Integrand size = 32, antiderivative size = 78 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx=-\frac {2 \cot ^5(e+f x)}{5 a^3 c f}+\frac {\csc (e+f x)}{a^3 c f}-\frac {\csc ^3(e+f x)}{a^3 c f}+\frac {2 \csc ^5(e+f x)}{5 a^3 c f} \] Output:
-2/5*cot(f*x+e)^5/a^3/c/f+csc(f*x+e)/a^3/c/f-csc(f*x+e)^3/a^3/c/f+2/5*csc( f*x+e)^5/a^3/c/f
Time = 0.43 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.86 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx=\frac {\left (-2+\sec (e+f x)+4 \sec ^2(e+f x)+2 \sec ^3(e+f x)\right ) \tan (e+f x)}{5 a^3 c f (-1+\sec (e+f x)) (1+\sec (e+f x))^3} \] Input:
Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])),x]
Output:
((-2 + Sec[e + f*x] + 4*Sec[e + f*x]^2 + 2*Sec[e + f*x]^3)*Tan[e + f*x])/( 5*a^3*c*f*(-1 + Sec[e + f*x])*(1 + Sec[e + f*x])^3)
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.
\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a)^3 (c-c \sec (e+f x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4446 |
\(\displaystyle -\frac {\int \left (c^2 \csc (e+f x) \cot ^5(e+f x)-2 c^2 \csc ^2(e+f x) \cot ^4(e+f x)+c^2 \csc ^3(e+f x) \cot ^3(e+f x)\right )dx}{a^3 c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {2 c^2 \cot ^5(e+f x)}{5 f}-\frac {2 c^2 \csc ^5(e+f x)}{5 f}+\frac {c^2 \csc ^3(e+f x)}{f}-\frac {c^2 \csc (e+f x)}{f}}{a^3 c^3}\) |
Input:
Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])),x]
Output:
-(((2*c^2*Cot[e + f*x]^5)/(5*f) - (c^2*Csc[e + f*x])/f + (c^2*Csc[e + f*x] ^3)/f - (2*c^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 = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.76
method | result | size |
parallelrisch | \(\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+5 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )+15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{40 f \,a^{3} c}\) | \(59\) |
derivativedivides | \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}}{8 f \,a^{3} c}\) | \(61\) |
default | \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}}{8 f \,a^{3} c}\) | \(61\) |
risch | \(\frac {2 i \left (5 \,{\mathrm e}^{5 i \left (f x +e \right )}+10 \,{\mathrm e}^{4 i \left (f x +e \right )}+10 \,{\mathrm e}^{3 i \left (f x +e \right )}-3 \,{\mathrm e}^{i \left (f x +e \right )}-2\right )}{5 f \,a^{3} c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}\) | \(85\) |
norman | \(\frac {\frac {1}{8 f a c}+\frac {3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{8 f a c}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{8 f a c}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{40 f a c}}{a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\) | \(94\) |
Input:
int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x,method=_RETURNVERBOSE )
Output:
1/40*(tan(1/2*f*x+1/2*e)^5-5*tan(1/2*f*x+1/2*e)^3+5*cot(1/2*f*x+1/2*e)+15* tan(1/2*f*x+1/2*e))/f/a^3/c
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx=-\frac {2 \, \cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{2} - 4 \, \cos \left (f x + e\right ) - 2}{5 \, {\left (a^{3} c f \cos \left (f x + e\right )^{2} + 2 \, a^{3} c f \cos \left (f x + e\right ) + a^{3} c f\right )} \sin \left (f x + e\right )} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),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^3*c*f*co s(f*x + e)^2 + 2*a^3*c*f*cos(f*x + e) + a^3*c*f)*sin(f*x + e))
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx=- \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} + 2 \sec ^{3}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} - 1}\, dx}{a^{3} c} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e)),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**3*c)
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.22 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx=\frac {\frac {\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3} c} + \frac {5 \, {\left (\cos \left (f x + e\right ) + 1\right )}}{a^{3} c \sin \left (f x + e\right )}}{40 \, f} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="max ima")
Output:
1/40*((15*sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^3*c) + 5*(cos(f*x + e) + 1)/(a^3*c*sin(f*x + e)))/f
Time = 0.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx=\frac {\frac {5}{a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \frac {a^{12} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 5 \, a^{12} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, a^{12} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{15} c^{5}}}{40 \, f} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="gia c")
Output:
1/40*(5/(a^3*c*tan(1/2*f*x + 1/2*e)) + (a^12*c^4*tan(1/2*f*x + 1/2*e)^5 - 5*a^12*c^4*tan(1/2*f*x + 1/2*e)^3 + 15*a^12*c^4*tan(1/2*f*x + 1/2*e))/(a^1 5*c^5))/f
Time = 10.35 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx=-\frac {16\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-28\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+8\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}{40\,a^3\,c\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )} \] Input:
int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))),x)
Output:
-(8*cos(e/2 + (f*x)/2)^2 - 28*cos(e/2 + (f*x)/2)^4 + 16*cos(e/2 + (f*x)/2) ^6 - 1)/(40*a^3*c*f*cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2))
Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))} \, dx=\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5}{40 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a^{3} c f} \] Input:
int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x)
Output:
(tan((e + f*x)/2)**6 - 5*tan((e + f*x)/2)**4 + 15*tan((e + f*x)/2)**2 + 5) /(40*tan((e + f*x)/2)*a**3*c*f)