\(\int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [119]

Optimal result

Integrand size = 33, antiderivative size = 144 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 a (7 A+6 B) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a B \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt {a+a \sec (c+d x)}}-\frac {4 (7 A+6 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 (7 A+6 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 a d} \] Output:

2/15*a*(7*A+6*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/7*a*B*sec(d*x+c)^3* 
tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/105*(7*A+6*B)*(a+a*sec(d*x+c))^(1/2) 
*tan(d*x+c)/d+2/35*(7*A+6*B)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.56 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 a \left (8 (7 A+6 B)+4 (7 A+6 B) \sec (c+d x)+3 (7 A+6 B) \sec ^2(c+d x)+15 B \sec ^3(c+d x)\right ) \tan (c+d x)}{105 d \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(2*a*(8*(7*A + 6*B) + 4*(7*A + 6*B)*Sec[c + d*x] + 3*(7*A + 6*B)*Sec[c + d 
*x]^2 + 15*B*Sec[c + d*x]^3)*Tan[c + d*x])/(105*d*Sqrt[a*(1 + Sec[c + d*x] 
)])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{7} (7 A+6 B) \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+6 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4287

\(\displaystyle \frac {1}{7} (7 A+6 B) \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} (7 A+6 B) \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+6 B) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {1}{7} (7 A+6 B) \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+6 B) \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {1}{7} (7 A+6 B) \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\)

Input:

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

Output:

(2*a*B*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + ((7*A 
 + 6*B)*((2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a*d) + ((14*a^2*Ta 
n[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
 

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]
 

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {2 \left (\cos \left (d x +c \right ) \left (56 \cos \left (d x +c \right )^{2}+28 \cos \left (d x +c \right )+21\right ) A +\left (48 \cos \left (d x +c \right )^{3}+24 \cos \left (d x +c \right )^{2}+18 \cos \left (d x +c \right )+15\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (1+\cos \left (d x +c \right )\right )}\)	\(103\)
parts	\(\frac {A \left (16 \sin \left (d x +c \right )+8 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right )}+\frac {2 B \left (16 \cos \left (d x +c \right )^{3}+8 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{35 d \left (1+\cos \left (d x +c \right )\right )}\)	\(134\)

Input:

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

2/105/d*(cos(d*x+c)*(56*cos(d*x+c)^2+28*cos(d*x+c)+21)*A+(48*cos(d*x+c)^3+ 
24*cos(d*x+c)^2+18*cos(d*x+c)+15)*B)*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c 
))*tan(d*x+c)*sec(d*x+c)^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.73 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (8 \, {\left (7 \, A + 6 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (7 \, A + 6 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 6 \, B\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="fricas")
 

Output:

2/105*(8*(7*A + 6*B)*cos(d*x + c)^3 + 4*(7*A + 6*B)*cos(d*x + c)^2 + 3*(7* 
A + 6*B)*cos(d*x + c) + 15*B)*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]

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

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

Output:

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

Maxima [F]

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

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="maxima")
 

Output:

-8/105*((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*A*sin(4*d*x + 4*c) + (7*A + 6*B)*sin(2*d*x + 2*c))*cos(7/2*ar 
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (35*A*cos(4*d*x + 4*c) + 
7*(7*A + 6*B)*cos(2*d*x + 2*c) + 14*A + 12*B)*sin(7/2*arctan2(sin(2*d*x + 
2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) - 105*(((A + 2*B)*d*cos(2*d*x + 2*c) 
^4 + (A + 2*B)*d*sin(2*d*x + 2*c)^4 + 4*(A + 2*B)*d*cos(2*d*x + 2*c)^3 + 6 
*(A + 2*B)*d*cos(2*d*x + 2*c)^2 + 4*(A + 2*B)*d*cos(2*d*x + 2*c) + 2*((A + 
 2*B)*d*cos(2*d*x + 2*c)^2 + 2*(A + 2*B)*d*cos(2*d*x + 2*c) + (A + 2*B)*d) 
*sin(2*d*x + 2*c)^2 + (A + 2*B)*d)*integrate((((cos(10*d*x + 10*c)*cos(2*d 
*x + 2*c) + 4*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 6*cos(6*d*x + 6*c)*cos(2 
*d*x + 2*c) + 4*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + s 
in(10*d*x + 10*c)*sin(2*d*x + 2*c) + 4*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 
 6*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) 
 + sin(2*d*x + 2*c)^2)*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)) 
) + (cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 4*cos(2*d*x + 2*c)*sin(8*d*x + 
8*c) + 6*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 4*cos(2*d*x + 2*c)*sin(4*d*x 
+ 4*c) - cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 4*cos(8*d*x + 8*c)*sin(2*d* 
x + 2*c) - 6*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 4*cos(4*d*x + 4*c)*sin(2* 
d*x + 2*c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2* 
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*s...
 

Giac [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.54 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=-\frac {2 \, {\left (105 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (175 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 105 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (119 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 147 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (49 \, \sqrt {2} A a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 27 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)),x, algorith 
m="giac")
 

Output:

-2/105*(105*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*B*a^4*sgn(cos(d* 
x + c)) - (175*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 105*sqrt(2)*B*a^4*sgn(cos 
(d*x + c)) - (119*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 147*sqrt(2)*B*a^4*sgn( 
cos(d*x + c)) - (49*sqrt(2)*A*a^4*sgn(cos(d*x + c)) + 27*sqrt(2)*B*a^4*sgn 
(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d* 
x + 1/2*c)^2)*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 = 15.83 (sec) , antiderivative size = 407, normalized size of antiderivative = 2.83 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (-\frac {A\,8{}\mathrm {i}}{5\,d}+\frac {B\,16{}\mathrm {i}}{5\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {B\,16{}\mathrm {i}}{35\,d}+\frac {\left (56\,A+112\,B\right )\,1{}\mathrm {i}}{35\,d}\right )\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {A\,8{}\mathrm {i}}{7\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,8{}\mathrm {i}}{7\,d}-\frac {\left (8\,A+16\,B\right )\,1{}\mathrm {i}}{7\,d}\right )-\frac {\left (8\,A+16\,B\right )\,1{}\mathrm {i}}{7\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\left (\frac {A\,8{}\mathrm {i}}{3\,d}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (56\,A+48\,B\right )\,1{}\mathrm {i}}{105\,d}\right )\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (112\,A+96\,B\right )\,1{}\mathrm {i}}{105\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \] Input:

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2))/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)*((B*16i)/(5 
*d) - (A*8i)/(5*d) + exp(c*1i + d*x*1i)*((B*16i)/(35*d) + ((56*A + 112*B)* 
1i)/(35*d))))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a 
+ a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*8i)/(7*d) + 
 exp(c*1i + d*x*1i)*((A*8i)/(7*d) - ((8*A + 16*B)*1i)/(7*d)) - ((8*A + 16* 
B)*1i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) + ((( 
A*8i)/(3*d) - (exp(c*1i + d*x*1i)*(56*A + 48*B)*1i)/(105*d))*(a + a/(exp(- 
 c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1 
)*(exp(c*2i + d*x*2i) + 1)) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x 
*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(112*A + 96*B)*1i)/(105*d*(exp(c*1i 
+ d*x*1i) + 1))
                                                                                    
                                                                                    
 

Reduce [F]

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

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

Output:

sqrt(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)