3.322 \(\int \cos ^6(c+d x) (a+a \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=160 \[ -\frac{a^2 (9 B+10 C) \sin ^3(c+d x)}{15 d}+\frac{a^2 (9 B+10 C) \sin (c+d x)}{5 d}+\frac{a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{a^2 (6 B+7 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}+\frac{1}{8} a^2 x (6 B+7 C) \]

[Out]

(a^2*(6*B + 7*C)*x)/8 + (a^2*(9*B + 10*C)*Sin[c + d*x])/(5*d) + (a^2*(6*B + 7*C)*Cos[c + d*x]*Sin[c + d*x])/(8
*d) + (a^2*(6*B + 5*C)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (B*Cos[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])*Sin[c
+ d*x])/(5*d) - (a^2*(9*B + 10*C)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.329132, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4017, 3996, 3787, 2633, 2635, 8} \[ -\frac{a^2 (9 B+10 C) \sin ^3(c+d x)}{15 d}+\frac{a^2 (9 B+10 C) \sin (c+d x)}{5 d}+\frac{a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{a^2 (6 B+7 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}+\frac{1}{8} a^2 x (6 B+7 C) \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(6*B + 7*C)*x)/8 + (a^2*(9*B + 10*C)*Sin[c + d*x])/(5*d) + (a^2*(6*B + 7*C)*Cos[c + d*x]*Sin[c + d*x])/(8
*d) + (a^2*(6*B + 5*C)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (B*Cos[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])*Sin[c
+ d*x])/(5*d) - (a^2*(9*B + 10*C)*Sin[c + d*x]^3)/(15*d)

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(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) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, 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 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 ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^5(c+d x) (a+a \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac{B \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^4(c+d x) (a+a \sec (c+d x)) (a (6 B+5 C)+a (3 B+5 C) \sec (c+d x)) \, dx\\ &=\frac{a^2 (6 B+5 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{B \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}-\frac{1}{20} \int \cos ^3(c+d x) \left (-4 a^2 (9 B+10 C)-5 a^2 (6 B+7 C) \sec (c+d x)\right ) \, dx\\ &=\frac{a^2 (6 B+5 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{B \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{1}{4} \left (a^2 (6 B+7 C)\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{5} \left (a^2 (9 B+10 C)\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac{a^2 (6 B+7 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (6 B+5 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{B \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac{1}{8} \left (a^2 (6 B+7 C)\right ) \int 1 \, dx-\frac{\left (a^2 (9 B+10 C)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{1}{8} a^2 (6 B+7 C) x+\frac{a^2 (9 B+10 C) \sin (c+d x)}{5 d}+\frac{a^2 (6 B+7 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 (6 B+5 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{B \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}-\frac{a^2 (9 B+10 C) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.374035, size = 108, normalized size = 0.68 \[ \frac{a^2 (60 (11 B+12 C) \sin (c+d x)+240 (B+C) \sin (2 (c+d x))+90 B \sin (3 (c+d x))+30 B \sin (4 (c+d x))+6 B \sin (5 (c+d x))+360 B c+360 B d x+80 C \sin (3 (c+d x))+15 C \sin (4 (c+d x))+420 C d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(360*B*c + 360*B*d*x + 420*C*d*x + 60*(11*B + 12*C)*Sin[c + d*x] + 240*(B + C)*Sin[2*(c + d*x)] + 90*B*Si
n[3*(c + d*x)] + 80*C*Sin[3*(c + d*x)] + 30*B*Sin[4*(c + d*x)] + 15*C*Sin[4*(c + d*x)] + 6*B*Sin[5*(c + d*x)])
)/(480*d)

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Maple [A]  time = 0.099, size = 186, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{B{a}^{2}\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 ) }+{a}^{2}C \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 ) +2\,B{a}^{2} \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 ) +{\frac{2\,{a}^{2}C \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}^{2}C \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)^6*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*(1/5*B*a^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a^2*C*(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*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)+2/3*a^2*C*(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^2*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.947882, size = 240, normalized size = 1.5 \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 )} B a^{2} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B 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 )} B a^{2} - 320 \,{\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 )} C a^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(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))*B*a^2 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c
))*B*a^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 - 320*(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))*C*a^2 + 120*(2*d*x + 2*c + sin(2*d*x
+ 2*c))*C*a^2)/d

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Fricas [A]  time = 0.525635, size = 271, normalized size = 1.69 \begin{align*} \frac{15 \,{\left (6 \, B + 7 \, C\right )} a^{2} d x +{\left (24 \, B a^{2} \cos \left (d x + c\right )^{4} + 30 \,{\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (9 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \,{\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right ) + 16 \,{\left (9 \, B + 10 \, C\right )} a^{2}\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)^6*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(6*B + 7*C)*a^2*d*x + (24*B*a^2*cos(d*x + c)^4 + 30*(2*B + C)*a^2*cos(d*x + c)^3 + 8*(9*B + 10*C)*a^
2*cos(d*x + c)^2 + 15*(6*B + 7*C)*a^2*cos(d*x + c) + 16*(9*B + 10*C)*a^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)**6*(a+a*sec(d*x+c))**2*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.20324, size = 284, normalized size = 1.78 \begin{align*} \frac{15 \,{\left (6 \, B a^{2} + 7 \, C a^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (90 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 105 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 420 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 490 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 864 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 800 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 540 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 790 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 390 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 375 \, C a^{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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(6*B*a^2 + 7*C*a^2)*(d*x + c) + 2*(90*B*a^2*tan(1/2*d*x + 1/2*c)^9 + 105*C*a^2*tan(1/2*d*x + 1/2*c)^
9 + 420*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 490*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 864*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 8
00*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 540*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 790*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 390*B*
a^2*tan(1/2*d*x + 1/2*c) + 375*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d