Integrand size = 26, antiderivative size = 100 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {x}{a^3 c^2}+\frac {\cot (e+f x) (15-8 \sec (e+f x))}{15 a^3 c^2 f}-\frac {\cot ^3(e+f x) (5-4 \sec (e+f x))}{15 a^3 c^2 f}+\frac {\cot ^5(e+f x) (1-\sec (e+f x))}{5 a^3 c^2 f} \]
x/a^3/c^2+1/15*cot(f*x+e)*(15-8*sec(f*x+e))/a^3/c^2/f-1/15*cot(f*x+e)^3*(5 -4*sec(f*x+e))/a^3/c^2/f+1/5*cot(f*x+e)^5*(1-sec(f*x+e))/a^3/c^2/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {\cot ^5(e+f x) \left (3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(e+f x)\right )-15 \sec (e+f x)+20 \sec ^3(e+f x)-8 \sec ^5(e+f x)\right )}{15 a^3 c^2 f} \]
(Cot[e + f*x]^5*(3*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[e + f*x]^2] - 15* Sec[e + f*x] + 20*Sec[e + f*x]^3 - 8*Sec[e + f*x]^5))/(15*a^3*c^2*f)
Time = 0.54 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 4392, 3042, 4370, 25, 3042, 4370, 25, 3042, 4370, 24}
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.
| \(\displaystyle \int \frac {1}{(a \sec (e+f x)+a)^3 (c-c \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4392 |
| \(\displaystyle -\frac {\int \cot ^6(e+f x) (c-c \sec (e+f x))dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}{\cot \left (e+f x+\frac {\pi }{2}\right )^6}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 4370 |
| \(\displaystyle -\frac {\frac {1}{5} \int -\cot ^4(e+f x) (5 c-4 c \sec (e+f x))dx-\frac {\cot ^5(e+f x) (c-c \sec (e+f x))}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {1}{5} \int \cot ^4(e+f x) (5 c-4 c \sec (e+f x))dx-\frac {\cot ^5(e+f x) (c-c \sec (e+f x))}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {1}{5} \int \frac {5 c-4 c \csc \left (e+f x+\frac {\pi }{2}\right )}{\cot \left (e+f x+\frac {\pi }{2}\right )^4}dx-\frac {\cot ^5(e+f x) (c-c \sec (e+f x))}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 4370 |
| \(\displaystyle -\frac {\frac {1}{5} \left (\frac {\cot ^3(e+f x) (5 c-4 c \sec (e+f x))}{3 f}-\frac {1}{3} \int -\cot ^2(e+f x) (15 c-8 c \sec (e+f x))dx\right )-\frac {\cot ^5(e+f x) (c-c \sec (e+f x))}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {1}{5} \left (\frac {1}{3} \int \cot ^2(e+f x) (15 c-8 c \sec (e+f x))dx+\frac {\cot ^3(e+f x) (5 c-4 c \sec (e+f x))}{3 f}\right )-\frac {\cot ^5(e+f x) (c-c \sec (e+f x))}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {15 c-8 c \csc \left (e+f x+\frac {\pi }{2}\right )}{\cot \left (e+f x+\frac {\pi }{2}\right )^2}dx+\frac {\cot ^3(e+f x) (5 c-4 c \sec (e+f x))}{3 f}\right )-\frac {\cot ^5(e+f x) (c-c \sec (e+f x))}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 4370 |
| \(\displaystyle -\frac {\frac {1}{5} \left (\frac {1}{3} \left (\int -15 cdx-\frac {\cot (e+f x) (15 c-8 c \sec (e+f x))}{f}\right )+\frac {\cot ^3(e+f x) (5 c-4 c \sec (e+f x))}{3 f}\right )-\frac {\cot ^5(e+f x) (c-c \sec (e+f x))}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {\frac {1}{5} \left (\frac {\cot ^3(e+f x) (5 c-4 c \sec (e+f x))}{3 f}+\frac {1}{3} \left (-\frac {\cot (e+f x) (15 c-8 c \sec (e+f x))}{f}-15 c x\right )\right )-\frac {\cot ^5(e+f x) (c-c \sec (e+f x))}{5 f}}{a^3 c^3}\) |
-((-1/5*(Cot[e + f*x]^5*(c - c*Sec[e + f*x]))/f + ((Cot[e + f*x]^3*(5*c - 4*c*Sec[e + f*x]))/(3*f) + (-15*c*x - (Cot[e + f*x]*(15*c - 8*c*Sec[e + f* x]))/f)/3)/5)/(a^3*c^3))
3.1.37.3.1 Defintions of rubi rules used
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-(e*Cot[c + d*x])^(m + 1))*((a + b*Csc[c + d*x])/( d*e*(m + 1))), x] - Simp[1/(e^2*(m + 1)) Int[(e*Cot[c + d*x])^(m + 2)*(a* (m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && L tQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m Int[Cot[e + f*x]^(2*m)*( c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] && !( IntegerQ[n] && GtQ[m - n, 0])
Time = 0.64 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78
| method | result | size |
| parallelrisch | \(\frac {-3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-5 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+240 f x +90 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )-240 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{240 f \,a^{3} c^{2}}\) | \(78\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-16 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+32 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {6}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}}{16 f \,c^{2} a^{3}}\) | \(88\) |
| default | \(\frac {-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-16 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+32 \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {6}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}}{16 f \,c^{2} a^{3}}\) | \(88\) |
| risch | \(\frac {x}{a^{3} c^{2}}-\frac {2 i \left (15 \,{\mathrm e}^{7 i \left (f x +e \right )}-15 \,{\mathrm e}^{6 i \left (f x +e \right )}-65 \,{\mathrm e}^{5 i \left (f x +e \right )}-25 \,{\mathrm e}^{4 i \left (f x +e \right )}+73 \,{\mathrm e}^{3 i \left (f x +e \right )}+31 \,{\mathrm e}^{2 i \left (f x +e \right )}-31 \,{\mathrm e}^{i \left (f x +e \right )}-23\right )}{15 f \,c^{2} a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3}}\) | \(127\) |
| norman | \(\frac {\frac {x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{c a}-\frac {1}{48 a c f}+\frac {3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{8 a c f}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a c f}+\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{8 a c f}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{80 a c f}}{a^{2} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}\) | \(138\) |
1/240*(-3*tan(1/2*f*x+1/2*e)^5-5*cot(1/2*f*x+1/2*e)^3+30*tan(1/2*f*x+1/2*e )^3+240*f*x+90*cot(1/2*f*x+1/2*e)-240*tan(1/2*f*x+1/2*e))/f/a^3/c^2
Time = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.54 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {23 \, \cos \left (f x + e\right )^{4} + 8 \, \cos \left (f x + e\right )^{3} - 27 \, \cos \left (f x + e\right )^{2} + 15 \, {\left (f x \cos \left (f x + e\right )^{3} + f x \cos \left (f x + e\right )^{2} - f x \cos \left (f x + e\right ) - f x\right )} \sin \left (f x + e\right ) - 7 \, \cos \left (f x + e\right ) + 8}{15 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )} \]
1/15*(23*cos(f*x + e)^4 + 8*cos(f*x + e)^3 - 27*cos(f*x + e)^2 + 15*(f*x*c os(f*x + e)^3 + f*x*cos(f*x + e)^2 - f*x*cos(f*x + e) - f*x)*sin(f*x + e) - 7*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))
\[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {\int \frac {1}{\sec ^{5}{\left (e + f x \right )} + \sec ^{4}{\left (e + f x \right )} - 2 \sec ^{3}{\left (e + f x \right )} - 2 \sec ^{2}{\left (e + f x \right )} + \sec {\left (e + f x \right )} + 1}\, dx}{a^{3} c^{2}} \]
Integral(1/(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 = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=-\frac {\frac {3 \, {\left (\frac {80 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3} c^{2}} - \frac {480 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{3} c^{2}} - \frac {5 \, {\left (\frac {18 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{a^{3} c^{2} \sin \left (f x + e\right )^{3}}}{240 \, f} \]
-1/240*(3*(80*sin(f*x + e)/(cos(f*x + e) + 1) - 10*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^2) - 480*arctan (sin(f*x + e)/(cos(f*x + e) + 1))/(a^3*c^2) - 5*(18*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 = 0.35 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=\frac {\frac {240 \, {\left (f x + e\right )}}{a^{3} c^{2}} + \frac {5 \, {\left (18 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{a^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}} - \frac {3 \, {\left (a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 80 \, a^{12} c^{8} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{15} c^{10}}}{240 \, f} \]
1/240*(240*(f*x + e)/(a^3*c^2) + 5*(18*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 - 10*a^12*c ^8*tan(1/2*f*x + 1/2*e)^3 + 80*a^12*c^8*tan(1/2*f*x + 1/2*e))/(a^15*c^10)) /f
Time = 14.90 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.61 \[ \int \frac {1}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^2} \, dx=-\frac {5\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-30\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+240\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-90\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-240\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (e+f\,x\right )}{240\,a^3\,c^2\,f\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3} \]
-(5*cos(e/2 + (f*x)/2)^8 + 3*sin(e/2 + (f*x)/2)^8 - 30*cos(e/2 + (f*x)/2)^ 2*sin(e/2 + (f*x)/2)^6 + 240*cos(e/2 + (f*x)/2)^4*sin(e/2 + (f*x)/2)^4 - 9 0*cos(e/2 + (f*x)/2)^6*sin(e/2 + (f*x)/2)^2 - 240*cos(e/2 + (f*x)/2)^5*sin (e/2 + (f*x)/2)^3*(e + f*x))/(240*a^3*c^2*f*cos(e/2 + (f*x)/2)^5*sin(e/2 + (f*x)/2)^3)