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

Optimal. Leaf size=156 \[ -\frac{\sin ^3(c+d x) (4 a A+5 a C+5 b B)}{15 d}+\frac{\sin (c+d x) (4 a A+5 a C+5 b B)}{5 d}+\frac{\sin (c+d x) \cos (c+d x) (3 a B+3 A b+4 b C)}{8 d}+\frac{(a B+A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{1}{8} x (3 a B+3 A b+4 b C)+\frac{a A \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.234685, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4074, 4047, 2633, 4045, 2635, 8} \[ -\frac{\sin ^3(c+d x) (4 a A+5 a C+5 b B)}{15 d}+\frac{\sin (c+d x) (4 a A+5 a C+5 b B)}{5 d}+\frac{\sin (c+d x) \cos (c+d x) (3 a B+3 A b+4 b C)}{8 d}+\frac{(a B+A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{1}{8} x (3 a B+3 A b+4 b C)+\frac{a A \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 8

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

Rubi steps

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

Mathematica [A]  time = 0.456157, size = 117, normalized size = 0.75 \[ \frac{-160 \sin ^3(c+d x) (a (2 A+C)+b B)+480 \sin (c+d x) (a (A+C)+b B)+15 (4 (c+d x) (3 a B+3 A b+4 b C)+8 \sin (2 (c+d x)) (a B+A b+b C)+(a B+A b) \sin (4 (c+d x)))+96 a A \sin ^5(c+d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(480*(b*B + a*(A + C))*Sin[c + d*x] - 160*(b*B + a*(2*A + C))*Sin[c + d*x]^3 + 96*a*A*Sin[c + d*x]^5 + 15*(4*(
3*A*b + 3*a*B + 4*b*C)*(c + d*x) + 8*(A*b + a*B + b*C)*Sin[2*(c + d*x)] + (A*b + a*B)*Sin[4*(c + d*x)]))/(480*
d)

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Maple [A]  time = 0.073, size = 173, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{A\sin \left ( dx+c \right ) a}{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+Ab \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 ) +Ba \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{Bb \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{aC \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+Cb \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/5*A*a*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+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)+B*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)+1/3*B*b*(2+cos(d*x+c)^2)*sin(d
*x+c)+1/3*a*C*(2+cos(d*x+c)^2)*sin(d*x+c)+C*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 = 1.03971, size = 224, normalized size = 1.44 \begin{align*} \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 + 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 - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 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 )} A b - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 0.526962, size = 305, normalized size = 1.96 \begin{align*} \frac{15 \,{\left (3 \, B a +{\left (3 \, A + 4 \, C\right )} b\right )} d x +{\left (24 \, A a \cos \left (d x + c\right )^{4} + 30 \,{\left (B a + A b\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left ({\left (4 \, A + 5 \, C\right )} a + 5 \, B b\right )} \cos \left (d x + c\right )^{2} + 16 \,{\left (4 \, A + 5 \, C\right )} a + 80 \, B b + 15 \,{\left (3 \, B a +{\left (3 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(3*B*a + (3*A + 4*C)*b)*d*x + (24*A*a*cos(d*x + c)^4 + 30*(B*a + A*b)*cos(d*x + c)^3 + 8*((4*A + 5*C
)*a + 5*B*b)*cos(d*x + c)^2 + 16*(4*A + 5*C)*a + 80*B*b + 15*(3*B*a + (3*A + 4*C)*b)*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)**5*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.14913, size = 590, normalized size = 3.78 \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)^5*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(3*B*a + 3*A*b + 4*C*b)*(d*x + c) + 2*(120*A*a*tan(1/2*d*x + 1/2*c)^9 - 75*B*a*tan(1/2*d*x + 1/2*c)^
9 + 120*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*A*b*tan(1/2*d*x + 1/2*c)^9 + 120*B*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*b*t
an(1/2*d*x + 1/2*c)^9 + 160*A*a*tan(1/2*d*x + 1/2*c)^7 - 30*B*a*tan(1/2*d*x + 1/2*c)^7 + 320*C*a*tan(1/2*d*x +
1/2*c)^7 - 30*A*b*tan(1/2*d*x + 1/2*c)^7 + 320*B*b*tan(1/2*d*x + 1/2*c)^7 - 120*C*b*tan(1/2*d*x + 1/2*c)^7 +
464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*C*a*tan(1/2*d*x + 1/2*c)^5 + 400*B*b*tan(1/2*d*x + 1/2*c)^5 + 160*A*a*tan
(1/2*d*x + 1/2*c)^3 + 30*B*a*tan(1/2*d*x + 1/2*c)^3 + 320*C*a*tan(1/2*d*x + 1/2*c)^3 + 30*A*b*tan(1/2*d*x + 1/
2*c)^3 + 320*B*b*tan(1/2*d*x + 1/2*c)^3 + 120*C*b*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*tan(1/2*d*x + 1/2*c) + 75*B
*a*tan(1/2*d*x + 1/2*c) + 120*C*a*tan(1/2*d*x + 1/2*c) + 75*A*b*tan(1/2*d*x + 1/2*c) + 120*B*b*tan(1/2*d*x + 1
/2*c) + 60*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d