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

Optimal. Leaf size=92 \[ \frac{B \sin ^5(c+d x)}{5 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 C \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 C x}{8} \]

[Out]

(3*C*x)/8 + (B*Sin[c + d*x])/d + (3*C*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (C*Cos[c + d*x]^3*Sin[c + d*x])/(4*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 = 0.0756441, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4047, 2633, 12, 2635, 8} \[ \frac{B \sin ^5(c+d x)}{5 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 C \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 C x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*x)/8 + (B*Sin[c + d*x])/d + (3*C*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (C*Cos[c + d*x]^3*Sin[c + d*x])/(4*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

\begin{align*} \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=B \int \cos ^5(c+d x) \, dx+\int C \cos ^4(c+d x) \, dx\\ &=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{C \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin ^5(c+d x)}{5 d}+\frac{1}{4} (3 C) \int \cos ^2(c+d x) \, dx\\ &=\frac{B \sin (c+d x)}{d}+\frac{3 C \cos (c+d x) \sin (c+d x)}{8 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin ^5(c+d x)}{5 d}+\frac{1}{8} (3 C) \int 1 \, dx\\ &=\frac{3 C x}{8}+\frac{B \sin (c+d x)}{d}+\frac{3 C \cos (c+d x) \sin (c+d x)}{8 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 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 = 0.10447, size = 89, normalized size = 0.97 \[ \frac{B \sin ^5(c+d x)}{5 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{3 C (c+d x)}{8 d}+\frac{C \sin (2 (c+d x))}{4 d}+\frac{C \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(c + d*x))/(8*d) + (B*Sin[c + d*x])/d - (2*B*Sin[c + d*x]^3)/(3*d) + (B*Sin[c + d*x]^5)/(5*d) + (C*Sin[2*
(c + d*x)])/(4*d) + (C*Sin[4*(c + d*x)])/(32*d)

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Maple [A]  time = 0.108, size = 70, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\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*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*(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 = 0.937335, size = 93, normalized size = 1.01 \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 + 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}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(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 + 15*(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 = 0.494508, size = 173, normalized size = 1.88 \begin{align*} \frac{45 \, C d x +{\left (24 \, B \cos \left (d x + c\right )^{4} + 30 \, C \cos \left (d x + c\right )^{3} + 32 \, B \cos \left (d x + c\right )^{2} + 45 \, C \cos \left (d x + c\right ) + 64 \, B\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*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(45*C*d*x + (24*B*cos(d*x + c)^4 + 30*C*cos(d*x + c)^3 + 32*B*cos(d*x + c)^2 + 45*C*cos(d*x + c) + 64*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*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.16142, size = 208, normalized size = 2.26 \begin{align*} \frac{45 \,{\left (d x + c\right )} C + \frac{2 \,{\left (120 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 75 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 160 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 30 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 464 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 160 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 75 \, 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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(45*(d*x + c)*C + 2*(120*B*tan(1/2*d*x + 1/2*c)^9 - 75*C*tan(1/2*d*x + 1/2*c)^9 + 160*B*tan(1/2*d*x + 1/
2*c)^7 - 30*C*tan(1/2*d*x + 1/2*c)^7 + 464*B*tan(1/2*d*x + 1/2*c)^5 + 160*B*tan(1/2*d*x + 1/2*c)^3 + 30*C*tan(
1/2*d*x + 1/2*c)^3 + 120*B*tan(1/2*d*x + 1/2*c) + 75*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d