\(\int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [137]

Optimal result

Integrand size = 31, antiderivative size = 138 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {64 a^3 (7 A+5 B) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (7 A+5 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 A+5 B) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 B (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \] Output:

64/105*a^3*(7*A+5*B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/105*a^2*(7*A+5 
*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/35*a*(7*A+5*B)*(a+a*sec(d*x+c))^ 
(3/2)*tan(d*x+c)/d+2/7*B*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.64 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^2 \sqrt {a (1+\sec (c+d x))} \left ((301 A+230 B) \sin (c+d x)+\left (98 A+115 B+3 (7 A+20 B) \sec (c+d x)+15 B \sec ^2(c+d x)\right ) \tan (c+d x)\right )}{105 d (1+\cos (c+d x))} \] Input:

Integrate[Sec[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]
 

Output:

(2*a^2*Sqrt[a*(1 + Sec[c + d*x])]*((301*A + 230*B)*Sin[c + d*x] + (98*A + 
115*B + 3*(7*A + 20*B)*Sec[c + d*x] + 15*B*Sec[c + d*x]^2)*Tan[c + d*x]))/ 
(105*d*(1 + Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 4489, 3042, 4280, 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.

\(\displaystyle \int \sec (c+d x) (a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4489

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4280

\(\displaystyle \frac {1}{7} (7 A+5 B) \left (\frac {8}{5} a \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+5 B) \left (\frac {8}{5} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 4280

\(\displaystyle \frac {1}{7} (7 A+5 B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+5 B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 B \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 4279

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

Input:

Int[Sec[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]
 

Output:

(2*B*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((7*A + 5*B)*((2*a*( 
a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (8*a*((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
 

Defintions of rubi rules used

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)]
 

Maple [A] (verified)

Time = 27.71 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77

method	result	size
default	\(\frac {2 a^{2} \left (\cos \left (d x +c \right ) \left (301 \cos \left (d x +c \right )^{2}+98 \cos \left (d x +c \right )+21\right ) A +\left (230 \cos \left (d x +c \right )^{3}+115 \cos \left (d x +c \right )^{2}+60 \cos \left (d x +c \right )+15\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (1+\cos \left (d x +c \right )\right )}\)	\(106\)
parts	\(\frac {A \left (86 \sin \left (d x +c \right )+28 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right )}+\frac {2 B \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (1+\cos \left (d x +c \right )\right )}\)	\(140\)

Input:

int(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

2/105/d*a^2*(cos(d*x+c)*(301*cos(d*x+c)^2+98*cos(d*x+c)+21)*A+(230*cos(d*x 
+c)^3+115*cos(d*x+c)^2+60*cos(d*x+c)+15)*B)*(a*(1+sec(d*x+c)))^(1/2)/(1+co 
s(d*x+c))*tan(d*x+c)*sec(d*x+c)^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left ({\left (301 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (98 \, A + 115 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 20 \, B\right )} a^{2} \cos \left (d x + c\right ) + 15 \, B a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm= 
"fricas")
 

Output:

2/105*((301*A + 230*B)*a^2*cos(d*x + c)^3 + (98*A + 115*B)*a^2*cos(d*x + c 
)^2 + 3*(7*A + 20*B)*a^2*cos(d*x + c) + 15*B*a^2)*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]

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

integrate(sec(d*x+c)*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)
 

Output:

Integral((a*(sec(c + d*x) + 1))**(5/2)*(A + B*sec(c + d*x))*sec(c + d*x), 
x)
 

Maxima [F]

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

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm= 
"maxima")
 

Output:

-2/105*((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*(15*A*a^2*sin(6*d*x + 6*c) + 5*(17*A + 10*B)*a^2*sin(4*d*x + 4*c 
) + (113*A + 100*B)*a^2*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c) 
, cos(2*d*x + 2*c) + 1)) - (105*A*a^2*cos(6*d*x + 6*c) + 35*(17*A + 10*B)* 
a^2*cos(4*d*x + 4*c) + 7*(113*A + 100*B)*a^2*cos(2*d*x + 2*c) + (301*A + 2 
30*B)*a^2)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt( 
a) - 105*(((5*A + 2*B)*a^2*d*cos(2*d*x + 2*c)^4 + (5*A + 2*B)*a^2*d*sin(2* 
d*x + 2*c)^4 + 4*(5*A + 2*B)*a^2*d*cos(2*d*x + 2*c)^3 + 6*(5*A + 2*B)*a^2* 
d*cos(2*d*x + 2*c)^2 + 4*(5*A + 2*B)*a^2*d*cos(2*d*x + 2*c) + (5*A + 2*B)* 
a^2*d + 2*((5*A + 2*B)*a^2*d*cos(2*d*x + 2*c)^2 + 2*(5*A + 2*B)*a^2*d*cos( 
2*d*x + 2*c) + (5*A + 2*B)*a^2*d)*sin(2*d*x + 2*c)^2)*integrate((((cos(6*d 
*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d 
*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*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(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin 
(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*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(5 
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*s 
in(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*s 
in(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(7/2*arctan2(...
 

Giac [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.57 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {8 \, {\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (7 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \sqrt {2} {\left (7 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, \sqrt {2} {\left (7 \, A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \sqrt {2} {\left (A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

integrate(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm= 
"giac")
 

Output:

8/105*((4*(2*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 - 7*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 + 35*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 - 105*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)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d 
)
 

Mupad [B] (verification not implemented)

Time = 16.89 (sec) , antiderivative size = 590, normalized size of antiderivative = 4.28 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/cos(c + d*x),x)
 

Output:

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*a^2*2i) 
/d - (a^2*exp(c*1i + d*x*1i)*(301*A + 230*B)*2i)/(105*d)))/(exp(c*1i + d*x 
*1i) + 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)*((A*a^2*2i)/(7*d) + (a^2*(3*A + 4*B)*10i)/(7*d) - (a^ 
2*(5*A + 2*B)*2i)/(7*d) - (a^2*(11*A + 10*B)*2i)/(7*d)) + (A*a^2*2i)/(7*d) 
 + (a^2*(3*A + 4*B)*10i)/(7*d) - (a^2*(5*A + 2*B)*2i)/(7*d) - (a^2*(11*A + 
 10*B)*2i)/(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^2*(5*A + 2*B)*2i)/(5*d) - (a^2*(5*A + 9*B)*4i)/(5*d) + (a^2* 
(7*A - 8*B)*2i)/(35*d)) - (A*a^2*2i)/(5*d) - (a^2*(A + 2*B)*2i)/d + (a^2*( 
A + B)*4i)/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)*(exp(c*1i + d* 
x*1i)*((a^2*(5*A + 2*B)*2i)/(3*d) - (a^2*(63*A + 80*B)*2i)/(105*d)) - (A*a 
^2*2i)/(3*d) + (a^2*(9*A + 10*B)*2i)/(3*d)))/((exp(c*1i + d*x*1i) + 1)*(ex 
p(c*2i + d*x*2i) + 1))
 

Reduce [F]

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

int(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)
 

Output:

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