Integrand size = 32, antiderivative size = 78 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^3} \, dx=\frac {2 \cot ^5(e+f x)}{5 a c^3 f}+\frac {\csc (e+f x)}{a c^3 f}-\frac {\csc ^3(e+f x)}{a c^3 f}+\frac {2 \csc ^5(e+f x)}{5 a c^3 f} \] Output:
2/5*cot(f*x+e)^5/a/c^3/f+csc(f*x+e)/a/c^3/f-csc(f*x+e)^3/a/c^3/f+2/5*csc(f *x+e)^5/a/c^3/f
Time = 0.45 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^3} \, dx=\frac {(5-5 \cos (e+f x)+\cos (2 (e+f x))+\cos (3 (e+f x))) \csc ^5\left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {1}{2} (e+f x)\right )}{80 a c^3 f} \] Input:
Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^3),x]
Output:
((5 - 5*Cos[e + f*x] + Cos[2*(e + f*x)] + Cos[3*(e + f*x)])*Csc[(e + f*x)/ 2]^5*Sec[(e + f*x)/2])/(80*a*c^3*f)
Time = 0.40 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a) (c-c \sec (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 4446 |
\(\displaystyle -\frac {\int \left (a^2 \csc (e+f x) \cot ^5(e+f x)+2 a^2 \csc ^2(e+f x) \cot ^4(e+f x)+a^2 \csc ^3(e+f x) \cot ^3(e+f x)\right )dx}{a^3 c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {2 a^2 \cot ^5(e+f x)}{5 f}-\frac {2 a^2 \csc ^5(e+f x)}{5 f}+\frac {a^2 \csc ^3(e+f x)}{f}-\frac {a^2 \csc (e+f x)}{f}}{a^3 c^3}\) |
Input:
Int[Sec[e + f*x]/((a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^3),x]
Output:
-(((-2*a^2*Cot[e + f*x]^5)/(5*f) - (a^2*Csc[e + f*x])/f + (a^2*Csc[e + f*x ]^3)/f - (2*a^2*Csc[e + f*x]^5)/(5*f))/(a^3*c^3))
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 = 0.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77
method | result | size |
parallelrisch | \(\frac {\sec \left (\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \left (5+\cos \left (3 f x +3 e \right )+\cos \left (2 f x +2 e \right )-5 \cos \left (f x +e \right )\right )}{80 f a \,c^{3}}\) | \(60\) |
derivativedivides | \(\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}}{8 f a \,c^{3}}\) | \(61\) |
default | \(\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {3}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}}{8 f a \,c^{3}}\) | \(61\) |
risch | \(\frac {2 i \left (5 \,{\mathrm e}^{5 i \left (f x +e \right )}-10 \,{\mathrm e}^{4 i \left (f x +e \right )}+10 \,{\mathrm e}^{3 i \left (f x +e \right )}-3 \,{\mathrm e}^{i \left (f x +e \right )}+2\right )}{5 f a \,c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{5}}\) | \(85\) |
norman | \(\frac {\frac {1}{40 f a c}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{8 f a c}+\frac {3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{8 f a c}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{8 f a c}}{c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\) | \(94\) |
Input:
int(sec(f*x+e)/(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^3,x,method=_RETURNVERBOSE )
Output:
1/80/f/a/c^3*sec(1/2*f*x+1/2*e)*csc(1/2*f*x+1/2*e)^5*(5+cos(3*f*x+3*e)+cos (2*f*x+2*e)-5*cos(f*x+e))
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^3} \, dx=\frac {2 \, \cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - 4 \, \cos \left (f x + e\right ) + 2}{5 \, {\left (a c^{3} f \cos \left (f x + e\right )^{2} - 2 \, a c^{3} f \cos \left (f x + e\right ) + a c^{3} f\right )} \sin \left (f x + e\right )} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^3,x, algorithm="fri cas")
Output:
1/5*(2*cos(f*x + e)^3 + cos(f*x + e)^2 - 4*cos(f*x + e) + 2)/((a*c^3*f*cos (f*x + e)^2 - 2*a*c^3*f*cos(f*x + e) + a*c^3*f)*sin(f*x + e))
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^3} \, dx=- \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 2 \sec ^{3}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} - 1}\, dx}{a c^{3}} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))/(c-c*sec(f*x+e))**3,x)
Output:
-Integral(sec(e + f*x)/(sec(e + f*x)**4 - 2*sec(e + f*x)**3 + 2*sec(e + f* x) - 1), x)/(a*c**3)
Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.24 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^3} \, dx=-\frac {\frac {{\left (\frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{a c^{3} \sin \left (f x + e\right )^{5}} - \frac {5 \, \sin \left (f x + e\right )}{a c^{3} {\left (\cos \left (f x + e\right ) + 1\right )}}}{40 \, f} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^3,x, algorithm="max ima")
Output:
-1/40*((5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(a*c^3*sin(f*x + e)^5) - 5*sin(f*x + e)/(a*c^3*(cos(f*x + e) + 1)))/f
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^3} \, dx=\frac {\frac {5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a c^{3}} + \frac {15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1}{a c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{40 \, f} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^3,x, algorithm="gia c")
Output:
1/40*(5*tan(1/2*f*x + 1/2*e)/(a*c^3) + (15*tan(1/2*f*x + 1/2*e)^4 - 5*tan( 1/2*f*x + 1/2*e)^2 + 1)/(a*c^3*tan(1/2*f*x + 1/2*e)^5))/f
Time = 10.54 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.81 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^3} \, dx=\frac {5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1}{40\,a\,c^3\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \] Input:
int(1/(cos(e + f*x)*(a + a/cos(e + f*x))*(c - c/cos(e + f*x))^3),x)
Output:
(15*tan(e/2 + (f*x)/2)^4 - 5*tan(e/2 + (f*x)/2)^2 + 5*tan(e/2 + (f*x)/2)^6 + 1)/(40*a*c^3*f*tan(e/2 + (f*x)/2)^5)
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.81 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^3} \, dx=\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1}{40 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a \,c^{3} f} \] Input:
int(sec(f*x+e)/(a+a*sec(f*x+e))/(c-c*sec(f*x+e))^3,x)
Output:
(5*tan((e + f*x)/2)**6 + 15*tan((e + f*x)/2)**4 - 5*tan((e + f*x)/2)**2 + 1)/(40*tan((e + f*x)/2)**5*a*c**3*f)