Integrand size = 32, antiderivative size = 132 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=-\frac {a^3 \text {arctanh}(\sin (e+f x))}{c^3 f}-\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{5 f (c-c \sec (e+f x))^3}+\frac {2 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 c f (c-c \sec (e+f x))^2}-\frac {2 a^3 \tan (e+f x)}{f \left (c^3-c^3 \sec (e+f x)\right )} \] Output:
-a^3*arctanh(sin(f*x+e))/c^3/f-2/5*a*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c* sec(f*x+e))^3+2/3*(a^3+a^3*sec(f*x+e))*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^2-2 *a^3*tan(f*x+e)/f/(c^3-c^3*sec(f*x+e))
Time = 0.17 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=-\frac {a^3 \left (-\frac {26 \cot \left (\frac {1}{2} (e+f x)\right )}{15 f}+\frac {2 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^2\left (\frac {1}{2} (e+f x)\right )}{15 f}-\frac {2 \cot \left (\frac {1}{2} (e+f x)\right ) \csc ^4\left (\frac {1}{2} (e+f x)\right )}{5 f}-\frac {\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f}+\frac {\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{f}\right )}{c^3} \] Input:
Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^3,x]
Output:
-((a^3*((-26*Cot[(e + f*x)/2])/(15*f) + (2*Cot[(e + f*x)/2]*Csc[(e + f*x)/ 2]^2)/(15*f) - (2*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^4)/(5*f) - Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]/f + Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] /f))/c^3)
Time = 0.71 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4445, 3042, 4445, 3042, 4445, 3042, 4257}
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.
| \(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^3}{(c-c \sec (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle -\frac {a \int \frac {\sec (e+f x) (\sec (e+f x) a+a)^2}{(c-c \sec (e+f x))^2}dx}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^2}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3}\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle -\frac {a \left (-\frac {a \int \frac {\sec (e+f x) (\sec (e+f x) a+a)}{c-c \sec (e+f x)}dx}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^2}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \left (-\frac {a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )}{c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^2}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3}\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle -\frac {a \left (-\frac {a \left (-\frac {a \int \sec (e+f x)dx}{c}-\frac {2 a \tan (e+f x)}{f (c-c \sec (e+f x))}\right )}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^2}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \left (-\frac {a \left (-\frac {a \int \csc \left (e+f x+\frac {\pi }{2}\right )dx}{c}-\frac {2 a \tan (e+f x)}{f (c-c \sec (e+f x))}\right )}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^2}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {a \left (-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{3 f (c-c \sec (e+f x))^2}-\frac {a \left (-\frac {a \text {arctanh}(\sin (e+f x))}{c f}-\frac {2 a \tan (e+f x)}{f (c-c \sec (e+f x))}\right )}{c}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{5 f (c-c \sec (e+f x))^3}\) |
Input:
Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^3,x]
Output:
(-2*a*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*(c - c*Sec[e + f*x])^3) - (a*((-2*(a^2 + a^2*Sec[e + f*x])*Tan[e + f*x])/(3*f*(c - c*Sec[e + f*x])^2 ) - (a*(-((a*ArcTanh[Sin[e + f*x]])/(c*f)) - (2*a*Tan[e + f*x])/(f*(c - c* Sec[e + f*x]))))/c))/c
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
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]
Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.59
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f \,c^{3}}\) | \(78\) |
| default | \(\frac {2 a^{3} \left (-\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2}+\frac {\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f \,c^{3}}\) | \(78\) |
| parallelrisch | \(\frac {a^{3} \left (6 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+10 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+15 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )-15 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+30 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c^{3} f}\) | \(78\) |
| risch | \(\frac {4 i a^{3} \left (15 \,{\mathrm e}^{4 i \left (f x +e \right )}-30 \,{\mathrm e}^{3 i \left (f x +e \right )}+100 \,{\mathrm e}^{2 i \left (f x +e \right )}-50 \,{\mathrm e}^{i \left (f x +e \right )}+13\right )}{15 f \,c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{5}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{c^{3} f}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{c^{3} f}\) | \(120\) |
| norman | \(\frac {-\frac {2 a^{3}}{5 c f}+\frac {8 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 c f}-\frac {6 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 c f}+\frac {22 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{5 c f}-\frac {16 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 c f}+\frac {2 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{c f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{c^{3} f}-\frac {a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{c^{3} f}\) | \(199\) |
Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x,method=_RETURNVERBO SE)
Output:
2/f*a^3/c^3*(-1/2*ln(tan(1/2*f*x+1/2*e)+1)+1/2*ln(tan(1/2*f*x+1/2*e)-1)+1/ 5/tan(1/2*f*x+1/2*e)^5+1/3/tan(1/2*f*x+1/2*e)^3+1/tan(1/2*f*x+1/2*e))
Time = 0.12 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.33 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {52 \, a^{3} \cos \left (f x + e\right )^{3} - 44 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} \cos \left (f x + e\right ) + 92 \, a^{3} - 15 \, {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 15 \, {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right )}{30 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="f ricas")
Output:
1/30*(52*a^3*cos(f*x + e)^3 - 44*a^3*cos(f*x + e)^2 - 4*a^3*cos(f*x + e) + 92*a^3 - 15*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^3)*log(sin(f*x + e) + 1)*sin(f*x + e) + 15*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^3) *log(-sin(f*x + e) + 1)*sin(f*x + e))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos (f*x + e) + c^3*f)*sin(f*x + e))
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=- \frac {a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx\right )}{c^{3}} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**3,x)
Output:
-a**3*(Integral(sec(e + f*x)/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec( e + f*x) - 1), x) + Integral(3*sec(e + f*x)**2/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x) + Integral(3*sec(e + f*x)**3/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x) + Integral(sec(e + f *x)**4/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x))/c** 3
Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (131) = 262\).
Time = 0.05 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.34 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=-\frac {a^{3} {\left (\frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{3}} - \frac {{\left (\frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}\right )} - \frac {3 \, a^{3} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} + \frac {a^{3} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} + \frac {9 \, a^{3} {\left (\frac {5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="m axima")
Output:
-1/60*(a^3*(60*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c^3 - 60*log(sin(f *x + e)/(cos(f*x + e) + 1) - 1)/c^3 - (20*sin(f*x + e)^2/(cos(f*x + e) + 1 )^2 + 105*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3)*(cos(f*x + e) + 1)^5/(c ^3*sin(f*x + e)^5)) - 3*a^3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*s in(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) + a^3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 15*sin(f*x + e)^4/ (cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) + 9*a^ 3*(5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(c^3*si n(f*x + e)^5))/f
Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.77 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=-\frac {\frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{c^{3}} - \frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{c^{3}} - \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 5 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{3}\right )}}{c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{15 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x, algorithm="g iac")
Output:
-1/15*(15*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^3 - 15*a^3*log(abs(tan( 1/2*f*x + 1/2*e) - 1))/c^3 - 2*(15*a^3*tan(1/2*f*x + 1/2*e)^4 + 5*a^3*tan( 1/2*f*x + 1/2*e)^2 + 3*a^3)/(c^3*tan(1/2*f*x + 1/2*e)^5))/f
Time = 10.92 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.59 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {2\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\frac {2\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{3}+\frac {2\,a^3}{5}}{c^3\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}-\frac {2\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{c^3\,f} \] Input:
int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))^3),x)
Output:
((2*a^3*tan(e/2 + (f*x)/2)^2)/3 + 2*a^3*tan(e/2 + (f*x)/2)^4 + (2*a^3)/5)/ (c^3*f*tan(e/2 + (f*x)/2)^5) - (2*a^3*atanh(tan(e/2 + (f*x)/2)))/(c^3*f)
Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {a^{3} \left (15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+6\right )}{15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} c^{3} f} \] Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x)
Output:
(a**3*(15*log(tan((e + f*x)/2) - 1)*tan((e + f*x)/2)**5 - 15*log(tan((e + f*x)/2) + 1)*tan((e + f*x)/2)**5 + 30*tan((e + f*x)/2)**4 + 10*tan((e + f* x)/2)**2 + 6))/(15*tan((e + f*x)/2)**5*c**3*f)