\(\int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx\) [657]

Optimal result

Integrand size = 27, antiderivative size = 82 \[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\frac {2 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 \cos (c+d x)}}{\sqrt {\cos (c+d x)}}\right ),-\frac {1}{5}\right ) \sqrt {-\tan ^2(c+d x)}}{\sqrt {5} d \sqrt {-\cos (c+d x)}} \] Output:

2/5*cos(d*x+c)^(3/2)*csc(d*x+c)*EllipticF((3-2*cos(d*x+c))^(1/2)/cos(d*x+c 
)^(1/2),1/5*I*5^(1/2))*(-tan(d*x+c)^2)^(1/2)*5^(1/2)/d/(-cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.78 \[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\frac {4 \sqrt {\cot ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {(3-2 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {-\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {\cos (c+d x)}{-1+\cos (c+d x)}}}{\sqrt {3}}\right ),6\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{d \sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \] Input:

Integrate[1/(Sqrt[3 - 2*Cos[c + d*x]]*Sqrt[-Cos[c + d*x]]),x]
 

Output:

(4*Sqrt[Cot[(c + d*x)/2]^2]*Sqrt[(3 - 2*Cos[c + d*x])*Csc[(c + d*x)/2]^2]* 
Sqrt[-(Cos[c + d*x]*Csc[(c + d*x)/2]^2)]*Csc[c + d*x]*EllipticF[ArcSin[Sqr 
t[Cos[c + d*x]/(-1 + Cos[c + d*x])]/Sqrt[3]], 6]*Sin[(c + d*x)/2]^4)/(d*Sq 
rt[3 - 2*Cos[c + d*x]]*Sqrt[-Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3296, 3042, 3294}

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.

\(\displaystyle \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {3-2 \sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {-\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3296

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {\cos (c+d x)}}dx}{\sqrt {-\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {3-2 \sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {-\cos (c+d x)}}\)

\(\Big \downarrow \) 3294

\(\displaystyle \frac {2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {-\tan ^2(c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 \cos (c+d x)}}{\sqrt {\cos (c+d x)}}\right ),-\frac {1}{5}\right )}{\sqrt {5} d \sqrt {-\cos (c+d x)}}\)

Input:

Int[1/(Sqrt[3 - 2*Cos[c + d*x]]*Sqrt[-Cos[c + d*x]]),x]
 

Output:

(2*Cos[c + d*x]^(3/2)*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[3 - 2*Cos[c + d*x 
]]/Sqrt[Cos[c + d*x]]], -1/5]*Sqrt[-Tan[c + d*x]^2])/(Sqrt[5]*d*Sqrt[-Cos[ 
c + d*x]])
 

Defintions of rubi rules used

rule 3042

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

rule 3294

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*Sqrt[a^2]*(Sqrt[-Cot[e + f*x]^2]/(a*f*Sqr 
t[a^2 - b^2]*Cot[e + f*x]))*Rt[(a + b)/d, 2]*EllipticF[ArcSin[Sqrt[a + b*Si 
n[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] && GtQ[a^2 - b^2, 0] && PosQ[(a + b)/d] && GtQ[ 
a^2, 0]
 

rule 3296

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(-d)*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]] 
   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[(-d)*Sin[e + f*x]]), x], x] /; Free 
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && NegQ[(a + b)/d]
 

Maple [A] (verified)

Time = 9.46 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.29

method	result	size
default	\(\frac {2 i \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {5}, \frac {i \sqrt {5}}{5}\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {3-2 \cos \left (d x +c \right )}\, \sqrt {5}}{5 d \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )+1}}\, \sqrt {-\cos \left (d x +c \right )}}\)	\(106\)

Input:

int(1/(3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/5*I/d*EllipticF(I*(cot(d*x+c)-csc(d*x+c))*5^(1/2),1/5*I*5^(1/2))*2^(1/2) 
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/(-2*(-3+2*cos(d*x+c))/(cos(d*x+c)+1))^( 
1/2)*(3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2)*5^(1/2)
 

Fricas [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) + 3}} \,d x } \] Input:

integrate(1/(3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2),x, algorithm="frica 
s")
 

Output:

integral(sqrt(-cos(d*x + c))*sqrt(-2*cos(d*x + c) + 3)/(2*cos(d*x + c)^2 - 
 3*cos(d*x + c)), x)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {- \cos {\left (c + d x \right )}} \sqrt {3 - 2 \cos {\left (c + d x \right )}}}\, dx \] Input:

integrate(1/(3-2*cos(d*x+c))**(1/2)/(-cos(d*x+c))**(1/2),x)
 

Output:

Integral(1/(sqrt(-cos(c + d*x))*sqrt(3 - 2*cos(c + d*x))), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) + 3}} \,d x } \] Input:

integrate(1/(3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2),x, algorithm="maxim 
a")
 

Output:

integrate(1/(sqrt(-cos(d*x + c))*sqrt(-2*cos(d*x + c) + 3)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) + 3}} \,d x } \] Input:

integrate(1/(3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2),x, algorithm="giac" 
)
 

Output:

integrate(1/(sqrt(-cos(d*x + c))*sqrt(-2*cos(d*x + c) + 3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {-\cos \left (c+d\,x\right )}\,\sqrt {3-2\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int(1/((-cos(c + d*x))^(1/2)*(3 - 2*cos(c + d*x))^(1/2)),x)
 

Output:

int(1/((-cos(c + d*x))^(1/2)*(3 - 2*cos(c + d*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {-2 \cos \left (d x +c \right )+3}\, \sqrt {\cos \left (d x +c \right )}}{2 \cos \left (d x +c \right )^{2}-3 \cos \left (d x +c \right )}d x \right ) i \] Input:

int(1/(3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2),x)
 

Output:

int((sqrt( - 2*cos(c + d*x) + 3)*sqrt(cos(c + d*x)))/(2*cos(c + d*x)**2 - 
3*cos(c + d*x)),x)*i