Integrand size = 32, antiderivative size = 80 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=-\frac {\cot ^5(e+f x)}{5 a^3 c^2 f}+\frac {\csc (e+f x)}{a^3 c^2 f}-\frac {2 \csc ^3(e+f x)}{3 a^3 c^2 f}+\frac {\csc ^5(e+f x)}{5 a^3 c^2 f} \] Output:
-1/5*cot(f*x+e)^5/a^3/c^2/f+csc(f*x+e)/a^3/c^2/f-2/3*csc(f*x+e)^3/a^3/c^2/ f+1/5*csc(f*x+e)^5/a^3/c^2/f
Time = 0.64 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {\left (3-12 \sec (e+f x)-12 \sec ^2(e+f x)+8 \sec ^3(e+f x)+8 \sec ^4(e+f x)\right ) \tan (e+f x)}{15 a^3 c^2 f (-1+\sec (e+f x))^2 (1+\sec (e+f x))^3} \] Input:
Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^2),x]
Output:
((3 - 12*Sec[e + f*x] - 12*Sec[e + f*x]^2 + 8*Sec[e + f*x]^3 + 8*Sec[e + f *x]^4)*Tan[e + f*x])/(15*a^3*c^2*f*(-1 + Sec[e + f*x])^2*(1 + Sec[e + f*x] )^3)
Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.86, 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.
\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4446 |
\(\displaystyle -\frac {\int \left (c \cot ^5(e+f x) \csc (e+f x)-c \cot ^4(e+f x) \csc ^2(e+f x)\right )dx}{a^3 c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {c \cot ^5(e+f x)}{5 f}-\frac {c \csc ^5(e+f x)}{5 f}+\frac {2 c \csc ^3(e+f x)}{3 f}-\frac {c \csc (e+f x)}{f}}{a^3 c^3}\) |
Input:
Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^2),x]
Output:
-(((c*Cot[e + f*x]^5)/(5*f) - (c*Csc[e + f*x])/f + (2*c*Csc[e + f*x]^3)/(3 *f) - (c*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 = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-5 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+60 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )+90 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{240 f \,a^{3} c^{2}}\) | \(74\) |
derivativedivides | \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}}{16 f \,a^{3} c^{2}}\) | \(76\) |
default | \(\frac {\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}}{16 f \,a^{3} c^{2}}\) | \(76\) |
risch | \(\frac {2 i \left (15 \,{\mathrm e}^{7 i \left (f x +e \right )}+15 \,{\mathrm e}^{6 i \left (f x +e \right )}-5 \,{\mathrm e}^{5 i \left (f x +e \right )}-25 \,{\mathrm e}^{4 i \left (f x +e \right )}+13 \,{\mathrm e}^{3 i \left (f x +e \right )}+21 \,{\mathrm e}^{2 i \left (f x +e \right )}+9 \,{\mathrm e}^{i \left (f x +e \right )}-3\right )}{15 f \,a^{3} c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3}}\) | \(118\) |
norman | \(\frac {-\frac {1}{48 f a c}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{4 f a c}+\frac {3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{8 f a c}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{12 f a c}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{80 f a c}}{a^{2} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}\) | \(119\) |
Input:
int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x,method=_RETURNVERBO SE)
Output:
1/240*(3*tan(1/2*f*x+1/2*e)^5-5*cot(1/2*f*x+1/2*e)^3-20*tan(1/2*f*x+1/2*e) ^3+60*cot(1/2*f*x+1/2*e)+90*tan(1/2*f*x+1/2*e))/f/a^3/c^2
Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.36 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=-\frac {3 \, \cos \left (f x + e\right )^{4} - 12 \, \cos \left (f x + e\right )^{3} - 12 \, \cos \left (f x + e\right )^{2} + 8 \, \cos \left (f x + e\right ) + 8}{15 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x, algorithm="f ricas")
Output:
-1/15*(3*cos(f*x + e)^4 - 12*cos(f*x + e)^3 - 12*cos(f*x + e)^2 + 8*cos(f* x + e) + 8)/((a^3*c^2*f*cos(f*x + e)^3 + a^3*c^2*f*cos(f*x + e)^2 - a^3*c^ 2*f*cos(f*x + e) - a^3*c^2*f)*sin(f*x + e))
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} + \sec ^{4}{\left (e + f x \right )} - 2 \sec ^{3}{\left (e + f x \right )} - 2 \sec ^{2}{\left (e + f x \right )} + \sec {\left (e + f x \right )} + 1}\, dx}{a^{3} c^{2}} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**2,x)
Output:
Integral(sec(e + f*x)/(sec(e + f*x)**5 + sec(e + f*x)**4 - 2*sec(e + f*x)* *3 - 2*sec(e + f*x)**2 + sec(e + f*x) + 1), x)/(a**3*c**2)
Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.50 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {\frac {\frac {90 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3} c^{2}} + \frac {5 \, {\left (\frac {12 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{a^{3} c^{2} \sin \left (f x + e\right )^{3}}}{240 \, f} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x, algorithm="m axima")
Output:
1/240*((90*sin(f*x + e)/(cos(f*x + e) + 1) - 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^3*c^2) + 5*(12*sin(f *x + e)^2/(cos(f*x + e) + 1)^2 - 1)*(cos(f*x + e) + 1)^3/(a^3*c^2*sin(f*x + e)^3))/f
Time = 0.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {\frac {5 \, {\left (12 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{a^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}} + \frac {3 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 20 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 90 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{15} c^{10}}}{240 \, f} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x, algorithm="g iac")
Output:
1/240*(5*(12*tan(1/2*f*x + 1/2*e)^2 - 1)/(a^3*c^2*tan(1/2*f*x + 1/2*e)^3) + (3*a^12*c^8*tan(1/2*f*x + 1/2*e)^5 - 20*a^12*c^8*tan(1/2*f*x + 1/2*e)^3 + 90*a^12*c^8*tan(1/2*f*x + 1/2*e))/(a^15*c^10))/f
Time = 10.86 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.39 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {48\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-192\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+168\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-32\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3}{240\,a^3\,c^2\,f\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \] Input:
int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^2),x)
Output:
(168*cos(e/2 + (f*x)/2)^4 - 32*cos(e/2 + (f*x)/2)^2 - 192*cos(e/2 + (f*x)/ 2)^6 + 48*cos(e/2 + (f*x)/2)^8 + 3)/(240*a^3*c^2*f*(cos(e/2 + (f*x)/2)^5*s in(e/2 + (f*x)/2) - cos(e/2 + (f*x)/2)^7*sin(e/2 + (f*x)/2)))
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+90 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5}{240 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a^{3} c^{2} f} \] Input:
int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x)
Output:
(3*tan((e + f*x)/2)**8 - 20*tan((e + f*x)/2)**6 + 90*tan((e + f*x)/2)**4 + 60*tan((e + f*x)/2)**2 - 5)/(240*tan((e + f*x)/2)**3*a**3*c**2*f)