\(\int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx\) [753]

Optimal result

Integrand size = 25, antiderivative size = 129 \[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d \sqrt {\sec (c+d x)}} \] Output:

2*(a+b)^(1/2)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2) 
/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b 
))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/d/sec(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )}{d \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \] Input:

Integrate[Sqrt[Sec[c + d*x]]/Sqrt[a + b*Cos[c + d*x]],x]
 

Output:

(2*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSi 
n[Tan[(c + d*x)/2]], (-a + b)/(a + b)])/(d*Sqrt[Cos[c + d*x]/(1 + Cos[c + 
d*x])]*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4710, 3042, 3295}

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.

\(\displaystyle \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3295

\(\displaystyle \frac {2 \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d \sqrt {\sec (c+d x)}}\)

Input:

Int[Sqrt[Sec[c + d*x]]/Sqrt[a + b*Cos[c + d*x]],x]
 

Output:

(2*Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b 
*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt 
[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a* 
d*Sqrt[Sec[c + d*x]])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 4710

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 14.70 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.84

method	result	size
default	\(-\frac {2 \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}\, \operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) \cos \left (d x +c \right )}{d \sqrt {a +\cos \left (d x +c \right ) b}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(109\)

Input:

int(sec(d*x+c)^(1/2)/(a+cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/d*sec(d*x+c)^(1/2)/(a+cos(d*x+c)*b)^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c) 
+1)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*cos 
(d*x+c)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
 

Fricas [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(sec(d*x + c))/sqrt(b*cos(d*x + c) + a), x)
 

Sympy [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\sqrt {\sec {\left (c + d x \right )}}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \] Input:

integrate(sec(d*x+c)**(1/2)/(a+b*cos(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(sec(c + d*x))/sqrt(a + b*cos(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(sec(d*x + c))/sqrt(b*cos(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(sec(d*x + c))/sqrt(b*cos(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int((1/cos(c + d*x))^(1/2)/(a + b*cos(c + d*x))^(1/2),x)
 

Output:

int((1/cos(c + d*x))^(1/2)/(a + b*cos(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}}{\cos \left (d x +c \right ) b +a}d x \] Input:

int(sec(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1/2),x)
 

Output:

int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a))/(cos(c + d*x)*b + a),x)