\(\int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {\cos (c+d x)}} \, dx\) [651]

Optimal result

Integrand size = 25, antiderivative size = 82 \[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {\cos (c+d x)}} \, dx=-\frac {2 \sqrt {-\cos (c+d x)} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {-3-2 \cos (c+d x)}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right ) \sqrt {-\tan ^2(c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.87 \[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {\cos (c+d x)}} \, dx=\frac {4 \sqrt {\cot ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {-\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {(3+2 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {5}{3}} \sqrt {\frac {\cos (c+d x)}{-1+\cos (c+d x)}}\right ),\frac {6}{5}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{\sqrt {5} d \sqrt {-3-2 \cos (c+d x)} \sqrt {\cos (c+d x)}} \] 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[-(Cos[c + d*x]*Csc[(c + d*x)/2]^2)]*Sqrt[ 
(3 + 2*Cos[c + d*x])*Csc[(c + d*x)/2]^2]*Csc[c + d*x]*EllipticF[ArcSin[Sqr 
t[5/3]*Sqrt[Cos[c + d*x]/(-1 + Cos[c + d*x])]], 6/5]*Sin[(c + d*x)/2]^4)/( 
Sqrt[5]*d*Sqrt[-3 - 2*Cos[c + d*x]]*Sqrt[Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3296

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3294

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

Input:

Int[1/(Sqrt[-3 - 2*Cos[c + d*x]]*Sqrt[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[5]*Sqrt[-Cos[c + d*x]])], -5]*Sqrt[-Tan[c + 
 d*x]^2])/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 = 9.06 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {i \sqrt {2}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \operatorname {EllipticF}\left (\frac {i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {5}}{5}, i \sqrt {5}\right ) \sqrt {5}}{5 d \sqrt {-3-2 \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}}\)	\(114\)

Input:

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

Output:

1/5*I/d/(-3-2*cos(d*x+c))^(1/2)*2^(1/2)*10^(1/2)*((3+2*cos(d*x+c))/(cos(d* 
x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/cos(d*x+c)^(1/2)*(cos(d*x 
+c)+1)*EllipticF(1/5*I*(cot(d*x+c)-csc(d*x+c))*5^(1/2),I*5^(1/2))*5^(1/2)
 

Fricas [F]

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

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

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

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

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

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

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

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

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