\(\int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [127]

Optimal result

Integrand size = 33, antiderivative size = 189 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^2 (39 A+34 B) \tan (c+d x)}{45 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (9 A+10 B) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}-\frac {4 a (39 A+34 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a B \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d} \] Output:

2/45*a^2*(39*A+34*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/63*a^2*(9*A+10* 
B)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/315*a*(39*A+34*B)*(a 
+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/9*a*B*sec(d*x+c)^3*(a+a*sec(d*x+c))^(1 
/2)*tan(d*x+c)/d+2/105*(39*A+34*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.53 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^2 \left (8 (39 A+34 B)+4 (39 A+34 B) \sec (c+d x)+3 (39 A+34 B) \sec ^2(c+d x)+5 (9 A+17 B) \sec ^3(c+d x)+35 B \sec ^4(c+d x)\right ) \tan (c+d x)}{315 d \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(2*a^2*(8*(39*A + 34*B) + 4*(39*A + 34*B)*Sec[c + d*x] + 3*(39*A + 34*B)*S 
ec[c + d*x]^2 + 5*(9*A + 17*B)*Sec[c + d*x]^3 + 35*B*Sec[c + d*x]^4)*Tan[c 
 + d*x])/(315*d*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4506, 27, 3042, 4504, 3042, 4287, 27, 3042, 4489, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {2}{9} \int \frac {1}{2} \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (3 A+2 B)+a (9 A+10 B) \sec (c+d x))dx+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (3 A+2 B)+a (9 A+10 B) \sec (c+d x))dx+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a (3 A+2 B)+a (9 A+10 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (39 A+34 B) \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (39 A+34 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 4287

\(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (39 A+34 B) \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (39 A+34 B) \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (39 A+34 B) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (39 A+34 B) \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {3}{7} a (39 A+34 B) \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {1}{9} \left (\frac {2 a^2 (9 A+10 B) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{7} a (39 A+34 B) \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\)

Input:

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

Output:

(2*a*B*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + ((2*a 
^2*(9*A + 10*B)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]] 
) + (3*a*(39*A + 34*B)*((2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d 
) + ((14*a^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*Sqrt[a + 
a*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a)))/7)/9
 

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 4287

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-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*(b*( 
m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - 
b^2, 0] &&  !LtQ[m, -2^(-1)]
 

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 4504

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

rule 4506

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

Maple [A] (verified)

Time = 1.78 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {2 a \left (\cos \left (d x +c \right ) \left (312 \cos \left (d x +c \right )^{3}+156 \cos \left (d x +c \right )^{2}+117 \cos \left (d x +c \right )+45\right ) A +\left (272 \cos \left (d x +c \right )^{4}+136 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+85 \cos \left (d x +c \right )+35\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (1+\cos \left (d x +c \right )\right )}\)	\(124\)
parts	\(\frac {2 A a \left (104 \cos \left (d x +c \right )^{3}+52 \cos \left (d x +c \right )^{2}+39 \cos \left (d x +c \right )+15\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (1+\cos \left (d x +c \right )\right )}+\frac {2 B a \left (272 \cos \left (d x +c \right )^{4}+136 \cos \left (d x +c \right )^{3}+102 \cos \left (d x +c \right )^{2}+85 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (1+\cos \left (d x +c \right )\right )}\)	\(158\)

Input:

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

2/315/d*a*(cos(d*x+c)*(312*cos(d*x+c)^3+156*cos(d*x+c)^2+117*cos(d*x+c)+45 
)*A+(272*cos(d*x+c)^4+136*cos(d*x+c)^3+102*cos(d*x+c)^2+85*cos(d*x+c)+35)* 
B)*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*tan(d*x+c)*sec(d*x+c)^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.67 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (8 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{4} + 4 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, A + 17 \, B\right )} a \cos \left (d x + c\right ) + 35 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorith 
m="fricas")
 

Output:

2/315*(8*(39*A + 34*B)*a*cos(d*x + c)^4 + 4*(39*A + 34*B)*a*cos(d*x + c)^3 
 + 3*(39*A + 34*B)*a*cos(d*x + c)^2 + 5*(9*A + 17*B)*a*cos(d*x + c) + 35*B 
*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 
 + d*cos(d*x + c)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.37 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {4 \, {\left ({\left ({\left ({\left (2 \, \sqrt {2} {\left (57 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \sqrt {2} {\left (57 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 47 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 819 \, \sqrt {2} {\left (A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \sqrt {2} {\left (7 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, \sqrt {2} {\left (A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorith 
m="giac")
 

Output:

4/315*((((2*sqrt(2)*(57*A*a^6*sgn(cos(d*x + c)) + 47*B*a^6*sgn(cos(d*x + c 
)))*tan(1/2*d*x + 1/2*c)^2 - 9*sqrt(2)*(57*A*a^6*sgn(cos(d*x + c)) + 47*B* 
a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 819*sqrt(2)*(A*a^6*sgn(co 
s(d*x + c)) + B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 105*sqrt( 
2)*(7*A*a^6*sgn(cos(d*x + c)) + 5*B*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 
1/2*c)^2 + 315*sqrt(2)*(A*a^6*sgn(cos(d*x + c)) + B*a^6*sgn(cos(d*x + c))) 
)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d 
*x + 1/2*c)^2 + a)*d)
 

Mupad [B] (verification not implemented)

Time = 20.42 (sec) , antiderivative size = 596, normalized size of antiderivative = 3.15 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2))/cos(c + d*x)^3,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*(A + 4*B)*8i)/(7*d) - (a*(3*A + 2*B)*8i)/(7*d) + (B*a*32i)/(6 
3*d)) + (A*a*8i)/(7*d) - (a*(A + 2*B)*24i)/(7*d) - (B*a*32i)/(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*(3*A + 2*B 
)*8i)/(9*d) - (a*(2*A + 3*B)*16i)/(9*d) + (A*a*8i)/(9*d)) + (a*(2*A + 3*B) 
*16i)/(9*d) - (a*(3*A + 2*B)*8i)/(9*d) - (A*a*8i)/(9*d)))/((exp(c*1i + d*x 
*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + (((A*a*8i)/(3*d) - (a*exp(c*1i + d 
*x*1i)*(39*A + 34*B)*8i)/(315*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1 
i + d*x*1i)/2))^(1/2))/((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*1 
i + d*x*1i)*((a*(3*A + 2*B)*8i)/(5*d) + (a*(3*A + B)*16i)/(105*d)) - (A*a* 
8i)/(5*d) + (a*(A + 3*B)*16i)/(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)*(39*A + 34*B)*16i)/(315*d*(exp(c*1i + d*x*1i 
) + 1))
 

Reduce [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) a \right ) \] Input:

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

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**5,x)*b + int(sqrt(sec( 
c + d*x) + 1)*sec(c + d*x)**4,x)*a + int(sqrt(sec(c + d*x) + 1)*sec(c + d* 
x)**4,x)*b + int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)*a)