3.293 \(\int \cos ^5(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=180 \[ -\frac {\left (4 a^2 A+10 a b B+5 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (4 a^2 A+10 a b B+5 A b^2\right ) \sin (c+d x)}{5 d}+\frac {\left (3 a^2 B+6 a A b+4 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^2 B+6 a A b+4 b^2 B\right )+\frac {a^2 A \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {a (a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

1/8*(6*A*a*b+3*B*a^2+4*B*b^2)*x+1/5*(4*A*a^2+5*A*b^2+10*B*a*b)*sin(d*x+c)/d+1/8*(6*A*a*b+3*B*a^2+4*B*b^2)*cos(
d*x+c)*sin(d*x+c)/d+1/4*a*(2*A*b+B*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*a^2*A*cos(d*x+c)^4*sin(d*x+c)/d-1/15*(4*A*
a^2+5*A*b^2+10*B*a*b)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.27, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4024, 4047, 2633, 4045, 2635, 8} \[ -\frac {\left (4 a^2 A+10 a b B+5 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (4 a^2 A+10 a b B+5 A b^2\right ) \sin (c+d x)}{5 d}+\frac {\left (3 a^2 B+6 a A b+4 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^2 B+6 a A b+4 b^2 B\right )+\frac {a^2 A \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {a (a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]

[Out]

((6*a*A*b + 3*a^2*B + 4*b^2*B)*x)/8 + ((4*a^2*A + 5*A*b^2 + 10*a*b*B)*Sin[c + d*x])/(5*d) + ((6*a*A*b + 3*a^2*
B + 4*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*(2*A*b + a*B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a^2*A*C
os[c + d*x]^4*Sin[c + d*x])/(5*d) - ((4*a^2*A + 5*A*b^2 + 10*a*b*B)*Sin[c + d*x]^3)/(15*d)

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

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] && Integer
Q[2*n]

Rule 4024

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

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac {a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) \left (-5 a (2 A b+a B)+\left (A \left (-4 a^2-5 b^2\right )-10 a b B\right ) \sec (c+d x)-5 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) \left (-5 a (2 A b+a B)-5 b^2 B \sec ^2(c+d x)\right ) \, dx-\frac {1}{5} \left (-4 a^2 A-5 A b^2-10 a b B\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac {a (2 A b+a B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {1}{4} \left (-6 a A b-3 a^2 B-4 b^2 B\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (4 a^2 A+5 A b^2+10 a b B\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {\left (4 a^2 A+5 A b^2+10 a b B\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a A b+3 a^2 B+4 b^2 B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+a B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {\left (4 a^2 A+5 A b^2+10 a b B\right ) \sin ^3(c+d x)}{15 d}-\frac {1}{8} \left (-6 a A b-3 a^2 B-4 b^2 B\right ) \int 1 \, dx\\ &=\frac {1}{8} \left (6 a A b+3 a^2 B+4 b^2 B\right ) x+\frac {\left (4 a^2 A+5 A b^2+10 a b B\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a A b+3 a^2 B+4 b^2 B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+a B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^2 A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {\left (4 a^2 A+5 A b^2+10 a b B\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 146, normalized size = 0.81 \[ \frac {60 (c+d x) \left (3 a^2 B+6 a A b+4 b^2 B\right )+60 \left (5 a^2 A+12 a b B+6 A b^2\right ) \sin (c+d x)+120 \left (a^2 B+2 a A b+b^2 B\right ) \sin (2 (c+d x))+10 \left (5 a^2 A+8 a b B+4 A b^2\right ) \sin (3 (c+d x))+6 a^2 A \sin (5 (c+d x))+15 a (a B+2 A b) \sin (4 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]

[Out]

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

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fricas [A]  time = 0.48, size = 142, normalized size = 0.79 \[ \frac {15 \, {\left (3 \, B a^{2} + 6 \, A a b + 4 \, B b^{2}\right )} d x + {\left (24 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{3} + 64 \, A a^{2} + 160 \, B a b + 80 \, A b^{2} + 8 \, {\left (4 \, A a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{2} + 6 \, A a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(15*(3*B*a^2 + 6*A*a*b + 4*B*b^2)*d*x + (24*A*a^2*cos(d*x + c)^4 + 30*(B*a^2 + 2*A*a*b)*cos(d*x + c)^3 +
64*A*a^2 + 160*B*a*b + 80*A*b^2 + 8*(4*A*a^2 + 10*B*a*b + 5*A*b^2)*cos(d*x + c)^2 + 15*(3*B*a^2 + 6*A*a*b + 4
*B*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.64, size = 487, normalized size = 2.71 \[ \frac {15 \, {\left (3 \, B a^{2} + 6 \, A a b + 4 \, B b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 150 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 240 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 60 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 640 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 640 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 150 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 240 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, B 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 )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(15*(3*B*a^2 + 6*A*a*b + 4*B*b^2)*(d*x + c) + 2*(120*A*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*B*a^2*tan(1/2*d*x
+ 1/2*c)^9 - 150*A*a*b*tan(1/2*d*x + 1/2*c)^9 + 240*B*a*b*tan(1/2*d*x + 1/2*c)^9 + 120*A*b^2*tan(1/2*d*x + 1/
2*c)^9 - 60*B*b^2*tan(1/2*d*x + 1/2*c)^9 + 160*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 30*B*a^2*tan(1/2*d*x + 1/2*c)^7
- 60*A*a*b*tan(1/2*d*x + 1/2*c)^7 + 640*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 320*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*
B*b^2*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 400*A*b^2
*tan(1/2*d*x + 1/2*c)^5 + 160*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 60*A*a*b*tan(1/
2*d*x + 1/2*c)^3 + 640*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 320*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*B*b^2*tan(1/2*d*x
+ 1/2*c)^3 + 120*A*a^2*tan(1/2*d*x + 1/2*c) + 75*B*a^2*tan(1/2*d*x + 1/2*c) + 150*A*a*b*tan(1/2*d*x + 1/2*c)
+ 240*B*a*b*tan(1/2*d*x + 1/2*c) + 120*A*b^2*tan(1/2*d*x + 1/2*c) + 60*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*
x + 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 1.58, size = 184, normalized size = 1.02 \[ \frac {\frac {a^{2} A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 B a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{2} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)

[Out]

1/d*(1/5*a^2*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+B*a^2*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x
+c)+3/8*d*x+3/8*c)+2*A*a*b*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+2/3*B*a*b*(2+cos(d*x+c
)^2)*sin(d*x+c)+1/3*A*b^2*(2+cos(d*x+c)^2)*sin(d*x+c)+b^2*B*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 0.96, size = 176, normalized size = 0.98 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^2*(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^2 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c
) + 8*sin(2*d*x + 2*c))*B*a^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a*b - 320*(sin(d*
x + c)^3 - 3*sin(d*x + c))*B*a*b - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^2 + 120*(2*d*x + 2*c + sin(2*d*x
+ 2*c))*B*b^2)/d

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mupad [B]  time = 5.81, size = 307, normalized size = 1.71 \[ \frac {x\,\left (\frac {3\,B\,a^2}{4}+\frac {3\,A\,a\,b}{2}+B\,b^2\right )}{2}+\frac {\left (2\,A\,a^2+2\,A\,b^2-\frac {5\,B\,a^2}{4}-B\,b^2-\frac {5\,A\,a\,b}{2}+4\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,A\,a^2}{3}+\frac {16\,A\,b^2}{3}-\frac {B\,a^2}{2}-2\,B\,b^2-A\,a\,b+\frac {32\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^2}{15}+\frac {40\,B\,a\,b}{3}+\frac {20\,A\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,A\,a^2}{3}+\frac {16\,A\,b^2}{3}+\frac {B\,a^2}{2}+2\,B\,b^2+A\,a\,b+\frac {32\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+\frac {5\,B\,a^2}{4}+B\,b^2+\frac {5\,A\,a\,b}{2}+4\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^2,x)

[Out]

(x*((3*B*a^2)/4 + B*b^2 + (3*A*a*b)/2))/2 + (tan(c/2 + (d*x)/2)^5*((116*A*a^2)/15 + (20*A*b^2)/3 + (40*B*a*b)/
3) + tan(c/2 + (d*x)/2)^9*(2*A*a^2 + 2*A*b^2 - (5*B*a^2)/4 - B*b^2 - (5*A*a*b)/2 + 4*B*a*b) + tan(c/2 + (d*x)/
2)^3*((8*A*a^2)/3 + (16*A*b^2)/3 + (B*a^2)/2 + 2*B*b^2 + A*a*b + (32*B*a*b)/3) + tan(c/2 + (d*x)/2)^7*((8*A*a^
2)/3 + (16*A*b^2)/3 - (B*a^2)/2 - 2*B*b^2 - A*a*b + (32*B*a*b)/3) + tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + (5
*B*a^2)/4 + B*b^2 + (5*A*a*b)/2 + 4*B*a*b))/(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))

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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)**5*(a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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