\(\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {-3+2 \cos (c+d x)}} \, dx\) [650]

Optimal result

Integrand size = 25, antiderivative size = 84 \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {-3+2 \cos (c+d x)}} \, dx=-\frac {2 \sqrt {-\cos (c+d x)} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {-3+2 \cos (c+d x)}}{\sqrt {-\cos (c+d x)}}\right ),-\frac {1}{5}\right ) \sqrt {-\tan ^2(c+d x)}}{\sqrt {5} d} \] Output:

-2/5*(-cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*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
 

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {-3+2 \cos (c+d x)}} \, dx=\frac {2 \sqrt {\cos (c+d x)} \sqrt {\frac {-3+2 \cos (c+d x)}{-1+\cos (c+d x)}} \sqrt {-\cot ^2\left (\frac {1}{2} (c+d x)\right )} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {-3+2 \cos (c+d x)}{-1+\cos (c+d x)}}}{\sqrt {3}}\right ),\frac {6}{5}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {5} d \sqrt {\frac {\cos (c+d x)}{-1+\cos (c+d x)}} \sqrt {-3+2 \cos (c+d x)}} \] Input:

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

Output:

(2*Sqrt[Cos[c + d*x]]*Sqrt[(-3 + 2*Cos[c + d*x])/(-1 + Cos[c + d*x])]*Sqrt 
[-Cot[(c + d*x)/2]^2]*EllipticF[ArcSin[Sqrt[(-3 + 2*Cos[c + d*x])/(-1 + Co 
s[c + d*x])]/Sqrt[3]], 6/5]*Tan[(c + d*x)/2])/(Sqrt[5]*d*Sqrt[Cos[c + d*x] 
/(-1 + Cos[c + d*x])]*Sqrt[-3 + 2*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 84, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3296

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3294

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

Input:

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

Output:

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

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 = 9.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.33

method	result	size
default	\(-\frac {i \left (\cos \left (d x +c \right )+1\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}, \frac {i \sqrt {5}}{5}\right ) \sqrt {5}}{5 d \sqrt {\cos \left (d x +c \right )}\, \sqrt {-3+2 \cos \left (d x +c \right )}}\)	\(112\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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