\(\int \cos ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [62]

Optimal result

Integrand size = 29, antiderivative size = 88 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} (3 A+4 C) x+\frac {B \sin (c+d x)}{d}+\frac {(3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {B \sin ^3(c+d x)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {36 A c+48 c C+36 A d x+48 C d x+96 B \sin (c+d x)-32 B \sin ^3(c+d x)+24 (A+C) \sin (2 (c+d x))+3 A \sin (4 (c+d x))}{96 d} \] Input:

Integrate[Cos[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

(36*A*c + 48*c*C + 36*A*d*x + 48*C*d*x + 96*B*Sin[c + d*x] - 32*B*Sin[c + 
d*x]^3 + 24*(A + C)*Sin[2*(c + d*x)] + 3*A*Sin[4*(c + d*x)])/(96*d)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4535, 3042, 3113, 2009, 4533, 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.

\(\displaystyle \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

\(\displaystyle \int \cos ^4(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \cos ^3(c+d x)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3113

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 4533

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

Input:

Int[Cos[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 4533

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] + 
Simp[(C*m + A*(m + 1))/(b^2*m)   Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr 
eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
 

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 = 0.34 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72

method	result	size
parallelrisch	\(\frac {24 \left (A +C \right ) \sin \left (2 d x +2 c \right )+8 B \sin \left (3 d x +3 c \right )+3 A \sin \left (4 d x +4 c \right )+72 B \sin \left (d x +c \right )+36 x d \left (A +\frac {4 C}{3}\right )}{96 d}\)	\(63\)
risch	\(\frac {3 A x}{8}+\frac {C x}{2}+\frac {3 B \sin \left (d x +c \right )}{4 d}+\frac {A \sin \left (4 d x +4 c \right )}{32 d}+\frac {B \sin \left (3 d x +3 c \right )}{12 d}+\frac {A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C}{4 d}\)	\(82\)
derivativedivides	\(\frac {A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(84\)
default	\(\frac {A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(84\)
norman	\(\frac {\left (-\frac {3 A}{8}-\frac {C}{2}\right ) x +\left (-\frac {9 A}{8}-\frac {3 C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {3 A}{4}-C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3 A}{4}+C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3 A}{8}+\frac {C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {9 A}{8}+\frac {3 C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {2 \left (3 A -2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 \left (3 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {\left (3 A -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}-\frac {\left (5 A -8 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (5 A +8 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(259\)

Input:

int(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

1/96*(24*(A+C)*sin(2*d*x+2*c)+8*B*sin(3*d*x+3*c)+3*A*sin(4*d*x+4*c)+72*B*s 
in(d*x+c)+36*x*d*(A+4/3*C))/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.74 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, A + 4 \, C\right )} d x + {\left (6 \, A \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right ) + 16 \, B\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:

integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="frica 
s")
 

Output:

1/24*(3*(3*A + 4*C)*d*x + (6*A*cos(d*x + c)^3 + 8*B*cos(d*x + c)^2 + 3*(3* 
A + 4*C)*cos(d*x + c) + 16*B)*sin(d*x + c))/d
 

Sympy [F]

\[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}\, dx \] Input:

integrate(cos(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 

Output:

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{96 \, d} \] Input:

integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxim 
a")
 

Output:

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A - 32*(si 
n(d*x + c)^3 - 3*sin(d*x + c))*B + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C)/ 
d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (80) = 160\).

Time = 0.35 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.27 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (d x + c\right )} {\left (3 \, A + 4 \, C\right )} - \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \] Input:

integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac" 
)
 

Output:

1/24*(3*(d*x + c)*(3*A + 4*C) - 2*(15*A*tan(1/2*d*x + 1/2*c)^7 - 24*B*tan( 
1/2*d*x + 1/2*c)^7 + 12*C*tan(1/2*d*x + 1/2*c)^7 - 9*A*tan(1/2*d*x + 1/2*c 
)^5 - 40*B*tan(1/2*d*x + 1/2*c)^5 + 12*C*tan(1/2*d*x + 1/2*c)^5 + 9*A*tan( 
1/2*d*x + 1/2*c)^3 - 40*B*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2* 
c)^3 - 15*A*tan(1/2*d*x + 1/2*c) - 24*B*tan(1/2*d*x + 1/2*c) - 12*C*tan(1/ 
2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
 

Mupad [B] (verification not implemented)

Time = 11.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,A\,x}{8}+\frac {C\,x}{2}+\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,B\,\sin \left (c+d\,x\right )}{4\,d} \] Input:

int(cos(c + d*x)^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94 \[ \int \cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -8 \sin \left (d x +c \right )^{3} b +24 \sin \left (d x +c \right ) b +9 a d x +12 c d x}{24 d} \] Input:

int(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

( - 6*cos(c + d*x)*sin(c + d*x)**3*a + 15*cos(c + d*x)*sin(c + d*x)*a + 12 
*cos(c + d*x)*sin(c + d*x)*c - 8*sin(c + d*x)**3*b + 24*sin(c + d*x)*b + 9 
*a*d*x + 12*c*d*x)/(24*d)