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

Optimal. Leaf size=215 \[ -\frac {\sin ^3(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac {\sin (c+d x) \left (a^2 (4 A+5 C)+10 a b B+b^2 (4 A+5 C)\right )}{5 d}+\frac {\sin (c+d x) \cos (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac {1}{8} x \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

[Out]

1/8*(6*A*a*b+3*B*a^2+4*B*b^2+8*C*a*b)*x+1/5*(10*a*b*B+a^2*(4*A+5*C)+b^2*(4*A+5*C))*sin(d*x+c)/d+1/8*(6*A*a*b+3
*B*a^2+4*B*b^2+8*C*a*b)*cos(d*x+c)*sin(d*x+c)/d+1/20*a*(2*A*b+5*B*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*A*cos(d*x+c
)^4*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-1/15*(2*A*b^2+10*a*b*B+a^2*(4*A+5*C))*sin(d*x+c)^3/d

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Rubi [A]  time = 0.51, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4094, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac {\sin ^3(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{15 d}+\frac {\sin (c+d x) \left (a^2 (4 A+5 C)+10 a b B+b^2 (4 A+5 C)\right )}{5 d}+\frac {\sin (c+d x) \cos (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )}{8 d}+\frac {1}{8} x \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*x)/8 + ((10*a*b*B + a^2*(4*A + 5*C) + b^2*(4*A + 5*C))*Sin[c + d*x])/
(5*d) + ((6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*(2*A*b + 5*a*B)*Cos[c +
d*x]^3*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) - ((2*A*b^2 + 10*a
*b*B + a^2*(4*A + 5*C))*Sin[c + d*x]^3)/(15*d)

Rule 8

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

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 3013

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 4044

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 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]

Rule 4094

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*} \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \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 (2 A b+5 a B+(4 a A+5 b B+5 a C) \sec (c+d x)+b (2 A+5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \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 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )-5 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x)-4 b^2 (2 A+5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \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 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )-4 b^2 (2 A+5 C) \sec ^2(c+d x)\right ) \, dx-\frac {1}{4} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \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+5 C)-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \cos ^2(c+d x)\right ) \, dx-\frac {1}{8} \left (-6 a A b-3 a^2 B-4 b^2 B-8 a b C\right ) \int 1 \, dx\\ &=\frac {1}{8} \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) x+\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {\operatorname {Subst}\left (\int \left (-4 b^2 (2 A+5 C)-4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right )+4 \left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac {1}{8} \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) x+\frac {\left (10 a b B+a^2 (4 A+5 C)+b^2 (4 A+5 C)\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.77, size = 169, normalized size = 0.79 \[ \frac {60 (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )+60 \sin (c+d x) \left (a^2 (5 A+6 C)+12 a b B+2 b^2 (3 A+4 C)\right )+120 \sin (2 (c+d x)) \left (a^2 B+2 a b (A+C)+b^2 B\right )+10 \sin (3 (c+d x)) \left (a^2 (5 A+4 C)+8 a b B+4 A b^2\right )+6 a^2 A \sin (5 (c+d x))+15 a (a B+2 A b) \sin (4 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*(6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*(c + d*x) + 60*(12*a*b*B + 2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Si
n[c + d*x] + 120*(a^2*B + b^2*B + 2*a*b*(A + C))*Sin[2*(c + d*x)] + 10*(4*A*b^2 + 8*a*b*B + a^2*(5*A + 4*C))*S
in[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 = 0.55, size = 171, normalized size = 0.80 \[ \frac {15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} d x + {\left (24 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (4 \, A + 5 \, C\right )} a^{2} + 160 \, B a b + 40 \, {\left (2 \, A + 3 \, C\right )} b^{2} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

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)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(3*B*a^2 + 2*(3*A + 4*C)*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 + 16*(4*A + 5*C)*a^2 + 160*B*a*b + 40*(2*A + 3*C)*b^2 + 8*((4*A + 5*C)*a^2 + 10*B*a*b + 5*A*b^2)*cos(
d*x + c)^2 + 15*(3*B*a^2 + 2*(3*A + 4*C)*a*b + 4*B*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.31, size = 720, normalized size = 3.35 \[ \text {result too large to display} \]

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)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(3*B*a^2 + 6*A*a*b + 8*C*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*t
an(1/2*d*x + 1/2*c)^9 + 120*C*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*C*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 + 120*C*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 + 320*C*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 -
240*C*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 + 480*
C*b^2*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 400*C*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 + 720*C*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 + 320*C*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 + 240*C*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 + 480*C*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) + 120*C*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*C*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) + 120*C*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 = 1.57, size = 244, normalized size = 1.13 \[ \frac {\frac {a^{2} A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 B a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 C a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{2} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} C \sin \left (d x +c \right )}{d} \]

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)+C*sec(d*x+c)^2),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)+1/3*a^2*C*(2+cos(d*x+c)^2)*sin(d*x+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)+2*C*a*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*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)+b^2*C*sin(d*x+c))

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maxima [A]  time = 0.35, size = 233, normalized size = 1.08 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 480 \, C b^{2} \sin \left (d x + c\right )}{480 \, d} \]

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)+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))*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 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*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 + 240*(2*d*x + 2*c + sin(2*d*x
+ 2*c))*C*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 + 480
*C*b^2*sin(d*x + c))/d

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mupad [B]  time = 4.55, size = 256, normalized size = 1.19 \[ \frac {\frac {25\,A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,A\,a^2\,\sin \left (5\,c+5\,d\,x\right )}{2}+30\,B\,a^2\,\sin \left (2\,c+2\,d\,x\right )+10\,A\,b^2\,\sin \left (3\,c+3\,d\,x\right )+\frac {15\,B\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{4}+30\,B\,b^2\,\sin \left (2\,c+2\,d\,x\right )+10\,C\,a^2\,\sin \left (3\,c+3\,d\,x\right )+75\,A\,a^2\,\sin \left (c+d\,x\right )+90\,A\,b^2\,\sin \left (c+d\,x\right )+90\,C\,a^2\,\sin \left (c+d\,x\right )+120\,C\,b^2\,\sin \left (c+d\,x\right )+60\,A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+\frac {15\,A\,a\,b\,\sin \left (4\,c+4\,d\,x\right )}{2}+20\,B\,a\,b\,\sin \left (3\,c+3\,d\,x\right )+60\,C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+45\,B\,a^2\,d\,x+60\,B\,b^2\,d\,x+180\,B\,a\,b\,\sin \left (c+d\,x\right )+90\,A\,a\,b\,d\,x+120\,C\,a\,b\,d\,x}{120\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((25*A*a^2*sin(3*c + 3*d*x))/2 + (3*A*a^2*sin(5*c + 5*d*x))/2 + 30*B*a^2*sin(2*c + 2*d*x) + 10*A*b^2*sin(3*c +
3*d*x) + (15*B*a^2*sin(4*c + 4*d*x))/4 + 30*B*b^2*sin(2*c + 2*d*x) + 10*C*a^2*sin(3*c + 3*d*x) + 75*A*a^2*sin
(c + d*x) + 90*A*b^2*sin(c + d*x) + 90*C*a^2*sin(c + d*x) + 120*C*b^2*sin(c + d*x) + 60*A*a*b*sin(2*c + 2*d*x)
+ (15*A*a*b*sin(4*c + 4*d*x))/2 + 20*B*a*b*sin(3*c + 3*d*x) + 60*C*a*b*sin(2*c + 2*d*x) + 45*B*a^2*d*x + 60*B
*b^2*d*x + 180*B*a*b*sin(c + d*x) + 90*A*a*b*d*x + 120*C*a*b*d*x)/(120*d)

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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)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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