3.1.99 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [99]

3.1.99.1 Optimal result

Integrand size = 33, antiderivative size = 136 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a^2 (7 A+12 C) x+\frac {a^2 (7 A+12 C) \sin (c+d x)}{6 d}+\frac {a^2 (7 A+12 C) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{4 d} \]

output

1/8*a^2*(7*A+12*C)*x+1/6*a^2*(7*A+12*C)*sin(d*x+c)/d+1/24*a^2*(7*A+12*C)*c 
os(d*x+c)*sin(d*x+c)/d+1/6*A*cos(d*x+c)^2*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+ 
1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^2*sin(d*x+c)/d
 

3.1.99.2 Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.54 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (84 A d x+144 C d x+48 (3 A+4 C) \sin (c+d x)+24 (2 A+C) \sin (2 (c+d x))+16 A \sin (3 (c+d x))+3 A \sin (4 (c+d x)))}{96 d} \]

input

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

output

(a^2*(84*A*d*x + 144*C*d*x + 48*(3*A + 4*C)*Sin[c + d*x] + 24*(2*A + C)*Si 
n[2*(c + d*x)] + 16*A*Sin[3*(c + d*x)] + 3*A*Sin[4*(c + d*x)]))/(96*d)
 

3.1.99.3 Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4575, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4575

\(\displaystyle \frac {\int \cos ^3(c+d x) (\sec (c+d x) a+a)^2 (2 a A+a (A+4 C) \sec (c+d x))dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

\(\displaystyle \frac {\frac {1}{3} a (7 A+12 C) \int \cos ^2(c+d x) (\sec (c+d x) a+a)^2dx+\frac {2 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} a (7 A+12 C) \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {2 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\)

\(\Big \downarrow \) 4275

\(\displaystyle \frac {\frac {1}{3} a (7 A+12 C) \left (2 a^2 \int \cos (c+d x)dx+\int \cos ^2(c+d x) \left (\sec ^2(c+d x) a^2+a^2\right )dx\right )+\frac {2 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} a (7 A+12 C) \left (2 a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )+\frac {2 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\)

\(\Big \downarrow \) 3117

\(\displaystyle \frac {\frac {1}{3} a (7 A+12 C) \left (\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {2 a^2 \sin (c+d x)}{d}\right )+\frac {2 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {\frac {1}{3} a (7 A+12 C) \left (\frac {3 a^2 \int 1dx}{2}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {2 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {1}{3} a (7 A+12 C) \left (\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 x}{2}\right )+\frac {2 a A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\)

input

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

output

(A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(4*d) + ((2*a*A*Cos 
[c + d*x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + (a*(7*A + 12*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)/(4*a)
 

3.1.99.3.1 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 4575

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

3.1.99.4 Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.54

method	result	size
parallelrisch	\(\frac {a^{2} \left (\left (16 A +8 C \right ) \sin \left (2 d x +2 c \right )+\frac {16 A \sin \left (3 d x +3 c \right )}{3}+A \sin \left (4 d x +4 c \right )+\left (48 A +64 C \right ) \sin \left (d x +c \right )+28 x d \left (A +\frac {12 C}{7}\right )\right )}{32 d}\)	\(73\)
risch	\(\frac {7 a^{2} A x}{8}+\frac {3 a^{2} x C}{2}+\frac {3 \sin \left (d x +c \right ) a^{2} A}{2 d}+\frac {2 \sin \left (d x +c \right ) C \,a^{2}}{d}+\frac {a^{2} A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{2} A \sin \left (3 d x +3 c \right )}{6 d}+\frac {a^{2} A \sin \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{4 d}\)	\(118\)
derivativedivides	\(\frac {a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a^{2} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 C \,a^{2} \sin \left (d x +c \right )+a^{2} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{2} \left (d x +c \right )}{d}\)	\(142\)
default	\(\frac {a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a^{2} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 C \,a^{2} \sin \left (d x +c \right )+a^{2} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{2} \left (d x +c \right )}{d}\)	\(142\)
norman	\(\frac {\frac {a^{2} \left (3 A -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a^{2} \left (7 A +12 C \right ) x}{8}-\frac {5 a^{2} \left (5 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{2} \left (7 A +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {a^{2} \left (7 A +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}-\frac {a^{2} \left (7 A +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {3 a^{2} \left (7 A +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {3 a^{2} \left (7 A +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}-\frac {3 a^{2} \left (7 A +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}-\frac {3 a^{2} \left (7 A +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {a^{2} \left (7 A +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {a^{2} \left (7 A +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}-\frac {a^{2} \left (53 A -156 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {a^{2} \left (71 A +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {a^{2} \left (85 A +132 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}\)	\(393\)

input

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x,method=_RETURNVER 
BOSE)
 

output

1/32*a^2*((16*A+8*C)*sin(2*d*x+2*c)+16/3*A*sin(3*d*x+3*c)+A*sin(4*d*x+4*c) 
+(48*A+64*C)*sin(d*x+c)+28*x*d*(A+12/7*C))/d
 

3.1.99.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.63 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (7 \, A + 12 \, C\right )} a^{2} d x + {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 16 \, A a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) + 16 \, {\left (2 \, A + 3 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \]

input

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

output

1/24*(3*(7*A + 12*C)*a^2*d*x + (6*A*a^2*cos(d*x + c)^3 + 16*A*a^2*cos(d*x 
+ c)^2 + 3*(7*A + 4*C)*a^2*cos(d*x + c) + 16*(2*A + 3*C)*a^2)*sin(d*x + c) 
)/d
 

3.1.99.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.1.99.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 96 \, {\left (d x + c\right )} C a^{2} - 192 \, C a^{2} \sin \left (d x + c\right )}{96 \, d} \]

input

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

output

-1/96*(64*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 3*(12*d*x + 12*c + sin 
(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 
2*c))*A*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 - 96*(d*x + c)*C*a 
^2 - 192*C*a^2*sin(d*x + c))/d
 

3.1.99.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.29 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (7 \, A a^{2} + 12 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (21 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 77 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 132 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 156 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

input

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

output

1/24*(3*(7*A*a^2 + 12*C*a^2)*(d*x + c) + 2*(21*A*a^2*tan(1/2*d*x + 1/2*c)^ 
7 + 36*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 77*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 13 
2*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 156*C*a 
^2*tan(1/2*d*x + 1/2*c)^3 + 75*A*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*tan(1 
/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
 

3.1.99.9 Mupad [B] (verification not implemented)

Time = 15.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.86 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {7\,A\,a^2\,x}{8}+\frac {3\,C\,a^2\,x}{2}+\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {A\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

input

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

output

(7*A*a^2*x)/8 + (3*C*a^2*x)/2 + (3*A*a^2*sin(c + d*x))/(2*d) + (2*C*a^2*si 
n(c + d*x))/d + (A*a^2*sin(2*c + 2*d*x))/(2*d) + (A*a^2*sin(3*c + 3*d*x))/ 
(6*d) + (A*a^2*sin(4*c + 4*d*x))/(32*d) + (C*a^2*sin(2*c + 2*d*x))/(4*d)