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

Optimal. Leaf size=161 \[ -\frac{\left (a^2 (4 A+5 C)+2 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (a^2+b^2\right ) (4 A+5 C) \sin (c+d x)}{5 d}+\frac{a b (3 A+4 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{a A b \sin (c+d x) \cos ^3(c+d x)}{10 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}+\frac{1}{4} a b x (3 A+4 C) \]

[Out]

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

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Rubi [A]  time = 0.399549, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4095, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac{\left (a^2 (4 A+5 C)+2 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (a^2+b^2\right ) (4 A+5 C) \sin (c+d x)}{5 d}+\frac{a b (3 A+4 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{a A b \sin (c+d x) \cos ^3(c+d x)}{10 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}+\frac{1}{4} a b x (3 A+4 C) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4095

Int[((A_.) + 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] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - 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, 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 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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.45228, size = 126, normalized size = 0.78 \[ \frac{30 \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \sin (c+d x)+5 \left (a^2 (5 A+4 C)+4 A b^2\right ) \sin (3 (c+d x))+3 a^2 A \sin (5 (c+d x))+60 a b (3 A+4 C) (c+d x)+120 a b (A+C) \sin (2 (c+d x))+15 a A b \sin (4 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a*b*(3*A + 4*C)*(c + d*x) + 30*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Sin[c + d*x] + 120*a*b*(A + C)*Sin[2*
(c + d*x)] + 5*(4*A*b^2 + a^2*(5*A + 4*C))*Sin[3*(c + d*x)] + 15*a*A*b*Sin[4*(c + d*x)] + 3*a^2*A*Sin[5*(c + d
*x)])/(240*d)

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Maple [A]  time = 0.077, size = 158, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}A\sin \left ( dx+c \right ) }{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 ) }+{\frac{{a}^{2}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,Aab \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +2\,abC \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{\frac{A{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{b}^{2}C\sin \left ( dx+c \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))^2*(A+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)+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*a*b*C*(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*C*sin(d*x+c))

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Maxima [A]  time = 0.989982, size = 208, normalized size = 1.29 \begin{align*} \frac{16 \,{\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} - 80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C 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 )} A a b + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b - 80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 240 \, C b^{2} \sin \left (d x + c\right )}{240 \, 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))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/240*(16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^2 - 80*(sin(d*x + c)^3 - 3*sin(d*x + c)
)*C*a^2 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a*b + 120*(2*d*x + 2*c + sin(2*d*x + 2*
c))*C*a*b - 80*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^2 + 240*C*b^2*sin(d*x + c))/d

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Fricas [A]  time = 0.513828, size = 300, normalized size = 1.86 \begin{align*} \frac{15 \,{\left (3 \, A + 4 \, C\right )} a b d x +{\left (12 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \, A a b \cos \left (d x + c\right )^{3} + 15 \,{\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right ) + 8 \,{\left (4 \, A + 5 \, C\right )} a^{2} + 20 \,{\left (2 \, A + 3 \, C\right )} b^{2} + 4 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, 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))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/60*(15*(3*A + 4*C)*a*b*d*x + (12*A*a^2*cos(d*x + c)^4 + 30*A*a*b*cos(d*x + c)^3 + 15*(3*A + 4*C)*a*b*cos(d*x
+ c) + 8*(4*A + 5*C)*a^2 + 20*(2*A + 3*C)*b^2 + 4*((4*A + 5*C)*a^2 + 5*A*b^2)*cos(d*x + c)^2)*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))**2*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.19396, size = 672, normalized size = 4.17 \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))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/60*(15*(3*A*a*b + 4*C*a*b)*(d*x + c) + 2*(60*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9
- 75*A*a*b*tan(1/2*d*x + 1/2*c)^9 - 60*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*A*b^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*b
^2*tan(1/2*d*x + 1/2*c)^9 + 80*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 160*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 30*A*a*b*tan(
1/2*d*x + 1/2*c)^7 - 120*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 160*A*b^2*tan(1/2*d*x + 1/2*c)^7 + 240*C*b^2*tan(1/2*d
*x + 1/2*c)^7 + 232*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*A*b^2*tan(1/2*d*x +
1/2*c)^5 + 360*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 80*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 160*C*a^2*tan(1/2*d*x + 1/2*c)
^3 + 30*A*a*b*tan(1/2*d*x + 1/2*c)^3 + 120*C*a*b*tan(1/2*d*x + 1/2*c)^3 + 160*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 2
40*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 60*A*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*tan(1/2*d*x + 1/2*c) + 75*A*a*b*tan
(1/2*d*x + 1/2*c) + 60*C*a*b*tan(1/2*d*x + 1/2*c) + 60*A*b^2*tan(1/2*d*x + 1/2*c) + 60*C*b^2*tan(1/2*d*x + 1/2
*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d