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

Optimal. Leaf size=132 \[ \frac{(5 A+6 C) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{(5 A+6 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{A \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{1}{16} x (5 A+6 C)+\frac{B \sin ^5(c+d x)}{5 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d} \]

[Out]

((5*A + 6*C)*x)/16 + (B*Sin[c + d*x])/d + ((5*A + 6*C)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((5*A + 6*C)*Cos[c
+ d*x]^3*Sin[c + d*x])/(24*d) + (A*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*B*Sin[c + d*x]^3)/(3*d) + (B*Sin[c
+ d*x]^5)/(5*d)

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Rubi [A]  time = 0.0991188, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {4047, 2633, 4045, 2635, 8} \[ \frac{(5 A+6 C) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{(5 A+6 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{A \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{1}{16} x (5 A+6 C)+\frac{B \sin ^5(c+d x)}{5 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*A + 6*C)*x)/16 + (B*Sin[c + d*x])/d + ((5*A + 6*C)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((5*A + 6*C)*Cos[c
+ d*x]^3*Sin[c + d*x])/(24*d) + (A*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*B*Sin[c + d*x]^3)/(3*d) + (B*Sin[c
+ d*x]^5)/(5*d)

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 ^6(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=B \int \cos ^5(c+d x) \, dx+\int \cos ^6(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{6} (5 A+6 C) \int \cos ^4(c+d x) \, dx-\frac{B \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{B \sin (c+d x)}{d}+\frac{(5 A+6 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{A \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin ^5(c+d x)}{5 d}+\frac{1}{8} (5 A+6 C) \int \cos ^2(c+d x) \, dx\\ &=\frac{B \sin (c+d x)}{d}+\frac{(5 A+6 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{(5 A+6 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{A \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin ^5(c+d x)}{5 d}+\frac{1}{16} (5 A+6 C) \int 1 \, dx\\ &=\frac{1}{16} (5 A+6 C) x+\frac{B \sin (c+d x)}{d}+\frac{(5 A+6 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{(5 A+6 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{A \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.268643, size = 102, normalized size = 0.77 \[ \frac{5 ((45 A+48 C) \sin (2 (c+d x))+(9 A+6 C) \sin (4 (c+d x))+A \sin (6 (c+d x))+60 A c+60 A d x+72 c C+72 C d x)+192 B \sin ^5(c+d x)-640 B \sin ^3(c+d x)+960 B \sin (c+d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(960*B*Sin[c + d*x] - 640*B*Sin[c + d*x]^3 + 192*B*Sin[c + d*x]^5 + 5*(60*A*c + 72*c*C + 60*A*d*x + 72*C*d*x +
(45*A + 48*C)*Sin[2*(c + d*x)] + (9*A + 6*C)*Sin[4*(c + d*x)] + A*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.058, size = 115, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( A \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{B\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 ) }+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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*(A*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+1/5*B*(8/3+cos(d*x+c)^
4+4/3*cos(d*x+c)^2)*sin(d*x+c)+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))

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Maxima [A]  time = 0.935274, size = 155, normalized size = 1.17 \begin{align*} -\frac{5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A - 64 \,{\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 - 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 )} C}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*A - 64*(3*sin(d*x
+ c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B - 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C
)/d

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Fricas [A]  time = 0.508536, size = 244, normalized size = 1.85 \begin{align*} \frac{15 \,{\left (5 \, A + 6 \, C\right )} d x +{\left (40 \, A \cos \left (d x + c\right )^{5} + 48 \, B \cos \left (d x + c\right )^{4} + 10 \,{\left (5 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{3} + 64 \, B \cos \left (d x + c\right )^{2} + 15 \,{\left (5 \, A + 6 \, C\right )} \cos \left (d x + c\right ) + 128 \, B\right )} \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)^6*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(5*A + 6*C)*d*x + (40*A*cos(d*x + c)^5 + 48*B*cos(d*x + c)^4 + 10*(5*A + 6*C)*cos(d*x + c)^3 + 64*B*
cos(d*x + c)^2 + 15*(5*A + 6*C)*cos(d*x + c) + 128*B)*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+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.19038, size = 383, normalized size = 2.9 \begin{align*} \frac{15 \,{\left (d x + c\right )}{\left (5 \, A + 6 \, C\right )} - \frac{2 \,{\left (165 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 240 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 150 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 25 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 560 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 210 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 450 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1248 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 60 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 450 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1248 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 60 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 25 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 560 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 210 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 165 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 240 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 150 \, C \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 )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(d*x + c)*(5*A + 6*C) - 2*(165*A*tan(1/2*d*x + 1/2*c)^11 - 240*B*tan(1/2*d*x + 1/2*c)^11 + 150*C*tan
(1/2*d*x + 1/2*c)^11 - 25*A*tan(1/2*d*x + 1/2*c)^9 - 560*B*tan(1/2*d*x + 1/2*c)^9 + 210*C*tan(1/2*d*x + 1/2*c)
^9 + 450*A*tan(1/2*d*x + 1/2*c)^7 - 1248*B*tan(1/2*d*x + 1/2*c)^7 + 60*C*tan(1/2*d*x + 1/2*c)^7 - 450*A*tan(1/
2*d*x + 1/2*c)^5 - 1248*B*tan(1/2*d*x + 1/2*c)^5 - 60*C*tan(1/2*d*x + 1/2*c)^5 + 25*A*tan(1/2*d*x + 1/2*c)^3 -
560*B*tan(1/2*d*x + 1/2*c)^3 - 210*C*tan(1/2*d*x + 1/2*c)^3 - 165*A*tan(1/2*d*x + 1/2*c) - 240*B*tan(1/2*d*x
+ 1/2*c) - 150*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d