3.876 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=175 \[ \frac {\sin (c+d x) \left (2 a^2 B+4 a A b+6 a b C+3 b^2 B\right )}{3 d}+\frac {\sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)+8 a b B+2 A b^2\right )}{8 d}+\frac {1}{8} x \left (a^2 (3 A+4 C)+8 a b B+4 b^2 (A+2 C)\right )+\frac {a (2 a B+A b) \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.45, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4094, 4074, 4047, 2637, 4045, 8} \[ \frac {\sin (c+d x) \left (2 a^2 B+4 a A b+6 a b C+3 b^2 B\right )}{3 d}+\frac {\sin (c+d x) \cos (c+d x) \left (a^2 (3 A+4 C)+8 a b B+2 A b^2\right )}{8 d}+\frac {1}{8} x \left (a^2 (3 A+4 C)+8 a b B+4 b^2 (A+2 C)\right )+\frac {a (2 a B+A b) \sin (c+d x) \cos ^2(c+d x)}{6 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.71, size = 134, normalized size = 0.77 \[ \frac {12 (c+d x) \left (a^2 (3 A+4 C)+8 a b B+4 b^2 (A+2 C)\right )+24 \sin (c+d x) \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right )+24 \sin (2 (c+d x)) \left (a^2 (A+C)+2 a b B+A b^2\right )+3 a^2 A \sin (4 (c+d x))+8 a (a B+2 A b) \sin (3 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*a*b*B + 4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*(c + d*x) + 24*(6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*Sin[c
+ d*x] + 24*(A*b^2 + 2*a*b*B + a^2*(A + C))*Sin[2*(c + d*x)] + 8*a*(2*A*b + a*B)*Sin[3*(c + d*x)] + 3*a^2*A*S
in[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.52, size = 134, normalized size = 0.77 \[ \frac {3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 8 \, B a b + 4 \, {\left (A + 2 \, C\right )} b^{2}\right )} d x + {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 16 \, B a^{2} + 16 \, {\left (2 \, A + 3 \, C\right )} a b + 24 \, B b^{2} + 8 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.29, size = 577, normalized size = 3.30 \[ \frac {3 \, {\left (3 \, A a^{2} + 4 \, C a^{2} + 8 \, B a b + 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} - 24 \, B 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} + 24 \, B 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} - 24 \, B 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} - 40 \, B 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} + 24 \, B 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} - 72 \, B 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} - 40 \, B 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} - 24 \, B 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} - 72 \, B 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 ) - 24 \, B 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 ) - 24 \, B 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 ) - 24 \, 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 )}^{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))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(3*A*a^2 + 4*C*a^2 + 8*B*a*b + 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 - 24*
B*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 + 24*B*a*b*ta
n(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 - 24*B*b^2*tan(1/2*d*
x + 1/2*c)^7 - 9*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 40*B*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 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 144*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 1
2*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 72*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^2*t
an(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 - 24*B*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*tan(1/2*d*x + 1/2*c)^3 - 72*B*b^2*tan(1/2*d*x + 1/
2*c)^3 - 15*A*a^2*tan(1/2*d*x + 1/2*c) - 24*B*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) - 24*B*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) - 24*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 1.34, size = 200, normalized size = 1.14 \[ \frac {a^{2} A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 A a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} B \sin \left (d x +c \right )+2 C a b \sin \left (d x +c \right )+b^{2} C \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 187, normalized size = 1.07 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b + 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} + 96 \, {\left (d x + c\right )} C b^{2} + 192 \, C a b \sin \left (d x + c\right ) + 96 \, B b^{2} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+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 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B
*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 + 48*(2*d*x + 2*
c + sin(2*d*x + 2*c))*B*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) + 96*B*b^2*sin(d*x + c))/d

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mupad [B]  time = 4.13, size = 214, normalized size = 1.22 \[ \frac {3\,A\,a^2\,x}{8}+\frac {A\,b^2\,x}{2}+\frac {C\,a^2\,x}{2}+C\,b^2\,x+\frac {3\,B\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b^2\,\sin \left (c+d\,x\right )}{d}+B\,a\,b\,x+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {A\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,A\,a\,b\,\sin \left (c+d\,x\right )}{2\,d}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {B\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + b/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/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 + (3*B*a^2*sin(c + d*x))/(4*d) + (B*b^2*sin(c + d*x))/d +
B*a*b*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) +
(B*a^2*sin(3*c + 3*d*x))/(12*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) + (B*a*b*sin(2*c + 2*d*x))/(2*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)**4*(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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