\(\int \sec (c+d x) (a+a \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [367]

Optimal result

Integrand size = 40, antiderivative size = 138 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {8 a^2 (21 B+19 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (21 B+19 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 B-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 a d} \] Output:

8/105*a^2*(21*B+19*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/105*a*(21*B+19 
*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/35*(7*B-2*C)*(a+a*sec(d*x+c))^(3 
/2)*tan(d*x+c)/d+2/7*C*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/a/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.59 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^2 \left (2 (63 B+52 C)+(63 B+52 C) \sec (c+d x)+3 (7 B+13 C) \sec ^2(c+d x)+15 C \sec ^3(c+d x)\right ) \tan (c+d x)}{105 d \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[Sec[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 

Output:

(2*a^2*(2*(63*B + 52*C) + (63*B + 52*C)*Sec[c + d*x] + 3*(7*B + 13*C)*Sec[ 
c + d*x]^2 + 15*C*Sec[c + d*x]^3)*Tan[c + d*x])/(105*d*Sqrt[a*(1 + Sec[c + 
 d*x])])
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.06, 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, 4498, 27, 3042, 4489, 3042, 4280, 3042, 4279}

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.

\(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a)^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4498

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (\sec (c+d x) a+a)^{3/2} (5 a C+a (7 B-2 C) \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) (\sec (c+d x) a+a)^{3/2} (5 a C+a (7 B-2 C) \sec (c+d x))dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (5 a C+a (7 B-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 4280

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {\frac {1}{5} a (21 B+19 C) \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a (7 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 a d}\)

Input:

Int[Sec[c + d*x]*(a + a*Sec[c + d*x])^(3/2)*(B*Sec[c + d*x] + C*Sec[c + d* 
x]^2),x]
 

Output:

(2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*a*d) + ((2*a*(7*B - 2*C)* 
(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (a*(21*B + 19*C)*((8*a^2* 
Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[a + a*Sec[c + d*x 
]]*Tan[c + d*x])/(3*d)))/5)/(7*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 4279

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free 
Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4280

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[(-b)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] 
+ Simp[a*((2*m - 1)/m)   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], 
 x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && Intege 
rQ[2*m]
 

rule 4489

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

rule 4498

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( 
csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* 
((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int 
[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) 
*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a 
*B, 0] &&  !LtQ[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 = 1.36 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {2 a \left (\cos \left (d x +c \right ) \left (126 \cos \left (d x +c \right )^{2}+63 \cos \left (d x +c \right )+21\right ) B +\left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\)	\(104\)
parts	\(\frac {B \left (12 \sin \left (d x +c \right )+6 \tan \left (d x +c \right )+2 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (5 \cos \left (d x +c \right )+5\right )}+\frac {2 C a \left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\)	\(136\)

Input:

int(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth 
od=_RETURNVERBOSE)
 

Output:

2/105/d*a*(cos(d*x+c)*(126*cos(d*x+c)^2+63*cos(d*x+c)+21)*B+(104*cos(d*x+c 
)^3+52*cos(d*x+c)^2+39*cos(d*x+c)+15)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x 
+c)+1)*tan(d*x+c)*sec(d*x+c)^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (2 \, {\left (63 \, B + 52 \, C\right )} a \cos \left (d x + c\right )^{3} + {\left (63 \, B + 52 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 13 \, C\right )} a \cos \left (d x + c\right ) + 15 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="fricas")
 

Output:

2/105*(2*(63*B + 52*C)*a*cos(d*x + c)^3 + (63*B + 52*C)*a*cos(d*x + c)^2 + 
 3*(7*B + 13*C)*a*cos(d*x + c) + 15*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
 + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
 

Sympy [F]

\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)*(a+a*sec(d*x+c))**(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2 
),x)
 

Output:

Integral((a*(sec(c + d*x) + 1))**(3/2)*(B + C*sec(c + d*x))*sec(c + d*x)** 
2, x)
 

Maxima [F]

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

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="maxima")
 

Output:

-4/105*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) 
^(1/4)*(7*(5*(3*B + 2*C)*a*sin(4*d*x + 4*c) + 2*(12*B + 13*C)*a*sin(2*d*x 
+ 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (35*(3* 
B + 2*C)*a*cos(4*d*x + 4*c) + 14*(12*B + 13*C)*a*cos(2*d*x + 2*c) + (63*B 
+ 52*C)*a)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt( 
a) - 105*((B*a*d*cos(2*d*x + 2*c)^4 + B*a*d*sin(2*d*x + 2*c)^4 + 4*B*a*d*c 
os(2*d*x + 2*c)^3 + 6*B*a*d*cos(2*d*x + 2*c)^2 + 4*B*a*d*cos(2*d*x + 2*c) 
+ B*a*d + 2*(B*a*d*cos(2*d*x + 2*c)^2 + 2*B*a*d*cos(2*d*x + 2*c) + B*a*d)* 
sin(2*d*x + 2*c)^2)*integrate((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2 
*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 3*cos( 
6*d*x + 6*c)*cos(2*d*x + 2*c) + 3*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos( 
2*d*x + 2*c)^2 + sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 3*sin(6*d*x + 6*c)*si 
n(2*d*x + 2*c) + 3*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2) 
*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)* 
sin(8*d*x + 8*c) + 3*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c 
)*sin(4*d*x + 4*c) - cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 3*cos(6*d*x + 6*c 
)*sin(2*d*x + 2*c) - 3*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*sin(7/2*arctan2( 
sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 3*cos(2*d*x + 
 2*c)*sin(6*d*x + 6*c) + 3*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(8*d*...
 

Giac [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.56 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left ({\left ({\left (2 \, \sqrt {2} {\left (21 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 19 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \sqrt {2} {\left (21 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 19 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 70 \, \sqrt {2} {\left (3 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \sqrt {2} {\left (B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="giac")
 

Output:

4/105*(((2*sqrt(2)*(21*B*a^5*sgn(cos(d*x + c)) + 19*C*a^5*sgn(cos(d*x + c) 
))*tan(1/2*d*x + 1/2*c)^2 - 7*sqrt(2)*(21*B*a^5*sgn(cos(d*x + c)) + 19*C*a 
^5*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 70*sqrt(2)*(3*B*a^5*sgn(co 
s(d*x + c)) + 2*C*a^5*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 105*sqr 
t(2)*(B*a^5*sgn(cos(d*x + c)) + C*a^5*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/ 
2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a) 
*d)
 

Mupad [B] (verification not implemented)

Time = 16.53 (sec) , antiderivative size = 479, normalized size of antiderivative = 3.47 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + 
 d*x*1i)*((a*(3*B + 2*C)*4i)/(7*d) - (a*(2*B + 3*C)*8i)/(7*d) + (B*a*4i)/( 
7*d)) - (a*(2*B + 3*C)*8i)/(7*d) + (a*(3*B + 2*C)*4i)/(7*d) + (B*a*4i)/(7* 
d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) - ((a + a/(exp( 
- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((a* 
(7*B + 13*C)*8i)/(105*d) - (B*a*4i)/(3*d)) - (a*(3*B + 2*C)*4i)/(3*d)))/(( 
exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) + ((a + a/(exp(- c*1i - 
d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((C*a*16i)/(3 
5*d) - (B*a*4i)/(5*d) + (a*(B + 2*C)*12i)/(5*d)) - (a*(3*B + 2*C)*4i)/(5*d 
) + (a*(B + 4*C)*4i)/(5*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) 
 + 1)^2) - (a*exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i 
 + d*x*1i)/2))^(1/2)*(63*B + 52*C)*4i)/(105*d*(exp(c*1i + d*x*1i) + 1))
 

Reduce [F]

\[ \int \sec (c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) b \right ) \] Input:

int(sec(d*x+c)*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4,x)*c + int(sqrt(sec( 
c + d*x) + 1)*sec(c + d*x)**3,x)*b + int(sqrt(sec(c + d*x) + 1)*sec(c + d* 
x)**3,x)*c + int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*b)