Integrand size = 36, antiderivative size = 95 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\tan (e+f x)}{2 f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {\text {arctanh}(\cos (e+f x)) \tan (e+f x)}{2 a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
1/2*tan(f*x+e)/f/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(1/2)-1/2*arctanh (cos(f*x+e))*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)
Time = 0.35 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.63 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx=-\frac {(-1+\text {arctanh}(\sec (e+f x)) (1+\sec (e+f x))) \tan (e+f x)}{2 f (a (1+\sec (e+f x)))^{3/2} \sqrt {c-c \sec (e+f x)}} \]
-1/2*((-1 + ArcTanh[Sec[e + f*x]]*(1 + Sec[e + f*x]))*Tan[e + f*x])/(f*(a* (1 + Sec[e + f*x]))^(3/2)*Sqrt[c - c*Sec[e + f*x]])
Time = 0.64 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4448, 3042, 4447, 25, 3042, 4257}
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.
\(\displaystyle \int \frac {\sec (e+f x)}{(a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2} \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4448 |
\(\displaystyle \frac {\int \frac {\sec (e+f x)}{\sqrt {\sec (e+f x) a+a} \sqrt {c-c \sec (e+f x)}}dx}{2 a}+\frac {\tan (e+f x)}{2 f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a} \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {\tan (e+f x)}{2 f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 4447 |
\(\displaystyle \frac {\tan (e+f x)}{2 f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x) \int -\csc (e+f x)dx}{2 a \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan (e+f x) \int \csc (e+f x)dx}{2 a \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{2 f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (e+f x) \int \csc (e+f x)dx}{2 a \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{2 f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\tan (e+f x)}{2 f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x) \text {arctanh}(\cos (e+f x))}{2 a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
Tan[e + f*x]/(2*f*(a + a*Sec[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x]]) - ( ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(2*a*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])
3.2.42.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(m_), x_Symbol] :> Simp[((-a)*c)^(m + 1 /2)*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) In t[Csc[e + f*x]*Cot[e + f*x]^(2*m), x], x] /; FreeQ[{a, b, c, d, e, f}, x] & & EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m + 1/2]
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 = 3.58 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.05
method | result | size |
default | \(\frac {\sin \left (f x +e \right ) \left (2 \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+\cos \left (f x +e \right )-1\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}}{4 f \,a^{2} \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}}\) | \(100\) |
risch | \(\frac {i \left (\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{3 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+{\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )-{\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-{\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+{\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-2 \,{\mathrm e}^{i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )}-\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )\right )}{2 a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) f}\) | \(280\) |
1/4/f/a^2*sin(f*x+e)*(2*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e))+2*ln(-cot(f* x+e)+csc(f*x+e))+cos(f*x+e)-1)*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)+1)^2/( -c*(sec(f*x+e)-1))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (83) = 166\).
Time = 0.32 (sec) , antiderivative size = 380, normalized size of antiderivative = 4.00 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx=\left [-\frac {\sqrt {-a c} {\left (\cos \left (f x + e\right ) + 1\right )} \log \left (-\frac {4 \, {\left (2 \, \sqrt {-a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + {\left (a c \cos \left (f x + e\right )^{2} + a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}, \frac {\sqrt {a c} {\left (\cos \left (f x + e\right ) + 1\right )} \arctan \left (\frac {\sqrt {a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{a c \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}\right ] \]
[-1/4*(sqrt(-a*c)*(cos(f*x + e) + 1)*log(-4*(2*sqrt(-a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)^2 + (a*c*cos(f*x + e)^2 + a*c)*sin(f*x + e))/((cos(f*x + e)^2 - 1)*sin( f*x + e)))*sin(f*x + e) - 2*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(( c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e))/((a^2*c*f*cos(f*x + e) + a ^2*c*f)*sin(f*x + e)), 1/2*(sqrt(a*c)*(cos(f*x + e) + 1)*arctan(sqrt(a*c)* sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(a*c*sin(f*x + e)))*sin(f*x + e) + sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e))/((a^2*c*f*cos (f*x + e) + a^2*c*f)*sin(f*x + e))]
\[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {\sec {\left (e + f x \right )}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (83) = 166\).
Time = 0.38 (sec) , antiderivative size = 397, normalized size of antiderivative = 4.18 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx=-\frac {{\left ({\left (2 \, {\left (2 \, \cos \left (f x + e\right ) + 1\right )} \cos \left (2 \, f x + 2 \, e\right ) + \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (f x + e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, \sin \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) + 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) - {\left (2 \, {\left (2 \, \cos \left (f x + e\right ) + 1\right )} \cos \left (2 \, f x + 2 \, e\right ) + \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (f x + e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, \sin \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) + 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) - 2 \, \cos \left (f x + e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 2 \, \cos \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{2 \, {\left (a^{2} c \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} c \cos \left (f x + e\right )^{2} + a^{2} c \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} c \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, a^{2} c \sin \left (f x + e\right )^{2} + 4 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c + 2 \, {\left (2 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c\right )} \cos \left (2 \, f x + 2 \, e\right )\right )} f} \]
-1/2*((2*(2*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + 4*co s(f*x + e)^2 + sin(2*f*x + 2*e)^2 + 4*sin(2*f*x + 2*e)*sin(f*x + e) + 4*si n(f*x + e)^2 + 4*cos(f*x + e) + 1)*arctan2(sin(f*x + e), cos(f*x + e) + 1) - (2*(2*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + 4*cos(f *x + e)^2 + sin(2*f*x + 2*e)^2 + 4*sin(2*f*x + 2*e)*sin(f*x + e) + 4*sin(f *x + e)^2 + 4*cos(f*x + e) + 1)*arctan2(sin(f*x + e), cos(f*x + e) - 1) - 2*cos(f*x + e)*sin(2*f*x + 2*e) + 2*cos(2*f*x + 2*e)*sin(f*x + e) + 2*sin( f*x + e))*sqrt(a)*sqrt(c)/((a^2*c*cos(2*f*x + 2*e)^2 + 4*a^2*c*cos(f*x + e )^2 + a^2*c*sin(2*f*x + 2*e)^2 + 4*a^2*c*sin(2*f*x + 2*e)*sin(f*x + e) + 4 *a^2*c*sin(f*x + e)^2 + 4*a^2*c*cos(f*x + e) + a^2*c + 2*(2*a^2*c*cos(f*x + e) + a^2*c)*cos(2*f*x + 2*e))*f)
Time = 1.90 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx=-\frac {c^{2} {\left (\frac {\log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{c} - \frac {\log \left ({\left | c \right |}\right )}{c} - \frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}{c^{2}}\right )}}{4 \, \sqrt {-a c} a f {\left | c \right |} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]
-1/4*c^2*(log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/c - log(abs(c))/c - (c*tan(1/ 2*f*x + 1/2*e)^2 - c)/c^2)/(sqrt(-a*c)*a*f*abs(c)*sgn(tan(1/2*f*x + 1/2*e) ^3 + tan(1/2*f*x + 1/2*e)))
Timed out. \[ \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {1}{\cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \]