Integrand size = 34, antiderivative size = 138 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{2 \sqrt {2} a^2 \sqrt {c} f}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{2 f \left (a^2+a^2 \sec (e+f x)\right ) \sqrt {c-c \sec (e+f x)}} \]
-1/4*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/a^2/f*2 ^(1/2)/c^(1/2)+1/3*tan(f*x+e)/f/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(1/2)+ 1/2*tan(f*x+e)/f/(a^2+a^2*sec(f*x+e))/(c-c*sec(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.32 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.43 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \tan (e+f x)}{3 f (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \]
(Hypergeometric2F1[-3/2, 1, -1/2, (1 + Sec[e + f*x])/2]*Tan[e + f*x])/(3*f *(a + a*Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])
Time = 0.72 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3042, 4448, 3042, 4448, 3042, 4282, 216}
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.
| \(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2 \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4448 |
| \(\displaystyle \frac {\int \frac {\sec (e+f x)}{(\sec (e+f x) a+a) \sqrt {c-c \sec (e+f x)}}dx}{2 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right ) \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 4448 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}}dx}{2 a}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}}{2 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}}{2 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}-\frac {\int \frac {1}{\frac {c^2 \tan ^2(e+f x)}{c-c \sec (e+f x)}+2 c}d\frac {c \tan (e+f x)}{\sqrt {c-c \sec (e+f x)}}}{a f}}{2 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}-\frac {\arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {2} a \sqrt {c} f}}{2 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}\) |
Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]]) + (-(Ar cTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])]/(Sqrt[2]*a *Sqrt[c]*f)) + Tan[e + f*x]/(f*(a + a*Sec[e + f*x])*Sqrt[c - c*Sec[e + f*x ]]))/(2*a)
3.1.97.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*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_)]*(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, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ((ILtQ[m, 0] && ILtQ[n - 1/2, 0]) || (IL tQ[m - 1/2, 0] && ILtQ[n - 1/2, 0] && LtQ[m, n]))
Time = 3.50 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.33
| method | result | size |
| default | \(\frac {\sqrt {2}\, \left (-\left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{\frac {3}{2}}+3 \arctan \left (\frac {1}{\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\right )+3 \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right ) \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{12 a^{2} f \sqrt {\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\) | \(184\) |
1/12/a^2/f*2^(1/2)/(c*(1-cos(f*x+e))^2/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)*c sc(f*x+e)^2)^(1/2)*(-((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(3/2)+3*arctan(1/(( 1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2))+3*((1-cos(f*x+e))^2*csc(f*x+e)^2-1) ^(1/2))/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*(-cot(f*x+e)+csc(f*x+e))
Time = 0.32 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.40 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \, dx=\left [-\frac {3 \, \sqrt {2} \sqrt {-c} {\left (\cos \left (f x + e\right ) + 1\right )} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (5 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{24 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}, \frac {3 \, \sqrt {2} \sqrt {c} {\left (\cos \left (f x + e\right ) + 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (5 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{12 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}\right ] \]
[-1/24*(3*sqrt(2)*sqrt(-c)*(cos(f*x + e) + 1)*log((2*sqrt(2)*(cos(f*x + e) ^2 + cos(f*x + e))*sqrt(-c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) + (3*c *cos(f*x + e) + c)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin(f* x + e) + 4*(5*cos(f*x + e)^2 + 3*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/c os(f*x + e)))/((a^2*c*f*cos(f*x + e) + a^2*c*f)*sin(f*x + e)), 1/12*(3*sqr t(2)*sqrt(c)*(cos(f*x + e) + 1)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/c os(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e) - 2*(5*cos( f*x + e)^2 + 3*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^ 2*c*f*cos(f*x + e) + a^2*c*f)*sin(f*x + e))]
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\int \frac {\sec {\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} + 2 \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx}{a^{2}} \]
Integral(sec(e + f*x)/(sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**2 + 2*sqrt( -c*sec(e + f*x) + c)*sec(e + f*x) + sqrt(-c*sec(e + f*x) + c)), x)/a**2
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2} \sqrt {-c \sec \left (f x + e\right ) + c}} \,d x } \]
Time = 0.57 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.67 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\sqrt {2} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right )}{\sqrt {c}} + \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} c^{4} - 3 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c^{5}}{c^{6}}\right )}}{12 \, a^{2} f} \]
1/12*sqrt(2)*(3*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/sqrt(c) + ((c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c^4 - 3*sqrt(c*tan(1/2*f*x + 1/2* e)^2 - c)*c^5)/c^6)/(a^2*f)
Timed out. \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {1}{\cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \]