Integrand size = 32, antiderivative size = 59 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx=\frac {\csc (e+f x)}{a^3 c^3 f}-\frac {2 \csc ^3(e+f x)}{3 a^3 c^3 f}+\frac {\csc ^5(e+f x)}{5 a^3 c^3 f} \] Output:
csc(f*x+e)/a^3/c^3/f-2/3*csc(f*x+e)^3/a^3/c^3/f+1/5*csc(f*x+e)^5/a^3/c^3/f
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx=-\frac {-\frac {\csc (e+f x)}{f}+\frac {2 \csc ^3(e+f x)}{3 f}-\frac {\csc ^5(e+f x)}{5 f}}{a^3 c^3} \] Input:
Integrate[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^3),x]
Output:
-((-(Csc[e + f*x]/f) + (2*Csc[e + f*x]^3)/(3*f) - Csc[e + f*x]^5/(5*f))/(a ^3*c^3))
Time = 0.36 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 4446, 3042, 25, 3086, 210, 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.
\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 4446 |
\(\displaystyle -\frac {\int \cot ^5(e+f x) \csc (e+f x)dx}{a^3 c^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int -\sec \left (e+f x-\frac {\pi }{2}\right ) \tan \left (e+f x-\frac {\pi }{2}\right )^5dx}{a^3 c^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 e-\pi )+f x\right ) \tan \left (\frac {1}{2} (2 e-\pi )+f x\right )^5dx}{a^3 c^3}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \left (\csc ^2(e+f x)-1\right )^2d\csc (e+f x)}{a^3 c^3 f}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle \frac {\int \left (\csc ^4(e+f x)-2 \csc ^2(e+f x)+1\right )d\csc (e+f x)}{a^3 c^3 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} \csc ^5(e+f x)-\frac {2}{3} \csc ^3(e+f x)+\csc (e+f x)}{a^3 c^3 f}\) |
Input:
Int[Sec[e + f*x]/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^3),x]
Output:
(Csc[e + f*x] - (2*Csc[e + f*x]^3)/3 + Csc[e + f*x]^5/5)/(a^3*c^3*f)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
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 = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69
method | result | size |
default | \(-\frac {-\frac {\csc \left (f x +e \right )^{5}}{5}+\frac {2 \csc \left (f x +e \right )^{3}}{3}-\csc \left (f x +e \right )}{a^{3} c^{3} f}\) | \(41\) |
parallelrisch | \(\frac {\left (15 \cos \left (4 f x +4 e \right )-20 \cos \left (2 f x +2 e \right )+29\right ) \sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \csc \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3840 f \,a^{3} c^{3}}\) | \(58\) |
risch | \(\frac {2 i \left (15 \,{\mathrm e}^{9 i \left (f x +e \right )}-20 \,{\mathrm e}^{7 i \left (f x +e \right )}+58 \,{\mathrm e}^{5 i \left (f x +e \right )}-20 \,{\mathrm e}^{3 i \left (f x +e \right )}+15 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{15 f \,a^{3} c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{5}}\) | \(95\) |
norman | \(\frac {\frac {1}{160 f a c}-\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{96 f a c}+\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{16 f a c}+\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{16 f a c}-\frac {5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{96 f a c}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{160 f a c}}{a^{2} c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\) | \(141\) |
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:
-1/a^3/c^3/f*(-1/5*csc(f*x+e)^5+2/3*csc(f*x+e)^3-csc(f*x+e))
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx=\frac {15 \, \cos \left (f x + e\right )^{4} - 20 \, \cos \left (f x + e\right )^{2} + 8}{15 \, {\left (a^{3} c^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} c^{3} f \cos \left (f x + e\right )^{2} + a^{3} c^{3} f\right )} \sin \left (f x + e\right )} \] 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:
1/15*(15*cos(f*x + e)^4 - 20*cos(f*x + e)^2 + 8)/((a^3*c^3*f*cos(f*x + e)^ 4 - 2*a^3*c^3*f*cos(f*x + e)^2 + a^3*c^3*f)*sin(f*x + e))
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx=- \frac {\int \frac {\sec {\left (e + f x \right )}}{\sec ^{6}{\left (e + f x \right )} - 3 \sec ^{4}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} - 1}\, dx}{a^{3} c^{3}} \] Input:
integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**3,x)
Output:
-Integral(sec(e + f*x)/(sec(e + f*x)**6 - 3*sec(e + f*x)**4 + 3*sec(e + f* x)**2 - 1), x)/(a**3*c**3)
Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx=\frac {15 \, \sin \left (f x + e\right )^{4} - 10 \, \sin \left (f x + e\right )^{2} + 3}{15 \, a^{3} c^{3} f \sin \left (f x + e\right )^{5}} \] 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:
1/15*(15*sin(f*x + e)^4 - 10*sin(f*x + e)^2 + 3)/(a^3*c^3*f*sin(f*x + e)^5 )
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx=\frac {15 \, \sin \left (f x + e\right )^{4} - 10 \, \sin \left (f x + e\right )^{2} + 3}{15 \, a^{3} c^{3} f \sin \left (f x + e\right )^{5}} \] 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*sin(f*x + e)^4 - 10*sin(f*x + e)^2 + 3)/(a^3*c^3*f*sin(f*x + e)^5 )
Time = 10.73 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.64 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx=\frac {{\sin \left (e+f\,x\right )}^4-\frac {2\,{\sin \left (e+f\,x\right )}^2}{3}+\frac {1}{5}}{a^3\,c^3\,f\,{\sin \left (e+f\,x\right )}^5} \] Input:
int(1/(cos(e + f*x)*(a + a/cos(e + f*x))^3*(c - c/cos(e + f*x))^3),x)
Output:
(sin(e + f*x)^4 - (2*sin(e + f*x)^2)/3 + 1/5)/(a^3*c^3*f*sin(e + f*x)^5)
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^3} \, dx=\frac {15 \sin \left (f x +e \right )^{4}-10 \sin \left (f x +e \right )^{2}+3}{15 \sin \left (f x +e \right )^{5} a^{3} c^{3} f} \] Input:
int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^3,x)
Output:
(15*sin(e + f*x)**4 - 10*sin(e + f*x)**2 + 3)/(15*sin(e + f*x)**5*a**3*c** 3*f)