Integrand size = 26, antiderivative size = 102 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {a^3 x}{c^3}-\frac {8 a^3 \tan (e+f x)}{5 c^3 f (1-\sec (e+f x))^3}+\frac {4 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))^2}-\frac {26 a^3 \tan (e+f x)}{15 c^3 f (1-\sec (e+f x))} \]
a^3*x/c^3-8/5*a^3*tan(f*x+e)/c^3/f/(1-sec(f*x+e))^3+4/15*a^3*tan(f*x+e)/c^ 3/f/(1-sec(f*x+e))^2-26/15*a^3*tan(f*x+e)/c^3/f/(1-sec(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.52 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {2 a^3 \cot ^5\left (\frac {e}{2}+\frac {f x}{2}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5 c^3 f} \]
(2*a^3*Cot[e/2 + (f*x)/2]^5*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[e/2 + (f *x)/2]^2])/(5*c^3*f)
Time = 0.59 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3042, 4391, 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.
\(\displaystyle \int \frac {(a \sec (e+f x)+a)^3}{(c-c \sec (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 4391 |
\(\displaystyle \frac {\int \left (\frac {\sec ^3(e+f x) a^3}{(1-\sec (e+f x))^3}+\frac {3 \sec ^2(e+f x) a^3}{(1-\sec (e+f x))^3}+\frac {3 \sec (e+f x) a^3}{(1-\sec (e+f x))^3}+\frac {a^3}{(1-\sec (e+f x))^3}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {26 a^3 \tan (e+f x)}{15 f (1-\sec (e+f x))}+\frac {4 a^3 \tan (e+f x)}{15 f (1-\sec (e+f x))^2}-\frac {8 a^3 \tan (e+f x)}{5 f (1-\sec (e+f x))^3}+a^3 x}{c^3}\) |
(a^3*x - (8*a^3*Tan[e + f*x])/(5*f*(1 - Sec[e + f*x])^3) + (4*a^3*Tan[e + f*x])/(15*f*(1 - Sec[e + f*x])^2) - (26*a^3*Tan[e + f*x])/(15*f*(1 - Sec[e + f*x])))/c^3
3.1.18.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_), x_Symbol] :> Simp[c^n Int[ExpandTrig[(1 + (d/c)*csc[e + f*x])^n, (a + b*csc[e + f*x])^m, x], x], x] /; FreeQ[{a, b, c, d, e, f, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0] && ILtQ[n, 0] && LtQ[m + n, 2]
Time = 0.55 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.53
method | result | size |
parallelrisch | \(\frac {a^{3} \left (6 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-10 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+15 f x +30 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c^{3} f}\) | \(54\) |
derivativedivides | \(\frac {2 a^{3} \left (\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f \,c^{3}}\) | \(60\) |
default | \(\frac {2 a^{3} \left (\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {1}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f \,c^{3}}\) | \(60\) |
risch | \(\frac {a^{3} x}{c^{3}}+\frac {4 i a^{3} \left (45 \,{\mathrm e}^{4 i \left (f x +e \right )}-90 \,{\mathrm e}^{3 i \left (f x +e \right )}+140 \,{\mathrm e}^{2 i \left (f x +e \right )}-70 \,{\mathrm e}^{i \left (f x +e \right )}+23\right )}{15 f \,c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{5}}\) | \(81\) |
norman | \(\frac {\frac {a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{c}+\frac {a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{c}+\frac {2 a^{3}}{5 c f}-\frac {22 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 c f}+\frac {56 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{15 c f}-\frac {14 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 c f}+\frac {2 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{c f}-\frac {2 a^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{c}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2} c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\) | \(189\) |
1/15*a^3*(6*cot(1/2*f*x+1/2*e)^5-10*cot(1/2*f*x+1/2*e)^3+15*f*x+30*cot(1/2 *f*x+1/2*e))/c^3/f
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.25 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {46 \, a^{3} \cos \left (f x + e\right )^{3} - 2 \, a^{3} \cos \left (f x + e\right )^{2} - 22 \, a^{3} \cos \left (f x + e\right ) + 26 \, a^{3} + 15 \, {\left (a^{3} f x \cos \left (f x + e\right )^{2} - 2 \, a^{3} f x \cos \left (f x + e\right ) + a^{3} f x\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \]
1/15*(46*a^3*cos(f*x + e)^3 - 2*a^3*cos(f*x + e)^2 - 22*a^3*cos(f*x + e) + 26*a^3 + 15*(a^3*f*x*cos(f*x + e)^2 - 2*a^3*f*x*cos(f*x + e) + a^3*f*x)*s in(f*x + e))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f* x + e))
\[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=- \frac {a^{3} \left (\int \frac {3 \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {1}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx\right )}{c^{3}} \]
-a**3*(Integral(3*sec(e + f*x)/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*se c(e + f*x) - 1), x) + Integral(3*sec(e + f*x)**2/(sec(e + f*x)**3 - 3*sec( e + f*x)**2 + 3*sec(e + f*x) - 1), x) + Integral(sec(e + f*x)**3/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x) + Integral(1/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x))/c**3
Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (90) = 180\).
Time = 0.29 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.76 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {a^{3} {\left (\frac {120 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{3}} - \frac {{\left (\frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}\right )} + \frac {a^{3} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac {3 \, a^{3} {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} - \frac {9 \, a^{3} {\left (\frac {5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, f} \]
1/60*(a^3*(120*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^3 - (20*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 105*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)* (cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5)) + a^3*(10*sin(f*x + e)^2/(cos(f *x + e) + 1)^2 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x + e)^5) - 3*a^3*(10*sin(f*x + e)^2/(cos(f*x + e) + 1) ^2 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) + 1)^5/(c^3 *sin(f*x + e)^5) - 9*a^3*(5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)*(cos( f*x + e) + 1)^5/(c^3*sin(f*x + e)^5))/f
Time = 0.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} a^{3}}{c^{3}} + \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 5 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{3}\right )}}{c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{15 \, f} \]
1/15*(15*(f*x + e)*a^3/c^3 + 2*(15*a^3*tan(1/2*f*x + 1/2*e)^4 - 5*a^3*tan( 1/2*f*x + 1/2*e)^2 + 3*a^3)/(c^3*tan(1/2*f*x + 1/2*e)^5))/f
Time = 14.50 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.94 \[ \int \frac {(a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^3} \, dx=\frac {a^3\,x}{c^3}+\frac {\frac {2\,a^3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{5}-\frac {2\,a^3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{3}+2\,a^3\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{c^3\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5} \]