3.1.34 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx\) [34]

3.1.34.1 Optimal result
3.1.34.2 Mathematica [C] (verified)
3.1.34.3 Rubi [A] (verified)
3.1.34.4 Maple [A] (verified)
3.1.34.5 Fricas [A] (verification not implemented)
3.1.34.6 Sympy [F]
3.1.34.7 Maxima [B] (verification not implemented)
3.1.34.8 Giac [A] (verification not implemented)
3.1.34.9 Mupad [B] (verification not implemented)

3.1.34.1 Optimal result

Integrand size = 32, antiderivative size = 121 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx=-\frac {35 c^4 \text {arctanh}(\sin (e+f x))}{2 a f}+\frac {28 c^4 \tan (e+f x)}{a f}-\frac {21 c^4 \sec (e+f x) \tan (e+f x)}{2 a f}+\frac {2 c (c-c \sec (e+f x))^3 \tan (e+f x)}{f (a+a \sec (e+f x))}+\frac {7 c^4 \tan ^3(e+f x)}{3 a f} \]

output
-35/2*c^4*arctanh(sin(f*x+e))/a/f+28*c^4*tan(f*x+e)/a/f-21/2*c^4*sec(f*x+e 
)*tan(f*x+e)/a/f+2*c*(c-c*sec(f*x+e))^3*tan(f*x+e)/f/(a+a*sec(f*x+e))+7/3* 
c^4*tan(f*x+e)^3/a/f
 
3.1.34.2 Mathematica [C] (verified)

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

Time = 0.99 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.44 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx=-\frac {16 c^4 \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \sqrt {2-2 \sec (e+f x)}}{a f} \]

input
Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^4)/(a + a*Sec[e + f*x]),x]
 
output
(-16*c^4*Cot[e + f*x]*Hypergeometric2F1[-7/2, -1/2, 1/2, (1 + Sec[e + f*x] 
)/2]*Sqrt[2 - 2*Sec[e + f*x]])/(a*f)
 
3.1.34.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 4445, 3042, 4278, 2009}

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.

\(\displaystyle \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a \sec (e+f x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4}{a \csc \left (e+f x+\frac {\pi }{2}\right )+a}dx\)

\(\Big \downarrow \) 4445

\(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^3}{f (a \sec (e+f x)+a)}-\frac {7 c \int \sec (e+f x) (c-c \sec (e+f x))^3dx}{a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4278

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^3}{f (a \sec (e+f x)+a)}-\frac {7 c \left (\frac {5 c^3 \text {arctanh}(\sin (e+f x))}{2 f}-\frac {c^3 \tan ^3(e+f x)}{3 f}-\frac {4 c^3 \tan (e+f x)}{f}+\frac {3 c^3 \tan (e+f x) \sec (e+f x)}{2 f}\right )}{a}\)

input
Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^4)/(a + a*Sec[e + f*x]),x]
 
output
(2*c*(c - c*Sec[e + f*x])^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])) - (7*c* 
((5*c^3*ArcTanh[Sin[e + f*x]])/(2*f) - (4*c^3*Tan[e + f*x])/f + (3*c^3*Sec 
[e + f*x]*Tan[e + f*x])/(2*f) - (c^3*Tan[e + f*x]^3)/(3*f)))/a
 

3.1.34.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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

rule 4278
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f 
*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I 
GtQ[m, 0] && RationalQ[n]
 

rule 4445
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + 
f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m + 1))), 
 x] - Simp[d*((2*n - 1)/(b*(2*m + 1)))   Int[Csc[e + f*x]*(a + b*Csc[e + f* 
x])^(m + 1)*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f 
}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && LtQ[m, -2^ 
(-1)] && IntegerQ[2*m]
 
3.1.34.4 Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.14

method result size
parallelrisch \(\frac {105 \left (\left (\cos \left (f x +e \right )+\frac {\cos \left (3 f x +3 e \right )}{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\left (-\cos \left (f x +e \right )-\frac {\cos \left (3 f x +3 e \right )}{3}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\frac {446 \left (\cos \left (f x +e \right )+\frac {55 \cos \left (2 f x +2 e \right )}{223}+\frac {83 \cos \left (3 f x +3 e \right )}{223}+\frac {59}{223}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{315}\right ) c^{4}}{2 a f \left (\cos \left (3 f x +3 e \right )+3 \cos \left (f x +e \right )\right )}\) \(138\)
derivativedivides \(\frac {16 c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{48 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {3}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {29}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {35 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{32}-\frac {1}{48 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {3}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {29}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {35 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{32}\right )}{f a}\) \(140\)
default \(\frac {16 c^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{48 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {3}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {29}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {35 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{32}-\frac {1}{48 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {3}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {29}{32 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {35 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{32}\right )}{f a}\) \(140\)
risch \(\frac {i c^{4} \left (111 \,{\mathrm e}^{6 i \left (f x +e \right )}+81 \,{\mathrm e}^{5 i \left (f x +e \right )}+354 \,{\mathrm e}^{4 i \left (f x +e \right )}+144 \,{\mathrm e}^{3 i \left (f x +e \right )}+417 \,{\mathrm e}^{2 i \left (f x +e \right )}+55 \,{\mathrm e}^{i \left (f x +e \right )}+166\right )}{3 a f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}+\frac {35 c^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 a f}-\frac {35 c^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 a f}\) \(156\)
norman \(\frac {\frac {35 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {385 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}+\frac {511 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 a f}-\frac {93 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}+\frac {16 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{4}}+\frac {35 c^{4} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 a f}-\frac {35 c^{4} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 a f}\) \(173\)

input
int(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e)),x,method=_RETURNVERBOSE 
)
 
output
105/2*((cos(f*x+e)+1/3*cos(3*f*x+3*e))*ln(tan(1/2*f*x+1/2*e)-1)+(-cos(f*x+ 
e)-1/3*cos(3*f*x+3*e))*ln(tan(1/2*f*x+1/2*e)+1)+446/315*(cos(f*x+e)+55/223 
*cos(2*f*x+2*e)+83/223*cos(3*f*x+3*e)+59/223)*tan(1/2*f*x+1/2*e))*c^4/a/f/ 
(cos(3*f*x+3*e)+3*cos(f*x+e))
 
3.1.34.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.26 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx=-\frac {105 \, {\left (c^{4} \cos \left (f x + e\right )^{4} + c^{4} \cos \left (f x + e\right )^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 105 \, {\left (c^{4} \cos \left (f x + e\right )^{4} + c^{4} \cos \left (f x + e\right )^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (166 \, c^{4} \cos \left (f x + e\right )^{3} + 55 \, c^{4} \cos \left (f x + e\right )^{2} - 13 \, c^{4} \cos \left (f x + e\right ) + 2 \, c^{4}\right )} \sin \left (f x + e\right )}{12 \, {\left (a f \cos \left (f x + e\right )^{4} + a f \cos \left (f x + e\right )^{3}\right )}} \]

input
integrate(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="fri 
cas")
 
output
-1/12*(105*(c^4*cos(f*x + e)^4 + c^4*cos(f*x + e)^3)*log(sin(f*x + e) + 1) 
 - 105*(c^4*cos(f*x + e)^4 + c^4*cos(f*x + e)^3)*log(-sin(f*x + e) + 1) - 
2*(166*c^4*cos(f*x + e)^3 + 55*c^4*cos(f*x + e)^2 - 13*c^4*cos(f*x + e) + 
2*c^4)*sin(f*x + e))/(a*f*cos(f*x + e)^4 + a*f*cos(f*x + e)^3)
 
3.1.34.6 Sympy [F]

\[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx=\frac {c^{4} \left (\int \frac {\sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {4 \sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {6 \sec ^{3}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {4 \sec ^{4}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{5}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx\right )}{a} \]

input
integrate(sec(f*x+e)*(c-c*sec(f*x+e))**4/(a+a*sec(f*x+e)),x)
 
output
c**4*(Integral(sec(e + f*x)/(sec(e + f*x) + 1), x) + Integral(-4*sec(e + f 
*x)**2/(sec(e + f*x) + 1), x) + Integral(6*sec(e + f*x)**3/(sec(e + f*x) + 
 1), x) + Integral(-4*sec(e + f*x)**4/(sec(e + f*x) + 1), x) + Integral(se 
c(e + f*x)**5/(sec(e + f*x) + 1), x))/a
 
3.1.34.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 591 vs. \(2 (116) = 232\).

Time = 0.21 (sec) , antiderivative size = 591, normalized size of antiderivative = 4.88 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx=\frac {c^{4} {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {16 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {a \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 12 \, c^{4} {\left (\frac {2 \, {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - 36 \, c^{4} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (f x + e\right )}{{\left (a - \frac {a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}} - \frac {\sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - 24 \, c^{4} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac {\sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + \frac {6 \, c^{4} \sin \left (f x + e\right )}{a {\left (\cos \left (f x + e\right ) + 1\right )}}}{6 \, f} \]

input
integrate(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="max 
ima")
 
output
1/6*(c^4*(2*(9*sin(f*x + e)/(cos(f*x + e) + 1) - 16*sin(f*x + e)^3/(cos(f* 
x + e) + 1)^3 + 15*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a - 3*a*sin(f*x + 
 e)^2/(cos(f*x + e) + 1)^2 + 3*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - a*s 
in(f*x + e)^6/(cos(f*x + e) + 1)^6) - 9*log(sin(f*x + e)/(cos(f*x + e) + 1 
) + 1)/a + 9*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a + 6*sin(f*x + e)/( 
a*(cos(f*x + e) + 1))) + 12*c^4*(2*(sin(f*x + e)/(cos(f*x + e) + 1) - 3*si 
n(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a - 2*a*sin(f*x + e)^2/(cos(f*x + e) + 
 1)^2 + a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4) - 3*log(sin(f*x + e)/(cos(f 
*x + e) + 1) + 1)/a + 3*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a + 2*sin 
(f*x + e)/(a*(cos(f*x + e) + 1))) - 36*c^4*(log(sin(f*x + e)/(cos(f*x + e) 
 + 1) + 1)/a - log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a - 2*sin(f*x + e) 
/((a - a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)) - sin(f* 
x + e)/(a*(cos(f*x + e) + 1))) - 24*c^4*(log(sin(f*x + e)/(cos(f*x + e) + 
1) + 1)/a - log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a - sin(f*x + e)/(a*( 
cos(f*x + e) + 1))) + 6*c^4*sin(f*x + e)/(a*(cos(f*x + e) + 1)))/f
 
3.1.34.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx=-\frac {\frac {105 \, c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac {105 \, c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac {96 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a} + \frac {2 \, {\left (87 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 136 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 57 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} a}}{6 \, f} \]

input
integrate(sec(f*x+e)*(c-c*sec(f*x+e))^4/(a+a*sec(f*x+e)),x, algorithm="gia 
c")
 
output
-1/6*(105*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a - 105*c^4*log(abs(tan(1 
/2*f*x + 1/2*e) - 1))/a - 96*c^4*tan(1/2*f*x + 1/2*e)/a + 2*(87*c^4*tan(1/ 
2*f*x + 1/2*e)^5 - 136*c^4*tan(1/2*f*x + 1/2*e)^3 + 57*c^4*tan(1/2*f*x + 1 
/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*a))/f
 
3.1.34.9 Mupad [B] (verification not implemented)

Time = 13.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.93 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^4}{a+a \sec (e+f x)} \, dx=\frac {16\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a\,f}-\frac {29\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-\frac {136\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}+19\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a\,f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}^3}-\frac {35\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a\,f} \]

input
int((c - c/cos(e + f*x))^4/(cos(e + f*x)*(a + a/cos(e + f*x))),x)
 
output
(16*c^4*tan(e/2 + (f*x)/2))/(a*f) - (29*c^4*tan(e/2 + (f*x)/2)^5 - (136*c^ 
4*tan(e/2 + (f*x)/2)^3)/3 + 19*c^4*tan(e/2 + (f*x)/2))/(a*f*(tan(e/2 + (f* 
x)/2)^2 - 1)^3) - (35*c^4*atanh(tan(e/2 + (f*x)/2)))/(a*f)