Integrand size = 32, antiderivative size = 120 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx=-\frac {\cot ^5(e+f x)}{5 a c^4 f}-\frac {4 \cot ^7(e+f x)}{7 a c^4 f}+\frac {\csc (e+f x)}{a c^4 f}-\frac {2 \csc ^3(e+f x)}{a c^4 f}+\frac {9 \csc ^5(e+f x)}{5 a c^4 f}-\frac {4 \csc ^7(e+f x)}{7 a c^4 f} \]
-1/5*cot(f*x+e)^5/a/c^4/f-4/7*cot(f*x+e)^7/a/c^4/f+csc(f*x+e)/a/c^4/f-2*cs c(f*x+e)^3/a/c^4/f+9/5*csc(f*x+e)^5/a/c^4/f-4/7*csc(f*x+e)^7/a/c^4/f
Time = 1.53 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx=\frac {\left (-13+4 \sec (e+f x)+20 \sec ^2(e+f x)-24 \sec ^3(e+f x)+8 \sec ^4(e+f x)\right ) \tan (e+f x)}{35 a c^4 f (-1+\sec (e+f x))^4 (1+\sec (e+f x))} \]
((-13 + 4*Sec[e + f*x] + 20*Sec[e + f*x]^2 - 24*Sec[e + f*x]^3 + 8*Sec[e + f*x]^4)*Tan[e + f*x])/(35*a*c^4*f*(-1 + Sec[e + f*x])^4*(1 + Sec[e + f*x] ))
Time = 0.47 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3042, 4446, 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.
\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a) (c-c \sec (e+f x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4}dx\) |
\(\Big \downarrow \) 4446 |
\(\displaystyle \frac {\int \left (a^3 \csc (e+f x) \cot ^7(e+f x)+3 a^3 \csc ^2(e+f x) \cot ^6(e+f x)+3 a^3 \csc ^3(e+f x) \cot ^5(e+f x)+a^3 \csc ^4(e+f x) \cot ^4(e+f x)\right )dx}{a^4 c^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {4 a^3 \cot ^7(e+f x)}{7 f}-\frac {a^3 \cot ^5(e+f x)}{5 f}-\frac {4 a^3 \csc ^7(e+f x)}{7 f}+\frac {9 a^3 \csc ^5(e+f x)}{5 f}-\frac {2 a^3 \csc ^3(e+f x)}{f}+\frac {a^3 \csc (e+f x)}{f}}{a^4 c^4}\) |
(-1/5*(a^3*Cot[e + f*x]^5)/f - (4*a^3*Cot[e + f*x]^7)/(7*f) + (a^3*Csc[e + f*x])/f - (2*a^3*Csc[e + f*x]^3)/f + (9*a^3*Csc[e + f*x]^5)/(5*f) - (4*a^ 3*Csc[e + f*x]^7)/(7*f))/(a^4*c^4)
3.1.41.3.1 Defintions of rubi rules used
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m Int[ExpandTrig[csc[e + f*x]*cot[e + f*x]^(2*m), (c + d*csc[e + f*x])^(n - m ), x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && Eq Q[a^2 - b^2, 0] && IntegersQ[m, n] && GeQ[n - m, 0] && GtQ[m*n, 0]
Time = 0.89 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {4}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}}{16 f a \,c^{4}}\) | \(74\) |
default | \(\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {4}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}}{16 f a \,c^{4}}\) | \(74\) |
parallelrisch | \(\frac {-5 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}+28 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-70 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+140 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )+35 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{560 f a \,c^{4}}\) | \(74\) |
norman | \(\frac {-\frac {1}{112 a c f}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{20 a c f}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{8 a c f}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{4 a c f}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{16 a c f}}{c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\) | \(116\) |
risch | \(\frac {2 i \left (35 \,{\mathrm e}^{7 i \left (f x +e \right )}-105 \,{\mathrm e}^{6 i \left (f x +e \right )}+175 \,{\mathrm e}^{5 i \left (f x +e \right )}-105 \,{\mathrm e}^{4 i \left (f x +e \right )}-7 \,{\mathrm e}^{3 i \left (f x +e \right )}+77 \,{\mathrm e}^{2 i \left (f x +e \right )}-43 \,{\mathrm e}^{i \left (f x +e \right )}+13\right )}{35 f a \,c^{4} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{7} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}\) | \(118\) |
1/16/f/a/c^4*(tan(1/2*f*x+1/2*e)+4/5/tan(1/2*f*x+1/2*e)^5-2/tan(1/2*f*x+1/ 2*e)^3+4/tan(1/2*f*x+1/2*e)-1/7/tan(1/2*f*x+1/2*e)^7)
Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.85 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx=\frac {13 \, \cos \left (f x + e\right )^{4} - 4 \, \cos \left (f x + e\right )^{3} - 20 \, \cos \left (f x + e\right )^{2} + 24 \, \cos \left (f x + e\right ) - 8}{35 \, {\left (a c^{4} f \cos \left (f x + e\right )^{3} - 3 \, a c^{4} f \cos \left (f x + e\right )^{2} + 3 \, a c^{4} f \cos \left (f x + e\right ) - a c^{4} f\right )} \sin \left (f x + e\right )} \]
1/35*(13*cos(f*x + e)^4 - 4*cos(f*x + e)^3 - 20*cos(f*x + e)^2 + 24*cos(f* x + e) - 8)/((a*c^4*f*cos(f*x + e)^3 - 3*a*c^4*f*cos(f*x + e)^2 + 3*a*c^4* f*cos(f*x + e) - a*c^4*f)*sin(f*x + e))
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx=\frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 3 \sec ^{4}{\left (e + f x \right )} + 2 \sec ^{3}{\left (e + f x \right )} + 2 \sec ^{2}{\left (e + f x \right )} - 3 \sec {\left (e + f x \right )} + 1}\, dx}{a c^{4}} \]
Integral(sec(e + f*x)/(sec(e + f*x)**5 - 3*sec(e + f*x)**4 + 2*sec(e + f*x )**3 + 2*sec(e + f*x)**2 - 3*sec(e + f*x) + 1), x)/(a*c**4)
Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx=\frac {\frac {{\left (\frac {28 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {70 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {140 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{a c^{4} \sin \left (f x + e\right )^{7}} + \frac {35 \, \sin \left (f x + e\right )}{a c^{4} {\left (\cos \left (f x + e\right ) + 1\right )}}}{560 \, f} \]
1/560*((28*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 70*sin(f*x + e)^4/(cos(f* x + e) + 1)^4 + 140*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 5)*(cos(f*x + e) + 1)^7/(a*c^4*sin(f*x + e)^7) + 35*sin(f*x + e)/(a*c^4*(cos(f*x + e) + 1) ))/f
Time = 0.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx=\frac {\frac {35 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a c^{4}} + \frac {140 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 70 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 28 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5}{a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}}}{560 \, f} \]
1/560*(35*tan(1/2*f*x + 1/2*e)/(a*c^4) + (140*tan(1/2*f*x + 1/2*e)^6 - 70* tan(1/2*f*x + 1/2*e)^4 + 28*tan(1/2*f*x + 1/2*e)^2 - 5)/(a*c^4*tan(1/2*f*x + 1/2*e)^7))/f
Time = 13.44 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^4} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{16\,a\,c^4\,f}+\frac {\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6}{4}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{8}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{20}-\frac {1}{112}}{a\,c^4\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7} \]