\(\int \cos ^5(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [63]

Optimal result

Integrand size = 29, antiderivative size = 98 \[ \int \cos ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 B x}{8}+\frac {(A+C) \sin (c+d x)}{d}+\frac {3 B \cos (c+d x) \sin (c+d x)}{8 d}+\frac {B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {(2 A+C) \sin ^3(c+d x)}{3 d}+\frac {A \sin ^5(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {4132, 2715, 8, 4129, 3092, 380} \[ \int \cos ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {(2 A+C) \sin ^3(c+d x)}{3 d}+\frac {(A+C) \sin (c+d x)}{d}+\frac {A \sin ^5(c+d x)}{5 d}+\frac {B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 B \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 B x}{8} \]

[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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 380

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]

Rule 2715

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] && Integ
erQ[2*n]

Rule 3092

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 4129

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 4132

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]

Rubi steps \begin{align*} \text {integral}& = 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 {\text {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 {\text {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] (verified)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {180 B c+180 B d x+60 (5 A+6 C) \sin (c+d x)+120 B \sin (2 (c+d x))+50 A \sin (3 (c+d x))+40 C \sin (3 (c+d x))+15 B \sin (4 (c+d x))+6 A \sin (5 (c+d x))}{480 d} \]

[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)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(\frac {\left (50 A +40 C \right ) \sin \left (3 d x +3 c \right )+120 B \sin \left (2 d x +2 c \right )+15 B \sin \left (4 d x +4 c \right )+6 A \sin \left (5 d x +5 c \right )+\left (300 A +360 C \right ) \sin \left (d x +c \right )+180 B x d}{480 d}\)	\(79\)
derivativedivides	\(\frac {\frac {A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+B \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(89\)
default	\(\frac {\frac {A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+B \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(89\)
risch	\(\frac {3 B x}{8}+\frac {5 A \sin \left (d x +c \right )}{8 d}+\frac {3 C \sin \left (d x +c \right )}{4 d}+\frac {A \sin \left (5 d x +5 c \right )}{80 d}+\frac {B \sin \left (4 d x +4 c \right )}{32 d}+\frac {5 A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) C}{12 d}+\frac {B \sin \left (2 d x +2 c \right )}{4 d}\)	\(105\)
norman	\(\frac {-\frac {3 B x}{8}-\frac {3 B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {15 B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {15 B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {3 B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2}+\frac {3 B x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}-\frac {\left (8 A -9 B +40 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {\left (8 A -5 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}-\frac {\left (8 A +5 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (8 A +9 B +40 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {\left (152 A -15 B +40 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 d}+\frac {\left (152 A +15 B +40 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(266\)

[In]

int(cos(d*x+c)^5*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/480*((50*A+40*C)*sin(3*d*x+3*c)+120*B*sin(2*d*x+2*c)+15*B*sin(4*d*x+4*c)+6*A*sin(5*d*x+5*c)+(300*A+360*C)*si
n(d*x+c)+180*B*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.74 \[ \int \cos ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, B d x + {\left (24 \, A \cos \left (d x + c\right )^{4} + 30 \, B \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, B \cos \left (d x + c\right ) + 64 \, A + 80 \, C\right )} \sin \left (d x + c\right )}{120 \, d} \]

[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

Sympy [F]

\[ \int \cos ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{5}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**5*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91 \[ \int \cos ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C}{480 \, d} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (88) = 176\).

Time = 0.31 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.27 \[ \int \cos ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, {\left (d x + c\right )} B + \frac {2 \, {\left (120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 14.92 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.06 \[ \int \cos ^5(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,B\,x}{8}+\frac {5\,A\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {A\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {B\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {5\,A\,\sin \left (c+d\,x\right )}{8\,d}+\frac {3\,C\,\sin \left (c+d\,x\right )}{4\,d} \]

[In]

int(cos(c + d*x)^5*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(3*B*x)/8 + (5*A*sin(3*c + 3*d*x))/(48*d) + (A*sin(5*c + 5*d*x))/(80*d) + (B*sin(2*c + 2*d*x))/(4*d) + (B*sin(
4*c + 4*d*x))/(32*d) + (C*sin(3*c + 3*d*x))/(12*d) + (5*A*sin(c + d*x))/(8*d) + (3*C*sin(c + d*x))/(4*d)