\(\int \frac {1}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx\) [754]

Optimal result

Integrand size = 25, antiderivative size = 136 \[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b d \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.71 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \left (\operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )\right ) \sqrt {1+\sec (c+d x)}}{d \sqrt {\frac {1}{1+\cos (c+d x)}} \sqrt {a+b \cos (c+d x)}} \] Input:

Integrate[1/(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 136, 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, 3288}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

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

Input:

Int[1/(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]),x]
 

Output:

(-2*Sqrt[a + b]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticPi[(a + b)/b, ArcS 
in[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)])/(b*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 3288

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c 
*((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti 
cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + 
 d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - 
 d^2, 0] && PosQ[(c + d)/b]
 

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 = 12.61 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {2 \left (-\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right )+2 \operatorname {EllipticPi}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {-\frac {a -b}{a +b}}\right )\right ) \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}}{d \sqrt {a +\cos \left (d x +c \right ) b}\, \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(139\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(1/(sqrt(b*cos(d*x + c) + a)*sqrt(sec(d*x + c))), x)
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt(a + b*cos(c + d*x))*sqrt(sec(c + d*x))), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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