\(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx\) [98]

Optimal result

Integrand size = 34, antiderivative size = 169 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx=-\frac {5 \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}-\frac {5 \tan (e+f x)}{8 a^2 f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {5 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}} \] Output:

-5/16*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))*2^(1/2 
)/a^2/c^(3/2)/f-5/8*tan(f*x+e)/a^2/f/(c-c*sec(f*x+e))^(3/2)+1/3*tan(f*x+e) 
/f/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2)+5/6*tan(f*x+e)/f/(a^2+a^2*sec 
(f*x+e))/(c-c*sec(f*x+e))^(3/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.43 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.38 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{6 a^2 c f (1+\sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \] Input:

Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)) 
,x]
 

Output:

(Hypergeometric2F1[-3/2, 2, -1/2, (1 + Sec[e + f*x])/2]*Tan[e + f*x])/(6*a 
^2*c*f*(1 + Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 4448, 3042, 4448, 3042, 4283, 3042, 4282, 216}

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.

\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4448

\(\displaystyle \frac {5 \int \frac {\sec (e+f x)}{(\sec (e+f x) a+a) (c-c \sec (e+f x))^{3/2}}dx}{6 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{6 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4448

\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}}dx}{2 a}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}\right )}{6 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{2 a}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}\right )}{6 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4283

\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}}dx}{4 c}-\frac {\tan (e+f x)}{2 f (c-c \sec (e+f x))^{3/2}}\right )}{2 a}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}\right )}{6 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4282

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}-\frac {\tan (e+f x)}{2 f (c-c \sec (e+f x))^{3/2}}\right )}{2 a}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}\right )}{6 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}\)

Input:

Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)),x]
 

Output:

Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)) + (5* 
(Tan[e + f*x]/(f*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)) + (3*(-1 
/2*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])]/(Sqrt 
[2]*c^(3/2)*f) - Tan[e + f*x]/(2*f*(c - c*Sec[e + f*x])^(3/2))))/(2*a)))/( 
6*a)
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

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

rule 4282

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4283

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/(a*f*(2*m + 1))), x] 
 + Simp[(m + 1)/(a*(2*m + 1))   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] && LtQ[m, -2^(-1) 
] && IntegerQ[2*m]
 

rule 4448

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[b*Cot[e + f*x]* 
(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[ 
(m + n + 1)/(a*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*( 
c + d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c 
 + a*d, 0] && EqQ[a^2 - b^2, 0] && ((ILtQ[m, 0] && ILtQ[n - 1/2, 0]) || (IL 
tQ[m - 1/2, 0] && ILtQ[n - 1/2, 0] && LtQ[m, n]))
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(967\) vs. \(2(146)=292\).

Time = 4.28 (sec) , antiderivative size = 968, normalized size of antiderivative = 5.73

method	result	size
default	\(\text {Expression too large to display}\)	\(968\)

Input:

int(sec(f*x+e)/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/a^2*(-1/96/f*2^(1/2)/((2*cos(1/2*f*x+1/2*e)^2-1)/(cos(1/2*f*x+1/2*e)+1)^ 
2)^(1/2)/(-c/(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)/c*(ln( 
2*(((2*cos(1/2*f*x+1/2*e)^2-1)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*cos(1/2*f*x 
+1/2*e)+((2*cos(1/2*f*x+1/2*e)^2-1)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)-2*cos( 
1/2*f*x+1/2*e)-1)/(cos(1/2*f*x+1/2*e)+1))*(117*cot(1/2*f*x+1/2*e)-117*csc( 
1/2*f*x+1/2*e))+arctanh((2*cos(1/2*f*x+1/2*e)-1)/(cos(1/2*f*x+1/2*e)+1)/(( 
2*cos(1/2*f*x+1/2*e)^2-1)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2))*(3*cot(1/2*f*x+ 
1/2*e)-3*csc(1/2*f*x+1/2*e))+ln(4*(((2*cos(1/2*f*x+1/2*e)^2-1)/(cos(1/2*f* 
x+1/2*e)+1)^2)^(1/2)*cos(1/2*f*x+1/2*e)+((2*cos(1/2*f*x+1/2*e)^2-1)/(cos(1 
/2*f*x+1/2*e)+1)^2)^(1/2)-2*cos(1/2*f*x+1/2*e)-1)/(cos(1/2*f*x+1/2*e)+1))* 
(-120*cot(1/2*f*x+1/2*e)+120*csc(1/2*f*x+1/2*e))+(14*cos(1/2*f*x+1/2*e)^4- 
12*cos(1/2*f*x+1/2*e)^2+4)*((2*cos(1/2*f*x+1/2*e)^2-1)/(cos(1/2*f*x+1/2*e) 
+1)^2)^(1/2)*sec(1/2*f*x+1/2*e)^3*csc(1/2*f*x+1/2*e))+1/16/f*2^(1/2)/((2*c 
os(1/2*f*x+1/2*e)^2-1)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)/(-c/(2*cos(1/2*f*x+ 
1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)/c*(ln(2*(((2*cos(1/2*f*x+1/2*e)^2- 
1)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)*cos(1/2*f*x+1/2*e)+((2*cos(1/2*f*x+1/2* 
e)^2-1)/(cos(1/2*f*x+1/2*e)+1)^2)^(1/2)-2*cos(1/2*f*x+1/2*e)-1)/(cos(1/2*f 
*x+1/2*e)+1))*(21*cot(1/2*f*x+1/2*e)-21*csc(1/2*f*x+1/2*e))+arctanh((2*cos 
(1/2*f*x+1/2*e)-1)/(cos(1/2*f*x+1/2*e)+1)/((2*cos(1/2*f*x+1/2*e)^2-1)/(cos 
(1/2*f*x+1/2*e)+1)^2)^(1/2))*(3*cot(1/2*f*x+1/2*e)-3*csc(1/2*f*x+1/2*e)...
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.18 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-c} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (13 \, \cos \left (f x + e\right )^{3} - 10 \, \cos \left (f x + e\right )^{2} - 15 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{96 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}, \frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (13 \, \cos \left (f x + e\right )^{3} - 10 \, \cos \left (f x + e\right )^{2} - 15 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{48 \, {\left (a^{2} c^{2} f \cos \left (f x + e\right )^{2} - a^{2} c^{2} f\right )} \sin \left (f x + e\right )}\right ] \] Input:

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2),x, algorith 
m="fricas")
 

Output:

[-1/96*(15*sqrt(2)*(cos(f*x + e)^2 - 1)*sqrt(-c)*log((2*sqrt(2)*(cos(f*x + 
 e)^2 + cos(f*x + e))*sqrt(-c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) + ( 
3*c*cos(f*x + e) + c)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin 
(f*x + e) + 4*(13*cos(f*x + e)^3 - 10*cos(f*x + e)^2 - 15*cos(f*x + e))*sq 
rt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^2*f*cos(f*x + e)^2 - a^2*c^ 
2*f)*sin(f*x + e)), 1/48*(15*sqrt(2)*(cos(f*x + e)^2 - 1)*sqrt(c)*arctan(s 
qrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f 
*x + e)))*sin(f*x + e) - 2*(13*cos(f*x + e)^3 - 10*cos(f*x + e)^2 - 15*cos 
(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^2*f*cos(f*x + 
e)^2 - a^2*c^2*f)*sin(f*x + e))]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\int \frac {\sec {\left (e + f x \right )}}{- c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{3}{\left (e + f x \right )} - c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} + c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx}{a^{2}} \] Input:

integrate(sec(f*x+e)/(a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**(3/2),x)
 

Output:

Integral(sec(e + f*x)/(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**3 - c*sq 
rt(-c*sec(e + f*x) + c)*sec(e + f*x)**2 + c*sqrt(-c*sec(e + f*x) + c)*sec( 
e + f*x) + c*sqrt(-c*sec(e + f*x) + c)), x)/a**2
 

Maxima [F]

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

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2),x, algorith 
m="maxima")
 

Output:

integrate(sec(f*x + e)/((a*sec(f*x + e) + a)^2*(-c*sec(f*x + e) + c)^(3/2) 
), x)
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.76 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (\frac {15 \, \arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {3 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}} + \frac {2 \, {\left ({\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} c^{6} - 6 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c^{7}\right )}}{c^{9}}\right )}}{48 \, a^{2} f} \] Input:

integrate(sec(f*x+e)/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2),x, algorith 
m="giac")
 

Output:

1/48*sqrt(2)*(15*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/c^(3/2 
) - 3*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/(c^2*tan(1/2*f*x + 1/2*e)^2) + 2* 
((c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c^6 - 6*sqrt(c*tan(1/2*f*x + 1/2*e)^ 
2 - c)*c^7)/c^9)/(a^2*f)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{\cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^(3/2)),x)
 

Output:

int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^2*(c - c/cos(e + f*x))^(3/2)), x)
 

Reduce [F]

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

int(sec(f*x+e)/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2),x)
 

Output:

(sqrt(c)*int((sqrt( - sec(e + f*x) + 1)*sec(e + f*x))/(sec(e + f*x)**4 - 2 
*sec(e + f*x)**2 + 1),x))/(a**2*c**2)