\(\int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [320]

Optimal result

Integrand size = 40, antiderivative size = 102 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^2 (2 B+3 C) x+\frac {2 a^2 (2 B+3 C) \sin (c+d x)}{3 d}+\frac {a^2 (2 B+3 C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sin (c+d x) \left (11 B+12 C+3 (2 B+C) \cos (c+d x)+B \cos (2 (c+d x))+\frac {6 (2 B+3 C) \arcsin \left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{\sqrt {\sin ^2(c+d x)}}\right )}{6 d} \] Input:

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

Output:

(a^2*Sin[c + d*x]*(11*B + 12*C + 3*(2*B + C)*Cos[c + d*x] + B*Cos[2*(c + d 
*x)] + (6*(2*B + 3*C)*ArcSin[Sqrt[Sin[(c + d*x)/2]^2]])/Sqrt[Sin[c + d*x]^ 
2]))/(6*d)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4560, 3042, 4501, 3042, 4275, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4275

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3117

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

\(\Big \downarrow \) 4533

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

\(\Big \downarrow \) 24

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

Input:

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

Output:

(B*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + ((2*B + 3*C 
)*((3*a^2*x)/2 + (2*a^2*Sin[c + d*x])/d + (a^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 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 4275

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^2, x_Symbol] :> Simp[2*a*(b/d)   Int[(d*Csc[e + f*x])^(n + 1), x], x] 
 + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, 
 e, f, n}, x]
 

rule 4501

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m 
 - b*B*n)/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] 
, x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a 
^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -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 4560

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.59

method	result	size
parallelrisch	\(\frac {a^{2} \left (\left (\frac {B}{2}+\frac {C}{4}\right ) \sin \left (2 d x +2 c \right )+\frac {B \sin \left (3 d x +3 c \right )}{12}+\left (\frac {7 B}{4}+2 C \right ) \sin \left (d x +c \right )+d x \left (B +\frac {3 C}{2}\right )\right )}{d}\)	\(60\)
risch	\(a^{2} B x +\frac {3 a^{2} x C}{2}+\frac {7 a^{2} B \sin \left (d x +c \right )}{4 d}+\frac {2 \sin \left (d x +c \right ) C \,a^{2}}{d}+\frac {B \,a^{2} \sin \left (3 d x +3 c \right )}{12 d}+\frac {B \,a^{2} \sin \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{4 d}\)	\(99\)
derivativedivides	\(\frac {B \,a^{2} \sin \left (d x +c \right )+C \,a^{2} \left (d x +c \right )+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C \,a^{2} \sin \left (d x +c \right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(116\)
default	\(\frac {B \,a^{2} \sin \left (d x +c \right )+C \,a^{2} \left (d x +c \right )+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C \,a^{2} \sin \left (d x +c \right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(116\)
norman	\(\frac {\frac {a^{2} \left (2 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {a^{2} \left (2 B +3 C \right ) x}{2}-\frac {4 a^{2} \left (2 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {13 a^{2} \left (2 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {2 a^{2} \left (2 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {a^{2} \left (2 B +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {3 a^{2} \left (2 B +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {3 a^{2} \left (2 B +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {3 a^{2} \left (2 B +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}-\frac {3 a^{2} \left (2 B +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2}+\frac {a^{2} \left (2 B +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2}+\frac {a^{2} \left (2 B +3 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{2}-\frac {a^{2} \left (6 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} \left (10 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {a^{2} \left (14 B +33 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}\)	\(391\)

Input:

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

Output:

a^2*((1/2*B+1/4*C)*sin(2*d*x+2*c)+1/12*B*sin(3*d*x+3*c)+(7/4*B+2*C)*sin(d* 
x+c)+d*x*(B+3/2*C))/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.69 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, B + 3 \, C\right )} a^{2} d x + {\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (5 \, B + 6 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:

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

Output:

1/6*(3*(2*B + 3*C)*a^2*d*x + (2*B*a^2*cos(d*x + c)^2 + 3*(2*B + C)*a^2*cos 
(d*x + c) + 2*(5*B + 6*C)*a^2)*sin(d*x + c))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.08 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 12 \, {\left (d x + c\right )} C a^{2} - 12 \, B a^{2} \sin \left (d x + c\right ) - 24 \, C a^{2} \sin \left (d x + c\right )}{12 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.39 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, B a^{2} + 3 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:

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

Output:

1/6*(3*(2*B*a^2 + 3*C*a^2)*(d*x + c) + 2*(6*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 
 9*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 16*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*C*a 
^2*tan(1/2*d*x + 1/2*c)^3 + 18*B*a^2*tan(1/2*d*x + 1/2*c) + 15*C*a^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 = 11.60 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.96 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=B\,a^2\,x+\frac {3\,C\,a^2\,x}{2}+\frac {7\,B\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \] Input:

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

Output:

B*a^2*x + (3*C*a^2*x)/2 + (7*B*a^2*sin(c + d*x))/(4*d) + (2*C*a^2*sin(c + 
d*x))/d + (B*a^2*sin(2*c + 2*d*x))/(2*d) + (B*a^2*sin(3*c + 3*d*x))/(12*d) 
 + (C*a^2*sin(2*c + 2*d*x))/(4*d)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{2} \left (6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -2 \sin \left (d x +c \right )^{3} b +12 \sin \left (d x +c \right ) b +12 \sin \left (d x +c \right ) c +6 b d x +9 c d x \right )}{6 d} \] Input:

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

Output:

(a**2*(6*cos(c + d*x)*sin(c + d*x)*b + 3*cos(c + d*x)*sin(c + d*x)*c - 2*s 
in(c + d*x)**3*b + 12*sin(c + d*x)*b + 12*sin(c + d*x)*c + 6*b*d*x + 9*c*d 
*x))/(6*d)