Integrand size = 32, antiderivative size = 77 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx=-\frac {2 \sqrt {2} a \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}+\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}} \]
-2*a*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))*2^(1/2) /f/c^(1/2)+2*a*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx=\frac {2 a \left (-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (e+f x)}}{\sqrt {2}}\right )+\sqrt {1+\sec (e+f x)}\right ) \tan (e+f x)}{f \sqrt {1+\sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
(2*a*(-(Sqrt[2]*ArcTanh[Sqrt[1 + Sec[e + f*x]]/Sqrt[2]]) + Sqrt[1 + Sec[e + f*x]])*Tan[e + f*x])/(f*Sqrt[1 + Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])
Time = 0.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 4444, 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 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4444 |
\(\displaystyle 2 a \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}}dx+\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx+\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle \frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}-\frac {4 a \int \frac {1}{\frac {c^2 \tan ^2(e+f x)}{c-c \sec (e+f x)}+2 c}d\frac {c \tan (e+f x)}{\sqrt {c-c \sec (e+f x)}}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}-\frac {2 \sqrt {2} a \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}\) |
(-2*Sqrt[2]*a*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f* x]])])/(Sqrt[c]*f) + (2*a*Tan[e + f*x])/(f*Sqrt[c - c*Sec[e + f*x]])
3.1.68.3.1 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_)]*(d_.) + (c_))^(n_.))/ Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*d*Cot[e + f*x]*((c + d*Csc[e + f*x])^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*c*((2*n - 1)/(2*n - 1)) Int[Csc[e + f*x]*((c + d*Csc[e + f*x ])^(n - 1)/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x ] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]
Time = 3.64 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {a \sqrt {2}\, \left (-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )+\sqrt {2}\right ) \tan \left (f x +e \right )}{f \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}}\) | \(80\) |
parts | \(\frac {a \sqrt {2}\, \sin \left (f x +e \right ) \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )}{f \left (\cos \left (f x +e \right )+1\right ) \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}+\frac {a \sqrt {2}\, \left (-\sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )+\sqrt {2}\right ) \tan \left (f x +e \right )}{f \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}}\) | \(164\) |
a/f*2^(1/2)*(-2*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*arctan(1/2*2^(1/2)/(-co s(f*x+e)/(cos(f*x+e)+1))^(1/2))+2^(1/2))/(-c*(sec(f*x+e)-1))^(1/2)*tan(f*x +e)
Time = 0.33 (sec) , antiderivative size = 272, normalized size of antiderivative = 3.53 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx=\left [\frac {\sqrt {2} a c \sqrt {-\frac {1}{c}} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{c}} - {\left (3 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (a \cos \left (f x + e\right ) + a\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{c f \sin \left (f x + e\right )}, \frac {2 \, {\left (\sqrt {2} a \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) + a\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}\right )}}{c f \sin \left (f x + e\right )}\right ] \]
[(sqrt(2)*a*c*sqrt(-1/c)*log(-(2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e))*s qrt((c*cos(f*x + e) - c)/cos(f*x + e))*sqrt(-1/c) - (3*cos(f*x + e) + 1)*s in(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin(f*x + e) - 2*(a*cos(f* x + e) + a)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/(c*f*sin(f*x + e)), 2 *(sqrt(2)*a*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) - (a*cos(f*x + e) + a)* sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/(c*f*sin(f*x + e))]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx=a \left (\int \frac {\sec {\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx\right ) \]
a*(Integral(sec(e + f*x)/sqrt(-c*sec(e + f*x) + c), x) + Integral(sec(e + f*x)**2/sqrt(-c*sec(e + f*x) + c), x))
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )} \sec \left (f x + e\right )}{\sqrt {-c \sec \left (f x + e\right ) + c}} \,d x } \]
Time = 0.95 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx=\frac {2 \, a {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right )}{\sqrt {c}} + \frac {\sqrt {2}}{\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}\right )}}{f} \]
2*a*(sqrt(2)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/sqrt(c) + sqrt(2)/sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c))/f
Timed out. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{\sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {a+\frac {a}{\cos \left (e+f\,x\right )}}{\cos \left (e+f\,x\right )\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \]