\(\int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx\) [64]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 32, antiderivative size = 163 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx=-\frac {256 c^4 (a+a \sec (e+f x)) \tan (e+f x)}{315 f \sqrt {c-c \sec (e+f x)}}-\frac {64 c^3 (a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{105 f}-\frac {8 c^2 (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{21 f}-\frac {2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{9 f} \] Output:

-256/315*c^4*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)-64/105*c 
^3*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f-8/21*c^2*(a+a*sec( 
f*x+e))*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f-2/9*c*(a+a*sec(f*x+e))*(c-c*se 
c(f*x+e))^(5/2)*tan(f*x+e)/f
 

Mathematica [A] (verified)

Time = 1.87 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.43 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx=\frac {2 a c^4 (1+\sec (e+f x)) \left (-319+321 \sec (e+f x)-165 \sec ^2(e+f x)+35 \sec ^3(e+f x)\right ) \tan (e+f x)}{315 f \sqrt {c-c \sec (e+f x)}} \] Input:

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(7/2),x]
 

Output:

(2*a*c^4*(1 + Sec[e + f*x])*(-319 + 321*Sec[e + f*x] - 165*Sec[e + f*x]^2 
+ 35*Sec[e + f*x]^3)*Tan[e + f*x])/(315*f*Sqrt[c - c*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4443, 3042, 4443, 3042, 4443, 3042, 4441}

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.

\(\displaystyle \int \sec (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{7/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{7/2}dx\)

\(\Big \downarrow \) 4443

\(\displaystyle \frac {4}{3} c \int \sec (e+f x) (\sec (e+f x) a+a) (c-c \sec (e+f x))^{5/2}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{3} c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f}\)

\(\Big \downarrow \) 4443

\(\displaystyle \frac {4}{3} c \left (\frac {8}{7} c \int \sec (e+f x) (\sec (e+f x) a+a) (c-c \sec (e+f x))^{3/2}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{3} c \left (\frac {8}{7} c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f}\)

\(\Big \downarrow \) 4443

\(\displaystyle \frac {4}{3} c \left (\frac {8}{7} c \left (\frac {4}{5} c \int \sec (e+f x) (\sec (e+f x) a+a) \sqrt {c-c \sec (e+f x)}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{3} c \left (\frac {8}{7} c \left (\frac {4}{5} c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right ) \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f}\)

\(\Big \downarrow \) 4441

\(\displaystyle \frac {4}{3} c \left (\frac {8}{7} c \left (-\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)}{15 f \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}{5 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{7 f}\right )-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f}\)

Input:

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(7/2),x]
 

Output:

(-2*c*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(9*f) 
+ (4*c*((-2*c*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x] 
)/(7*f) + (8*c*((-8*c^2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(15*f*Sqrt[c - 
c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x]]*Tan[ 
e + f*x])/(5*f)))/7))/3
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4441
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*Sq 
rt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Simp[2*a*c*Cot[e + f 
*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1)*Sqrt[c + d*Csc[e + f*x]])), x] / 
; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] 
 && NeQ[m, -2^(-1)]
 

rule 4443
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c 
sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[(-d)*Cot[e + f 
*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + 
 Simp[c*((2*n - 1)/(m + n))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(c + 
d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b 
*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0] &&  !LtQ[m, -2^(-1)] 
&&  !(IGtQ[m - 1/2, 0] && LtQ[m, n])
 
Maple [A] (verified)

Time = 4.81 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.47

method result size
default \(a \left (-\frac {16 \left (364 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-819 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+711 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-285 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+45\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, \sqrt {2}\, c^{3} \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{45 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{4}}+\frac {16 \left (177 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-301 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+175 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-35\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, \sqrt {2}\, c^{3} \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{35 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}\right )\) \(239\)
parts \(\frac {16 a \left (177 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-301 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+175 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-35\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, \sqrt {2}\, c^{3} \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{35 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}-\frac {16 a \left (364 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-819 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+711 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-285 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+45\right ) \sqrt {-\frac {c \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}}\, \sqrt {2}\, c^{3} \cot \left (\frac {f x}{2}+\frac {e}{2}\right )}{45 f \left (2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{4}}\) \(239\)

Input:

int(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(7/2),x,method=_RETURNVER 
BOSE)
 

Output:

a*(-16/45/f*(364*cos(1/2*f*x+1/2*e)^8-819*cos(1/2*f*x+1/2*e)^6+711*cos(1/2 
*f*x+1/2*e)^4-285*cos(1/2*f*x+1/2*e)^2+45)*(-c/(2*cos(1/2*f*x+1/2*e)^2-1)* 
sin(1/2*f*x+1/2*e)^2)^(1/2)*2^(1/2)*c^3/(2*cos(1/2*f*x+1/2*e)^2-1)^4*cot(1 
/2*f*x+1/2*e)+16/35/f*(177*cos(1/2*f*x+1/2*e)^6-301*cos(1/2*f*x+1/2*e)^4+1 
75*cos(1/2*f*x+1/2*e)^2-35)*(-c/(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2 
*e)^2)^(1/2)*2^(1/2)*c^3/(2*cos(1/2*f*x+1/2*e)^2-1)^3*cot(1/2*f*x+1/2*e))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.73 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx=\frac {2 \, {\left (319 \, a c^{3} \cos \left (f x + e\right )^{5} + 317 \, a c^{3} \cos \left (f x + e\right )^{4} - 158 \, a c^{3} \cos \left (f x + e\right )^{3} - 26 \, a c^{3} \cos \left (f x + e\right )^{2} + 95 \, a c^{3} \cos \left (f x + e\right ) - 35 \, a c^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{315 \, f \cos \left (f x + e\right )^{4} \sin \left (f x + e\right )} \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(7/2),x, algorithm= 
"fricas")
 

Output:

2/315*(319*a*c^3*cos(f*x + e)^5 + 317*a*c^3*cos(f*x + e)^4 - 158*a*c^3*cos 
(f*x + e)^3 - 26*a*c^3*cos(f*x + e)^2 + 95*a*c^3*cos(f*x + e) - 35*a*c^3)* 
sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^4*sin(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )} {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \sec \left (f x + e\right ) \,d x } \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(7/2),x, algorithm= 
"maxima")
 

Output:

-2/315*(315*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) 
+ 1)^(1/4)*(5*(a*c^3*f*cos(2*f*x + 2*e)^4 + a*c^3*f*sin(2*f*x + 2*e)^4 + 4 
*a*c^3*f*cos(2*f*x + 2*e)^3 + 6*a*c^3*f*cos(2*f*x + 2*e)^2 + 4*a*c^3*f*cos 
(2*f*x + 2*e) + a*c^3*f + 2*(a*c^3*f*cos(2*f*x + 2*e)^2 + 2*a*c^3*f*cos(2* 
f*x + 2*e) + a*c^3*f)*sin(2*f*x + 2*e)^2)*integrate((cos(2*f*x + 2*e)^2 + 
sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(((cos(6*f*x + 6*e)*cos 
(2*f*x + 2*e) + 2*cos(4*f*x + 4*e)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + 
 sin(6*f*x + 6*e)*sin(2*f*x + 2*e) + 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 
 sin(2*f*x + 2*e)^2)*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) 
+ (cos(2*f*x + 2*e)*sin(6*f*x + 6*e) + 2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) 
 - cos(6*f*x + 6*e)*sin(2*f*x + 2*e) - 2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e) 
)*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*cos(7/2*arctan2(si 
n(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - ((cos(2*f*x + 2*e)*sin(6*f*x + 6* 
e) + 2*cos(2*f*x + 2*e)*sin(4*f*x + 4*e) - cos(6*f*x + 6*e)*sin(2*f*x + 2* 
e) - 2*cos(4*f*x + 4*e)*sin(2*f*x + 2*e))*cos(9/2*arctan2(sin(2*f*x + 2*e) 
, cos(2*f*x + 2*e))) - (cos(6*f*x + 6*e)*cos(2*f*x + 2*e) + 2*cos(4*f*x + 
4*e)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + sin(6*f*x + 6*e)*sin(2*f*x + 
2*e) + 2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + sin(2*f*x + 2*e)^2)*sin(9/2*a 
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(7/2*arctan2(sin(2*f*x + 2 
*e), cos(2*f*x + 2*e) + 1)))/((cos(2*f*x + 2*e)^6 + sin(2*f*x + 2*e)^6 ...
 

Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.66 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx=\frac {32 \, \sqrt {2} {\left (105 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{3} c^{2} + 189 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} c^{3} + 135 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{4} + 35 \, c^{5}\right )} a c^{3}}{315 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {9}{2}} f} \] Input:

integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(7/2),x, algorithm= 
"giac")
 

Output:

32/315*sqrt(2)*(105*(c*tan(1/2*f*x + 1/2*e)^2 - c)^3*c^2 + 189*(c*tan(1/2* 
f*x + 1/2*e)^2 - c)^2*c^3 + 135*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^4 + 35*c^ 
5)*a*c^3/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(9/2)*f)
 

Mupad [B] (verification not implemented)

Time = 17.32 (sec) , antiderivative size = 483, normalized size of antiderivative = 2.96 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx =\text {Too large to display} \] Input:

int(((a + a/cos(e + f*x))*(c - c/cos(e + f*x))^(7/2))/cos(e + f*x),x)
 

Output:

((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a*c^3*2i) 
/f + (a*c^3*exp(e*1i + f*x*1i)*638i)/(315*f)))/(exp(e*1i + f*x*1i) - 1) - 
((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a*c^3*32i 
)/(9*f) + (a*c^3*exp(e*1i + f*x*1i)*32i)/(9*f)))/((exp(e*1i + f*x*1i) - 1) 
*(exp(e*2i + f*x*2i) + 1)^4) + ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i 
+ f*x*1i)/2))^(1/2)*((a*c^3*96i)/(7*f) + (a*c^3*exp(e*1i + f*x*1i)*32i)/(6 
3*f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^3) - ((c - c/(ex 
p(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((a*c^3*64i)/(5*f) - ( 
a*c^3*exp(e*1i + f*x*1i)*736i)/(105*f)))/((exp(e*1i + f*x*1i) - 1)*(exp(e* 
2i + f*x*2i) + 1)^2) + ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i 
)/2))^(1/2)*((a*c^3*8i)/(3*f) - (a*c^3*exp(e*1i + f*x*1i)*1256i)/(315*f))) 
/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1))
 

Reduce [F]

\[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{7/2} \, dx=\sqrt {c}\, a \,c^{3} \left (-\left (\int \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{5}d x \right )+2 \left (\int \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{4}d x \right )-2 \left (\int \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )^{2}d x \right )+\int \sqrt {-\sec \left (f x +e \right )+1}\, \sec \left (f x +e \right )d x \right ) \] Input:

int(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e))^(7/2),x)
 

Output:

sqrt(c)*a*c**3*( - int(sqrt( - sec(e + f*x) + 1)*sec(e + f*x)**5,x) + 2*in 
t(sqrt( - sec(e + f*x) + 1)*sec(e + f*x)**4,x) - 2*int(sqrt( - sec(e + f*x 
) + 1)*sec(e + f*x)**2,x) + int(sqrt( - sec(e + f*x) + 1)*sec(e + f*x),x))