Integrand size = 30, antiderivative size = 76 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {(c-c \sec (e+f x)) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {(c-c \sec (e+f x)) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2} \]
1/5*(c-c*sec(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^3+1/15*(c-c*sec(f*x+e)) *tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2
Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.57 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=-\frac {c (-1+\sec (e+f x)) (4+\sec (e+f x)) \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))^3} \]
Time = 0.38 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 4439, 3042, 4438}
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.
\(\displaystyle \int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(a \sec (e+f x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 4439 |
\(\displaystyle \frac {\int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(\sec (e+f x) a+a)^2}dx}{5 a}+\frac {\tan (e+f x) (c-c \sec (e+f x))}{5 f (a \sec (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {\tan (e+f x) (c-c \sec (e+f x))}{5 f (a \sec (e+f x)+a)^3}\) |
\(\Big \downarrow \) 4438 |
\(\displaystyle \frac {\tan (e+f x) (c-c \sec (e+f x))}{15 a f (a \sec (e+f x)+a)^2}+\frac {\tan (e+f x) (c-c \sec (e+f x))}{5 f (a \sec (e+f x)+a)^3}\) |
((c - c*Sec[e + f*x])*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + ((c - c *Sec[e + f*x])*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2)
3.1.57.3.1 Defintions of rubi rules used
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[b*Cot[e + f*x] *(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] /; Fre eQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] & & EqQ[m + n + 1, 0] && NeQ[2*m + 1, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[b*Cot[e + f*x] *(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp [(m + n + 1)/(a*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)* (c + d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ [b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[m + n + 1, 0] && NeQ[2*m + 1, 0 ] && !LtQ[n, 0] && !(IGtQ[n + 1/2, 0] && LtQ[n + 1/2, -(m + n)])
Time = 0.66 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.47
method | result | size |
parallelrisch | \(\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5\right )}{30 a^{3} f}\) | \(36\) |
derivativedivides | \(\frac {c \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}\right )}{2 f \,a^{3}}\) | \(37\) |
default | \(\frac {c \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}\right )}{2 f \,a^{3}}\) | \(37\) |
risch | \(\frac {2 i c \left (15 \,{\mathrm e}^{4 i \left (f x +e \right )}+15 \,{\mathrm e}^{3 i \left (f x +e \right )}+25 \,{\mathrm e}^{2 i \left (f x +e \right )}+5 \,{\mathrm e}^{i \left (f x +e \right )}+4\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\) | \(70\) |
norman | \(\frac {\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{6 a f}-\frac {4 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 a f}+\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{10 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) a^{2}}\) | \(81\) |
Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {{\left (4 \, c \cos \left (f x + e\right )^{2} - 3 \, c \cos \left (f x + e\right ) - c\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]
1/15*(4*c*cos(f*x + e)^2 - 3*c*cos(f*x + e) - c)*sin(f*x + e)/(a^3*f*cos(f *x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f)
\[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=- \frac {c \left (\int \left (- \frac {\sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{3}} \]
-c*(Integral(-sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**2/(sec(e + f*x)**3 + 3*sec(e + f* x)**2 + 3*sec(e + f*x) + 1), x))/a**3
Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.51 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, c {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]
1/60*(c*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*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 - 3*c*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3)/f
Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.49 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}{30 \, a^{3} f} \]
Time = 13.13 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.46 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-5\right )}{30\,a^3\,f} \]