\(\int \cos ^6(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [46]

Optimal result

Integrand size = 28, antiderivative size = 92 \[ \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C x}{8}+\frac {B \sin (c+d x)}{d}+\frac {3 C \cos (c+d x) \sin (c+d x)}{8 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 B \sin ^3(c+d x)}{3 d}+\frac {B \sin ^5(c+d x)}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C (c+d x)}{8 d}+\frac {B \sin (c+d x)}{d}-\frac {2 B \sin ^3(c+d x)}{3 d}+\frac {B \sin ^5(c+d x)}{5 d}+\frac {C \sin (2 (c+d x))}{4 d}+\frac {C \sin (4 (c+d x))}{32 d} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4535, 27, 3042, 3113, 2009, 3115, 3042, 3115, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^6}dx\)

\(\Big \downarrow \) 4535

\(\displaystyle B \int \cos ^5(c+d x)dx+\int C \cos ^4(c+d x)dx\)

\(\Big \downarrow \) 27

\(\displaystyle B \int \cos ^5(c+d x)dx+C \int \cos ^4(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle B \int \sin \left (c+d x+\frac {\pi }{2}\right )^5dx+C \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\)

\(\Big \downarrow \) 3113

\(\displaystyle C \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {B \int \left (\sin ^4(c+d x)-2 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle C \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {B \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle C \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {B \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle C \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {B \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3115

\(\displaystyle C \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {B \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 24

\(\displaystyle C \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {B \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3113

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] 
 && IGtQ[(n - 1)/2, 0]
 

rule 3115

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] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[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]
 

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(\frac {180 d x C +300 B \sin \left (d x +c \right )+6 B \sin \left (5 d x +5 c \right )+50 B \sin \left (3 d x +3 c \right )+15 \sin \left (4 d x +4 c \right ) C +120 \sin \left (2 d x +2 c \right ) C}{480 d}\)	\(69\)
derivativedivides	\(\frac {\frac {B \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(70\)
default	\(\frac {\frac {B \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(70\)
risch	\(\frac {3 C x}{8}+\frac {5 B \sin \left (d x +c \right )}{8 d}+\frac {B \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) C}{32 d}+\frac {5 B \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (2 d x +2 c \right ) C}{4 d}\)	\(78\)
norman	\(\frac {\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {3 C x}{8}-\frac {15 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}-\frac {27 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}-\frac {15 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}+\frac {15 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {27 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {15 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {3 C x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}+\frac {\left (8 B -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}-\frac {\left (8 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (16 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}-\frac {\left (16 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {\left (344 B -75 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d}+\frac {\left (344 B +75 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(294\)

Input:

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

Output:

1/480*(180*d*x*C+300*B*sin(d*x+c)+6*B*sin(5*d*x+5*c)+50*B*sin(3*d*x+3*c)+1 
5*sin(4*d*x+4*c)*C+120*sin(2*d*x+2*c)*C)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, C d x + {\left (24 \, B \cos \left (d x + c\right )^{4} + 30 \, C \cos \left (d x + c\right )^{3} + 32 \, B \cos \left (d x + c\right )^{2} + 45 \, C \cos \left (d x + c\right ) + 64 \, B\right )} \sin \left (d x + c\right )}{120 \, d} \] Input:

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

Output:

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
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**6*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C}{480 \, d} \] Input:

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

Output:

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
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.67 \[ \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, {\left (d x + c\right )} C + \frac {2 \, {\left (120 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \] Input:

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

Output:

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
 

Mupad [B] (verification not implemented)

Time = 16.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.23 \[ \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,C\,x}{8}+\frac {\left (2\,B-\frac {5\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,B}{3}-\frac {C}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {116\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\left (\frac {8\,B}{3}+\frac {C}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B+\frac {5\,C}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \] Input:

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

Output:

(3*C*x)/8 + (tan(c/2 + (d*x)/2)^3*((8*B)/3 + C/2) + tan(c/2 + (d*x)/2)^9*( 
2*B - (5*C)/4) + tan(c/2 + (d*x)/2)^7*((8*B)/3 - C/2) + (116*B*tan(c/2 + ( 
d*x)/2)^5)/15 + tan(c/2 + (d*x)/2)*(2*B + (5*C)/4))/(d*(tan(c/2 + (d*x)/2) 
^2 + 1)^5)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.80 \[ \int \cos ^6(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +24 \sin \left (d x +c \right )^{5} b -80 \sin \left (d x +c \right )^{3} b +120 \sin \left (d x +c \right ) b +45 c d x}{120 d} \] Input:

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

Output:

( - 30*cos(c + d*x)*sin(c + d*x)**3*c + 75*cos(c + d*x)*sin(c + d*x)*c + 2 
4*sin(c + d*x)**5*b - 80*sin(c + d*x)**3*b + 120*sin(c + d*x)*b + 45*c*d*x 
)/(120*d)