\(\int \frac {\cos ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\) [345]

Optimal result

Integrand size = 40, antiderivative size = 98 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {(2 B-C) x}{a^2}+\frac {2 (5 B-2 C) \sin (c+d x)}{3 a^2 d}-\frac {(2 B-C) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(B-C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(245\) vs. \(2(98)=196\).

Time = 0.81 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.50 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-18 (2 B-C) d x \cos \left (\frac {d x}{2}\right )-18 (2 B-C) d x \cos \left (c+\frac {d x}{2}\right )-12 B d x \cos \left (c+\frac {3 d x}{2}\right )+6 C d x \cos \left (c+\frac {3 d x}{2}\right )-12 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+6 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+66 B \sin \left (\frac {d x}{2}\right )-36 C \sin \left (\frac {d x}{2}\right )-30 B \sin \left (c+\frac {d x}{2}\right )+24 C \sin \left (c+\frac {d x}{2}\right )+41 B \sin \left (c+\frac {3 d x}{2}\right )-20 C \sin \left (c+\frac {3 d x}{2}\right )+9 B \sin \left (2 c+\frac {3 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )\right )}{12 a^2 d (1+\cos (c+d x))^2} \] Input:

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

Output:

(Cos[(c + d*x)/2]*Sec[c/2]*(-18*(2*B - C)*d*x*Cos[(d*x)/2] - 18*(2*B - C)* 
d*x*Cos[c + (d*x)/2] - 12*B*d*x*Cos[c + (3*d*x)/2] + 6*C*d*x*Cos[c + (3*d* 
x)/2] - 12*B*d*x*Cos[2*c + (3*d*x)/2] + 6*C*d*x*Cos[2*c + (3*d*x)/2] + 66* 
B*Sin[(d*x)/2] - 36*C*Sin[(d*x)/2] - 30*B*Sin[c + (d*x)/2] + 24*C*Sin[c + 
(d*x)/2] + 41*B*Sin[c + (3*d*x)/2] - 20*C*Sin[c + (3*d*x)/2] + 9*B*Sin[2*c 
 + (3*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2] + 3*B*Sin[3*c + (5*d*x)/2]))/(12* 
a^2*d*(1 + Cos[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.275, Rules used = {3042, 4560, 3042, 4508, 3042, 4508, 3042, 4274, 24, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \cos (c+d x) \left (2 a^2 (5 B-2 C)-3 a^2 (2 B-C) \sec (c+d x)\right )dx}{a^2}-\frac {3 (2 B-C) \sin (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 a^2 (5 B-2 C)-3 a^2 (2 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {3 (2 B-C) \sin (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {2 a^2 (5 B-2 C) \int \cos (c+d x)dx-3 a^2 (2 B-C) \int 1dx}{a^2}-\frac {3 (2 B-C) \sin (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {2 a^2 (5 B-2 C) \int \cos (c+d x)dx-3 a^2 x (2 B-C)}{a^2}-\frac {3 (2 B-C) \sin (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a^2 (5 B-2 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx-3 a^2 x (2 B-C)}{a^2}-\frac {3 (2 B-C) \sin (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3117

\(\displaystyle \frac {\frac {\frac {2 a^2 (5 B-2 C) \sin (c+d x)}{d}-3 a^2 x (2 B-C)}{a^2}-\frac {3 (2 B-C) \sin (c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((B - C)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + ((-3*(2*B - C)*Si 
n[c + d*x])/(d*(1 + Sec[c + d*x])) + (-3*a^2*(2*B - C)*x + (2*a^2*(5*B - 2 
*C)*Sin[c + d*x])/d)/a^2)/(3*a^2)
 

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 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 4508

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

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 = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-20 C +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (3 B \cos \left (2 d x +2 c \right )+28 B \cos \left (d x +c \right )+23 B +2 C \right )\right )-24 \left (B -\frac {C}{2}\right ) d x}{12 a^{2} d}\)	\(73\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {4 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-4 \left (2 B -C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(108\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {4 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-4 \left (2 B -C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(108\)
risch	\(-\frac {2 B x}{a^{2}}+\frac {x C}{a^{2}}-\frac {i B \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}+\frac {i B \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {2 i \left (9 B \,{\mathrm e}^{2 i \left (d x +c \right )}-6 C \,{\mathrm e}^{2 i \left (d x +c \right )}+15 B \,{\mathrm e}^{i \left (d x +c \right )}-9 C \,{\mathrm e}^{i \left (d x +c \right )}+8 B -5 C \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(130\)
norman	\(\frac {\frac {\left (2 B -C \right ) x}{a}+\frac {\left (2 B -C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {\left (2 B -C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {\left (2 B -C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {3 \left (3 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (7 B -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\left (7 B -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}+\frac {\left (14 B -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a}\)	\(245\)

Input:

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

Output:

1/12*(tan(1/2*d*x+1/2*c)*(-20*C+sec(1/2*d*x+1/2*c)^2*(3*B*cos(2*d*x+2*c)+2 
8*B*cos(d*x+c)+23*B+2*C))-24*(B-1/2*C)*d*x)/a^2/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {3 \, {\left (2 \, B - C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (2 \, B - C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (2 \, B - C\right )} d x - {\left (3 \, B \cos \left (d x + c\right )^{2} + {\left (14 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 10 \, B - 4 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

-1/3*(3*(2*B - C)*d*x*cos(d*x + c)^2 + 6*(2*B - C)*d*x*cos(d*x + c) + 3*(2 
*B - C)*d*x - (3*B*cos(d*x + c)^2 + (14*B - 5*C)*cos(d*x + c) + 10*B - 4*C 
)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:

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

Output:

(Integral(B*cos(c + d*x)**2*sec(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) 
 + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2/(sec(c + d*x)**2 + 
2*sec(c + d*x) + 1), x))/a**2
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (94) = 188\).

Time = 0.12 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.95 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \] Input:

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

Output:

1/6*(B*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) 
 + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x 
 + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) 
) - C*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 
 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2))/d
 

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {6 \, {\left (d x + c\right )} {\left (2 \, B - C\right )}}{a^{2}} - \frac {12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:

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

Output:

-1/6*(6*(d*x + c)*(2*B - C)/a^2 - 12*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x 
+ 1/2*c)^2 + 1)*a^2) + (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 
 1/2*c)^3 - 15*B*a^4*tan(1/2*d*x + 1/2*c) + 9*C*a^4*tan(1/2*d*x + 1/2*c))/ 
a^6)/d
 

Mupad [B] (verification not implemented)

Time = 12.17 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B-C}{a^2}+\frac {3\,B-C}{2\,a^2}\right )}{d}-\frac {x\,\left (2\,B-C\right )}{a^2}+\frac {2\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*((B - C)/a^2 + (3*B - C)/(2*a^2)))/d - (x*(2*B - C))/a 
^2 + (2*B*tan(c/2 + (d*x)/2))/(d*(a^2*tan(c/2 + (d*x)/2)^2 + a^2)) - (tan( 
c/2 + (d*x)/2)^3*(B - C))/(6*a^2*d)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b d x +6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} c d x +27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c -12 b d x +6 c d x}{6 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:

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

Output:

( - tan((c + d*x)/2)**5*b + tan((c + d*x)/2)**5*c + 14*tan((c + d*x)/2)**3 
*b - 8*tan((c + d*x)/2)**3*c - 12*tan((c + d*x)/2)**2*b*d*x + 6*tan((c + d 
*x)/2)**2*c*d*x + 27*tan((c + d*x)/2)*b - 9*tan((c + d*x)/2)*c - 12*b*d*x 
+ 6*c*d*x)/(6*a**2*d*(tan((c + d*x)/2)**2 + 1))