\(\int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [50]

Optimal result

Integrand size = 29, antiderivative size = 77 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {1}{2} a (A+B) x+\frac {a (2 A+3 B) \sin (c+d x)}{3 d}+\frac {a (A+B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a A \cos ^2(c+d x) \sin (c+d x)}{3 d} \] Output:

1/2*a*(A+B)*x+1/3*a*(2*A+3*B)*sin(d*x+c)/d+1/2*a*(A+B)*cos(d*x+c)*sin(d*x+ 
c)/d+1/3*a*A*cos(d*x+c)^2*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a (6 A c+6 B c+6 A d x+6 B d x+3 (3 A+4 B) \sin (c+d x)+3 (A+B) \sin (2 (c+d x))+A \sin (3 (c+d x)))}{12 d} \] Input:

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
 

Output:

(a*(6*A*c + 6*B*c + 6*A*d*x + 6*B*d*x + 3*(3*A + 4*B)*Sin[c + d*x] + 3*(A 
+ B)*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)]))/(12*d)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4484, 25, 3042, 4274, 3042, 3115, 24, 3117}

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.

\(\displaystyle \int \cos ^3(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4484

\(\displaystyle \frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) (3 a (A+B)+a (2 A+3 B) \sec (c+d x))dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) (3 a (A+B)+a (2 A+3 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^2(c+d x)}{3 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 3117

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

Input:

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
 

Output:

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

Defintions of rubi rules used

rule 24

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

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

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 3117

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; 
 FreeQ[{c, d}, x]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4484

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + 
f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n)   Int[(d*Csc[e + f*x])^( 
n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] 
/; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {a \left (\frac {\left (A +B \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {A \sin \left (3 d x +3 c \right )}{6}+\left (\frac {3 A}{2}+2 B \right ) \sin \left (d x +c \right )+\left (A +B \right ) x d \right )}{2 d}\)	\(54\)
derivativedivides	\(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \sin \left (d x +c \right )}{d}\)	\(85\)
default	\(\frac {\frac {a A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \sin \left (d x +c \right )}{d}\)	\(85\)
risch	\(\frac {a A x}{2}+\frac {a x B}{2}+\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) B a}{d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\)	\(85\)
norman	\(\frac {\frac {a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {a \left (A +B \right ) x}{2}-\frac {3 a \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {a \left (A +B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}+\frac {a \left (A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {a \left (5 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(176\)

Input:

int(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE 
)
 

Output:

1/2*a*(1/2*(A+B)*sin(2*d*x+2*c)+1/6*A*sin(3*d*x+3*c)+(3/2*A+2*B)*sin(d*x+c 
)+(A+B)*x*d)/d
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (A + B\right )} a d x + {\left (2 \, A a \cos \left (d x + c\right )^{2} + 3 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, {\left (2 \, A + 3 \, B\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fri 
cas")
 

Output:

1/6*(3*(A + B)*a*d*x + (2*A*a*cos(d*x + c)^2 + 3*(A + B)*a*cos(d*x + c) + 
2*(2*A + 3*B)*a)*sin(d*x + c))/d
 

Sympy [F]

\[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=a \left (\int A \cos ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
 

Output:

a*(Integral(A*cos(c + d*x)**3, x) + Integral(A*cos(c + d*x)**3*sec(c + d*x 
), x) + Integral(B*cos(c + d*x)**3*sec(c + d*x), x) + Integral(B*cos(c + d 
*x)**3*sec(c + d*x)**2, x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 12 \, B a \sin \left (d x + c\right )}{12 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="max 
ima")
 

Output:

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 3*(2*d*x + 2*c + sin(2*d* 
x + 2*c))*A*a - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a - 12*B*a*sin(d*x + 
c))/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.61 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (A a + B a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="gia 
c")
 

Output:

1/6*(3*(A*a + B*a)*(d*x + c) + 2*(3*A*a*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*tan 
(1/2*d*x + 1/2*c)^5 + 4*A*a*tan(1/2*d*x + 1/2*c)^3 + 12*B*a*tan(1/2*d*x + 
1/2*c)^3 + 9*A*a*tan(1/2*d*x + 1/2*c) + 9*B*a*tan(1/2*d*x + 1/2*c))/(tan(1 
/2*d*x + 1/2*c)^2 + 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 11.48 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {A\,a\,x}{2}+\frac {B\,a\,x}{2}+\frac {3\,A\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \] Input:

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + a/cos(c + d*x)),x)
                                                                                    
                                                                                    
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \left (3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -2 \sin \left (d x +c \right )^{3} a +6 \sin \left (d x +c \right ) a +6 \sin \left (d x +c \right ) b +3 a d x +3 b d x \right )}{6 d} \] Input:

int(cos(d*x+c)^3*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)),x)
 

Output:

(a*(3*cos(c + d*x)*sin(c + d*x)*a + 3*cos(c + d*x)*sin(c + d*x)*b - 2*sin( 
c + d*x)**3*a + 6*sin(c + d*x)*a + 6*sin(c + d*x)*b + 3*a*d*x + 3*b*d*x))/ 
(6*d)