\frac {1}{8} \left (3 a^3 A+12 a A b^2+12 a^2 b B+8 b^3 B\right ) x+\frac {\left (6 a^2 A b+3 A b^3+2 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 d}+\frac {a \left (3 a^2 A+10 A b^2+12 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (3 A b+2 a B) \cos ^2(c+d x) \sin (c+d x)}{6 d}+\frac {a A \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 140, normalized size = 0.78 \[ \frac {3 a^3 A \sin (4 (c+d x))+24 a \left (a^2 A+3 a b B+3 A b^2\right ) \sin (2 (c+d x))+8 a^2 (a B+3 A b) \sin (3 (c+d x))+12 (c+d x) \left (3 a^3 A+12 a^2 b B+12 a A b^2+8 b^3 B\right )+24 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \sin (c+d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(12*(3*a^3*A + 12*a*A*b^2 + 12*a^2*b*B + 8*b^3*B)*(c + d*x) + 24*(9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*
Sin[c + d*x] + 24*a*(a^2*A + 3*A*b^2 + 3*a*b*B)*Sin[2*(c + d*x)] + 8*a^2*(3*A*b + a*B)*Sin[3*(c + d*x)] + 3*a^
3*A*Sin[4*(c + d*x)])/(96*d)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 136, normalized size = 0.76 \[ \frac {3 \, {\left (3 \, A a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 8 \, B b^{3}\right )} d x + {\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 16 \, B a^{3} + 48 \, A a^{2} b + 72 \, B a b^{2} + 24 \, A b^{3} + 8 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{2} + 9 \, {\left (A a^{3} + 4 \, B a^{2} b + 4 \, A a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(3*(3*A*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 8*B*b^3)*d*x + (6*A*a^3*cos(d*x + c)^3 + 16*B*a^3 + 48*A*a^2*b +
72*B*a*b^2 + 24*A*b^3 + 8*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)^2 + 9*(A*a^3 + 4*B*a^2*b + 4*A*a*b^2)*cos(d*x + c))
*sin(d*x + c))/d

________________________________________________________________________________________

giac [B]  time = 0.32, size = 536, normalized size = 2.99 \[ \frac {3 \, {\left (3 \, A a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 8 \, B b^{3}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 120 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 216 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 216 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 36 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A 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 )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(3*A*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 8*B*b^3)*(d*x + c) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^
3*tan(1/2*d*x + 1/2*c)^7 - 72*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 36*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 36*A*a*b^2*
tan(1/2*d*x + 1/2*c)^7 - 72*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 9*A*a^3*tan(1/2
*d*x + 1/2*c)^5 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 120*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 36*B*a^2*b*tan(1/2*d*
x + 1/2*c)^5 + 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 216*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 72*A*b^3*tan(1/2*d*x +
1/2*c)^5 + 9*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 120*A*a^2*b*tan(1/2*d*x + 1/2*c
)^3 - 36*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 216*B*a*b^2*tan(1/2*d*x + 1/2*c)
^3 - 72*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^3*tan(1/2*d*x + 1/2*c) - 24*B*a^3*tan(1/2*d*x + 1/2*c) - 72*A*a^
2*b*tan(1/2*d*x + 1/2*c) - 36*B*a^2*b*tan(1/2*d*x + 1/2*c) - 36*A*a*b^2*tan(1/2*d*x + 1/2*c) - 72*B*a*b^2*tan(
1/2*d*x + 1/2*c) - 24*A*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

________________________________________________________________________________________

maple [A]  time = 1.62, size = 180, normalized size = 1.01 \[ \frac {A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \,a^{2} b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 A a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{3} \sin \left (d x +c \right )+3 B a \,b^{2} \sin \left (d x +c \right )+B \left (d x +c \right ) b^{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.67, size = 171, normalized size = 0.96 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 96 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} + 96 \, {\left (d x + c\right )} B b^{3} + 288 \, B a b^{2} \sin \left (d x + c\right ) + 96 \, A b^{3} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),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^3 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B
*a^3 - 96*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2*b + 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2*b + 72*(2*d*x
+ 2*c + sin(2*d*x + 2*c))*A*a*b^2 + 96*(d*x + c)*B*b^3 + 288*B*a*b^2*sin(d*x + c) + 96*A*b^3*sin(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 2.49, size = 202, normalized size = 1.13 \[ \frac {3\,A\,a^3\,x}{8}+B\,b^3\,x+\frac {3\,A\,a\,b^2\,x}{2}+\frac {3\,B\,a^2\,b\,x}{2}+\frac {A\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {3\,A\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {9\,A\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^3,x)

[Out]

(3*A*a^3*x)/8 + B*b^3*x + (3*A*a*b^2*x)/2 + (3*B*a^2*b*x)/2 + (A*b^3*sin(c + d*x))/d + (3*B*a^3*sin(c + d*x))/
(4*d) + (A*a^3*sin(2*c + 2*d*x))/(4*d) + (A*a^3*sin(4*c + 4*d*x))/(32*d) + (B*a^3*sin(3*c + 3*d*x))/(12*d) + (
3*A*a*b^2*sin(2*c + 2*d*x))/(4*d) + (A*a^2*b*sin(3*c + 3*d*x))/(4*d) + (3*B*a^2*b*sin(2*c + 2*d*x))/(4*d) + (9
*A*a^2*b*sin(c + d*x))/(4*d) + (3*B*a*b^2*sin(c + d*x))/d

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________