Integrand size = 35, antiderivative size = 61 \[ \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {d} f} \]
2*arctanh(a^(1/2)*d^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e ))^(1/2))*a^(1/2)/f/d^(1/2)
Time = 0.72 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.67 \[ \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {d+c \cos (e+f x)}}\right ) \sqrt {d+c \cos (e+f x)} \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (1+\sec (e+f x))}}{\sqrt {d} f \sqrt {c+d \sec (e+f x)}} \]
(Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[d]*Sin[(e + f*x)/2])/Sqrt[d + c*Cos[e + f*x ]]]*Sqrt[d + c*Cos[e + f*x]]*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])])/ (Sqrt[d]*f*Sqrt[c + d*Sec[e + f*x]])
Time = 0.33 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3042, 4468, 219}
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.
\(\displaystyle \int \frac {\sec (e+f x) \sqrt {a \sec (e+f x)+a}}{\sqrt {c+d \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a}}{\sqrt {c+d \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4468 |
\(\displaystyle -\frac {2 a \int \frac {1}{1-\frac {a d \tan ^2(e+f x)}{(\sec (e+f x) a+a) (c+d \sec (e+f x))}}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a} \sqrt {c+d \sec (e+f x)}}\right )}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {d} f}\) |
(2*Sqrt[a]*ArcTanh[(Sqrt[a]*Sqrt[d]*Tan[e + f*x])/(Sqrt[a + a*Sec[e + f*x] ]*Sqrt[c + d*Sec[e + f*x]])])/(Sqrt[d]*f)
3.3.34.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/Sq rt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Simp[-2*(b/f) Subs t[Int[1/(1 - b*d*x^2), x], x, Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(277\) vs. \(2(49)=98\).
Time = 5.51 (sec) , antiderivative size = 278, normalized size of antiderivative = 4.56
method | result | size |
default | \(\frac {\sqrt {2}\, \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \sqrt {c +d \sec \left (f x +e \right )}\, \left (\ln \left (\frac {2 \sqrt {2}\, \sqrt {-d}\, \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-2 \sin \left (f x +e \right ) c -2 \sin \left (f x +e \right ) d +2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}\right )-\ln \left (-\frac {2 \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-\sin \left (f x +e \right ) c -\sin \left (f x +e \right ) d -c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d \right )}{\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}\right )\right ) \cos \left (f x +e \right )}{f \sqrt {-d}\, \left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )+1}}}\) | \(278\) |
1/f*2^(1/2)/(-d)^(1/2)*(a*(sec(f*x+e)+1))^(1/2)*(c+d*sec(f*x+e))^(1/2)*(ln (2*(2^(1/2)*(-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+ e)-sin(f*x+e)*c-sin(f*x+e)*d+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(-cos(f*x+e)+1 +sin(f*x+e)))-ln(-2*(2^(1/2)*(-d)^(1/2)*(-2*(d+c*cos(f*x+e))/(cos(f*x+e)+1 ))^(1/2)*sin(f*x+e)-sin(f*x+e)*c-sin(f*x+e)*d-c*cos(f*x+e)+d*cos(f*x+e)+c- d)/(cos(f*x+e)-1+sin(f*x+e))))*cos(f*x+e)/(cos(f*x+e)+1)/(-2*(d+c*cos(f*x+ e))/(cos(f*x+e)+1))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (49) = 98\).
Time = 0.41 (sec) , antiderivative size = 307, normalized size of antiderivative = 5.03 \[ \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\left [\frac {\sqrt {\frac {a}{d}} \log \left (-\frac {8 \, a c d \cos \left (f x + e\right ) + {\left (a c^{2} - 6 \, a c d + a d^{2}\right )} \cos \left (f x + e\right )^{3} + 4 \, {\left (2 \, d^{2} \cos \left (f x + e\right ) + {\left (c d - d^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a}{d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + 8 \, a d^{2} + {\left (a c^{2} + 2 \, a c d - 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2}}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac {\sqrt {-\frac {a}{d}} \arctan \left (-\frac {2 \, d \sqrt {-\frac {a}{d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) + d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{{\left (a c - a d\right )} \cos \left (f x + e\right )^{2} + 2 \, a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}\right )}{f}\right ] \]
[1/2*sqrt(a/d)*log(-(8*a*c*d*cos(f*x + e) + (a*c^2 - 6*a*c*d + a*d^2)*cos( f*x + e)^3 + 4*(2*d^2*cos(f*x + e) + (c*d - d^2)*cos(f*x + e)^2)*sqrt(a/d) *sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*sin(f*x + e) + 8*a*d^2 + (a*c^2 + 2*a*c*d - 7*a*d^2)*cos(f*x + e)^2 )/(cos(f*x + e)^3 + cos(f*x + e)^2))/f, sqrt(-a/d)*arctan(-2*d*sqrt(-a/d)* sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) + d)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/((a*c - a*d)*cos(f*x + e)^2 + 2*a*d + (a*c + a*d)*cos(f*x + e)))/f]
\[ \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\int \frac {\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \sec {\left (e + f x \right )}}{\sqrt {c + d \sec {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\int { \frac {\sqrt {a \sec \left (f x + e\right ) + a} \sec \left (f x + e\right )}{\sqrt {d \sec \left (f x + e\right ) + c}} \,d x } \]
\[ \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\int { \frac {\sqrt {a \sec \left (f x + e\right ) + a} \sec \left (f x + e\right )}{\sqrt {d \sec \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx=\int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}}{\cos \left (e+f\,x\right )\,\sqrt {c+\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]