Integrand size = 34, antiderivative size = 89 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {2} a \sqrt {c} f}+\frac {\tan (e+f x)}{f (a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \]
-1/2*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/a/f*2^( 1/2)/c^(1/2)+tan(f*x+e)/f/(a+a*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 = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.58 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{a f \sqrt {c-c \sec (e+f x)}} \]
(Hypergeometric2F1[-1/2, 1, 1/2, (1 + Sec[e + f*x])/2]*Tan[(e + f*x)/2])/( a*f*Sqrt[c - c*Sec[e + f*x]])
Time = 0.46 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {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.
\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right ) \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4448 |
\(\displaystyle \frac {\int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}}dx}{2 a}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle \frac {\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}-\frac {\int \frac {1}{\frac {c^2 \tan ^2(e+f x)}{c-c \sec (e+f x)}+2 c}d\frac {c \tan (e+f x)}{\sqrt {c-c \sec (e+f x)}}}{a f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\tan (e+f x)}{f (a \sec (e+f x)+a) \sqrt {c-c \sec (e+f x)}}-\frac {\arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {2} a \sqrt {c} f}\) |
-(ArcTan[(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]])
3.1.90.3.1 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 = 3.00 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.26
method | result | size |
default | \(\frac {\sqrt {2}\, \sin \left (f x +e \right ) \left (\sqrt {2}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+\arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )\right )}{2 a f \left (\cos \left (f x +e \right )+1\right ) \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\) | \(112\) |
1/2/a/f*2^(1/2)*sin(f*x+e)*(2^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+arc tan(1/2*2^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)))/(cos(f*x+e)+1)/(-c*(s ec(f*x+e)-1))^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)
Time = 0.32 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.02 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \, dx=\left [\frac {\sqrt {2} c \sqrt {-\frac {1}{c}} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{c}} - {\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 4 \, \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, a c f \sin \left (f x + e\right )}, \frac {\sqrt {2} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, a c f \sin \left (f x + e\right )}\right ] \]
[1/4*(sqrt(2)*c*sqrt(-1/c)*log(-(2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e)) *sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sqrt(-1/c) - (3*cos(f*x + e) + 1) *sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin(f*x + e) - 4*sqrt((c *cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e))/(a*c*f*sin(f*x + e)), 1/2*( sqrt(2)*sqrt(c)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos (f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e) - 2*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e))/(a*c*f*sin(f*x + e))]
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\int \frac {\sec {\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx}{a} \]
Integral(sec(e + f*x)/(sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + sqrt(-c*se c(e + f*x) + c)), x)/a
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )}{{\left (a \sec \left (f x + e\right ) + a\right )} \sqrt {-c \sec \left (f x + e\right ) + c}} \,d x } \]
Time = 0.44 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\sqrt {2} {\left (\frac {\arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right )}{\sqrt {c}} - \frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{c}\right )}}{2 \, a f} \]
1/2*sqrt(2)*(arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/sqrt(c) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/c)/(a*f)
Timed out. \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) \sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {1}{\cos \left (e+f\,x\right )\,\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \]