Integrand size = 32, antiderivative size = 86 \[ \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=-\frac {2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {11 c \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac {4 c \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \]
-2/5*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^3+11/15*c*tan(f*x+e)/a/f/(a+a*sec(f*x +e))^2-4/15*c*tan(f*x+e)/f/(a^3+a^3*sec(f*x+e))
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.55 \[ \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {c \left (1+3 \sec (e+f x)-4 \sec ^2(e+f x)\right ) \tan (e+f x)}{15 a^3 f (1+\sec (e+f x))^3} \]
Time = 0.52 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 4496, 25, 3042, 4488, 3042, 4281}
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.
\(\displaystyle \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a \sec (e+f x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )^2 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 4496 |
\(\displaystyle -\frac {\int -\frac {\sec (e+f x) (6 a c-5 a c \sec (e+f x))}{(\sec (e+f x) a+a)^2}dx}{5 a^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sec (e+f x) (6 a c-5 a c \sec (e+f x))}{(\sec (e+f x) a+a)^2}dx}{5 a^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (6 a c-5 a c \csc \left (e+f x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
\(\Big \downarrow \) 4488 |
\(\displaystyle \frac {\frac {11 a c \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2}-\frac {4}{3} c \int \frac {\sec (e+f x)}{\sec (e+f x) a+a}dx}{5 a^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {11 a c \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2}-\frac {4}{3} c \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}dx}{5 a^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
\(\Big \downarrow \) 4281 |
\(\displaystyle \frac {\frac {11 a c \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2}-\frac {4 c \tan (e+f x)}{3 f (a \sec (e+f x)+a)}}{5 a^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
(-2*c*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + ((11*a*c*Tan[e + f*x])/ (3*f*(a + a*Sec[e + f*x])^2) - (4*c*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x] )))/(5*a^2)
3.2.73.3.1 Defintions of rubi rules used
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(a*b*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(b^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b *B*(2*m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && Ne Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.75 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.42
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 | \(\frac {2 i c \left (15 \,{\mathrm e}^{3 i \left (f x +e \right )}-5 \,{\mathrm e}^{2 i \left (f x +e \right )}+5 \,{\mathrm e}^{i \left (f x +e \right )}+1\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\) | \(59\) |
norman | \(\frac {-\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{6 a f}+\frac {7 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{30 a f}+\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{30 a f}-\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{10 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2} a^{2}}\) | \(101\) |
Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {{\left (c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) - 4 \, c\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]
1/15*(c*cos(f*x + e)^2 + 3*c*cos(f*x + e) - 4*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)
\[ \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=- \frac {c \left (\int \left (- \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{3}} \]
-c*(Integral(-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))/a**3
Time = 0.23 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.34 \[ \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=-\frac {\frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, c {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]
-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 = 0.43 \[ \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=-\frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}{30 \, a^{3} f} \]
Time = 13.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.41 \[ \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=-\frac {c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\right )}{30\,a^3\,f} \]