\(\int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [291]

Optimal result

Integrand size = 31, antiderivative size = 107 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {1}{2} \left (2 a A b+a^2 B+2 b^2 B\right ) x+\frac {\left (2 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{3 d}+\frac {a (2 A b+a B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^2 A \cos ^2(c+d x) \sin (c+d x)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {6 \left (2 a A b+a^2 B+2 b^2 B\right ) (c+d x)+3 \left (3 a^2 A+4 A b^2+8 a b B\right ) \sin (c+d x)+3 a (2 A b+a B) \sin (2 (c+d x))+a^2 A \sin (3 (c+d x))}{12 d} \] Input:

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
 

Output:

(6*(2*a*A*b + a^2*B + 2*b^2*B)*(c + d*x) + 3*(3*a^2*A + 4*A*b^2 + 8*a*b*B) 
*Sin[c + d*x] + 3*a*(2*A*b + a*B)*Sin[2*(c + d*x)] + a^2*A*Sin[3*(c + d*x) 
])/(12*d)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4512, 25, 3042, 4535, 3042, 3117, 4533, 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.

\(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4512

\(\displaystyle \frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) \left (3 b^2 B \sec ^2(c+d x)+\left (2 A a^2+6 b B a+3 A b^2\right ) \sec (c+d x)+3 a (2 A b+a B)\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) \left (3 b^2 B \sec ^2(c+d x)+\left (2 A a^2+6 b B a+3 A b^2\right ) \sec (c+d x)+3 a (2 A b+a B)\right )dx+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {3 b^2 B \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 A a^2+6 b B a+3 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 a (2 A b+a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {1}{3} \left (\left (2 a^2 A+6 a b B+3 A b^2\right ) \int \cos (c+d x)dx+\int \cos ^2(c+d x) \left (3 b^2 B \sec ^2(c+d x)+3 a (2 A b+a B)\right )dx\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\left (2 a^2 A+6 a b B+3 A b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+\int \frac {3 b^2 B \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 a (2 A b+a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}\)

\(\Big \downarrow \) 3117

\(\displaystyle \frac {1}{3} \left (\int \frac {3 b^2 B \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 a (2 A b+a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {\left (2 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x)}{d}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (a^2 B+2 a A b+2 b^2 B\right ) \int 1dx+\frac {\left (2 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x)}{d}+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{3} \left (\frac {\left (2 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x)}{d}+\frac {3}{2} x \left (a^2 B+2 a A b+2 b^2 B\right )+\frac {3 a (a B+2 A b) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a^2 A \sin (c+d x) \cos ^2(c+d x)}{3 d}\)

Input:

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x]),x]
 

Output:

(a^2*A*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((3*(2*a*A*b + a^2*B + 2*b^2*B 
)*x)/2 + ((2*a^2*A + 3*A*b^2 + 6*a*b*B)*Sin[c + d*x])/d + (3*a*(2*A*b + a* 
B)*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 3117

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

rule 4512

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^2*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a^2*A*Cos[ 
e + f*x]*((d*Csc[e + f*x])^(n + 1)/(d*f*n)), x] + Simp[1/(d*n)   Int[(d*Csc 
[e + f*x])^(n + 1)*(a*(2*A*b + a*B)*n + (2*a*b*B*n + A*(b^2*n + a^2*(n + 1) 
))*Csc[e + f*x] + b^2*B*n*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, 
A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
 

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 = 0.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(\frac {\left (6 A a b +3 B \,a^{2}\right ) \sin \left (2 d x +2 c \right )+A \,a^{2} \sin \left (3 d x +3 c \right )+\left (9 A \,a^{2}+12 A \,b^{2}+24 B a b \right ) \sin \left (d x +c \right )+12 x \left (A a b +\frac {1}{2} B \,a^{2}+b^{2} B \right ) d}{12 d}\)	\(88\)
derivativedivides	\(\frac {\frac {A \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 A a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{2} \sin \left (d x +c \right )+2 B a b \sin \left (d x +c \right )+b^{2} B \left (d x +c \right )}{d}\)	\(114\)
default	\(\frac {\frac {A \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 A a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{2} \sin \left (d x +c \right )+2 B a b \sin \left (d x +c \right )+b^{2} B \left (d x +c \right )}{d}\)	\(114\)
risch	\(A a b x +\frac {B \,a^{2} x}{2}+x \,b^{2} B +\frac {3 a^{2} A \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) A \,b^{2}}{d}+\frac {2 \sin \left (d x +c \right ) B a b}{d}+\frac {A \,a^{2} \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A a b}{2 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{4 d}\)	\(116\)
norman	\(\frac {\left (A a b +\frac {1}{2} B \,a^{2}+b^{2} B \right ) x +\left (A a b +\frac {1}{2} B \,a^{2}+b^{2} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (A a b +\frac {1}{2} B \,a^{2}+b^{2} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (A a b +\frac {1}{2} B \,a^{2}+b^{2} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-2 A a b -B \,a^{2}-2 b^{2} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-2 A a b -B \,a^{2}-2 b^{2} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (2 A \,a^{2}-2 A a b +2 A \,b^{2}-B \,a^{2}+4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {\left (2 A \,a^{2}+2 A a b +2 A \,b^{2}+B \,a^{2}+4 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 \left (A \,a^{2}-3 A \,b^{2}-6 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {2 a \left (4 A a -6 A b -3 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {2 a \left (4 A a +6 A b +3 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\)	\(378\)

Input:

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

1/12*((6*A*a*b+3*B*a^2)*sin(2*d*x+2*c)+A*a^2*sin(3*d*x+3*c)+(9*A*a^2+12*A* 
b^2+24*B*a*b)*sin(d*x+c)+12*x*(A*a*b+1/2*B*a^2+b^2*B)*d)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.79 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} d x + {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 4 \, A a^{2} + 12 \, B a b + 6 \, A b^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/6*(3*(B*a^2 + 2*A*a*b + 2*B*b^2)*d*x + (2*A*a^2*cos(d*x + c)^2 + 4*A*a^2 
 + 12*B*a*b + 6*A*b^2 + 3*(B*a^2 + 2*A*a*b)*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.01 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 12 \, {\left (d x + c\right )} B b^{2} - 24 \, B a b \sin \left (d x + c\right ) - 12 \, A b^{2} \sin \left (d x + c\right )}{12 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 3*(2*d*x + 2*c + sin(2* 
d*x + 2*c))*B*a^2 - 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b - 12*(d*x + c 
)*B*b^2 - 24*B*a*b*sin(d*x + c) - 12*A*b^2*sin(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (99) = 198\).

Time = 0.14 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.37 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {3 \, {\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="g 
iac")
 

Output:

1/6*(3*(B*a^2 + 2*A*a*b + 2*B*b^2)*(d*x + c) + 2*(6*A*a^2*tan(1/2*d*x + 1/ 
2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a*b*tan(1/2*d*x + 1/2*c)^5 + 
 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*a^ 
2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2*tan( 
1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*tan(1/2*d*x + 
1/2*c) + 6*A*a*b*tan(1/2*d*x + 1/2*c) + 12*B*a*b*tan(1/2*d*x + 1/2*c) + 6* 
A*b^2*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 = 11.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.07 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {B\,a^2\,x}{2}+B\,b^2\,x+\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,b^2\,\sin \left (c+d\,x\right )}{d}+A\,a\,b\,x+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,B\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \] Input:

int(cos(c + d*x)^3*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.70 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -2 \sin \left (d x +c \right )^{3} a^{3}+6 \sin \left (d x +c \right ) a^{3}+18 \sin \left (d x +c \right ) a \,b^{2}+9 a^{2} b d x +6 b^{3} d x}{6 d} \] Input:

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)
 

Output:

(9*cos(c + d*x)*sin(c + d*x)*a**2*b - 2*sin(c + d*x)**3*a**3 + 6*sin(c + d 
*x)*a**3 + 18*sin(c + d*x)*a*b**2 + 9*a**2*b*d*x + 6*b**3*d*x)/(6*d)