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

Optimal. Leaf size=98 \[ -\frac{(2 A+C) \sin ^3(c+d x)}{3 d}+\frac{(A+C) \sin (c+d x)}{d}+\frac{A \sin ^5(c+d x)}{5 d}+\frac{B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 B \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 B x}{8} \]

[Out]

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

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Rubi [A]  time = 0.115202, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4047, 2635, 8, 4044, 3013, 373} \[ -\frac{(2 A+C) \sin ^3(c+d x)}{3 d}+\frac{(A+C) \sin (c+d x)}{d}+\frac{A \sin ^5(c+d x)}{5 d}+\frac{B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 B \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 B x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.182186, size = 87, normalized size = 0.89 \[ \frac{60 (5 A+6 C) \sin (c+d x)+50 A \sin (3 (c+d x))+6 A \sin (5 (c+d x))+120 B \sin (2 (c+d x))+15 B \sin (4 (c+d x))+180 B c+180 B d x+40 C \sin (3 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(180*B*c + 180*B*d*x + 60*(5*A + 6*C)*Sin[c + d*x] + 120*B*Sin[2*(c + d*x)] + 50*A*Sin[3*(c + d*x)] + 40*C*Sin
[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 6*A*Sin[5*(c + d*x)])/(480*d)

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Maple [A]  time = 0.054, size = 89, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{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 ) }+B \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{C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \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)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.932985, size = 120, normalized size = 1.22 \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 + 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 - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C}{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)+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 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) +
8*sin(2*d*x + 2*c))*B - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C)/d

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Fricas [A]  time = 0.498851, size = 194, normalized size = 1.98 \begin{align*} \frac{45 \, B d x +{\left (24 \, A \cos \left (d x + c\right )^{4} + 30 \, B \cos \left (d x + c\right )^{3} + 8 \,{\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, B \cos \left (d x + c\right ) + 64 \, A + 80 \, C\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)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(45*B*d*x + (24*A*cos(d*x + c)^4 + 30*B*cos(d*x + c)^3 + 8*(4*A + 5*C)*cos(d*x + c)^2 + 45*B*cos(d*x + c
) + 64*A + 80*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)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.16485, size = 300, normalized size = 3.06 \begin{align*} \frac{45 \,{\left (d x + c\right )} B + \frac{2 \,{\left (120 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 75 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 120 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 160 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 30 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 320 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 464 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 400 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 160 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 320 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 75 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 120 \, 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)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(45*(d*x + c)*B + 2*(120*A*tan(1/2*d*x + 1/2*c)^9 - 75*B*tan(1/2*d*x + 1/2*c)^9 + 120*C*tan(1/2*d*x + 1/
2*c)^9 + 160*A*tan(1/2*d*x + 1/2*c)^7 - 30*B*tan(1/2*d*x + 1/2*c)^7 + 320*C*tan(1/2*d*x + 1/2*c)^7 + 464*A*tan
(1/2*d*x + 1/2*c)^5 + 400*C*tan(1/2*d*x + 1/2*c)^5 + 160*A*tan(1/2*d*x + 1/2*c)^3 + 30*B*tan(1/2*d*x + 1/2*c)^
3 + 320*C*tan(1/2*d*x + 1/2*c)^3 + 120*A*tan(1/2*d*x + 1/2*c) + 75*B*tan(1/2*d*x + 1/2*c) + 120*C*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d