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]

((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)

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Rubi [A]  time = 0.44836, antiderivative size = 175, normalized size of antiderivative = 1., 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 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]

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 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 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 8

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

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*}

Mathematica [A]  time = 0.730957, 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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Maple [A]  time = 0.072, size = 200, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{2\,Aab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+A{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,Bab \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{2}C \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +B{b}^{2}\sin \left ( dx+c \right ) +2\,abC\sin \left ( dx+c \right ) +{b}^{2}C \left ( dx+c \right ) \right ) } \end{align*}

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*b^2*sin(d*x+c)+2*a*b*C*sin(d*x+c)+
b^2*C*(d*x+c))

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Maxima [A]  time = 1.01152, size = 252, normalized size = 1.44 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 0.519028, size = 320, normalized size = 1.83 \begin{align*} \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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [B]  time = 1.21662, size = 779, normalized size = 4.45 \begin{align*} \text{result too large to display} \end{align*}

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