\(\int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {3-2 \cos (c+d x)}} \, dx\) [673]

Optimal result

Integrand size = 27, antiderivative size = 97 \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {3-2 \cos (c+d x)}} \, dx=\frac {3 \sqrt {-\cos (c+d x)} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (-\frac {1}{2},\arcsin \left (\frac {\sqrt {3-2 \cos (c+d x)}}{\sqrt {\cos (c+d x)}}\right ),-\frac {1}{5}\right ) \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{\sqrt {5} d} \] Output:

3/5*(-cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticPi((3-2*cos(d* 
x+c))^(1/2)/cos(d*x+c)^(1/2),-1/2,1/5*I*5^(1/2))*(1-sec(d*x+c))^(1/2)*(1+s 
ec(d*x+c))^(1/2)*5^(1/2)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {3-2 \cos (c+d x)}} \, dx=\frac {2 i \sqrt {-\cos (c+d x)} \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right ),-\frac {1}{5}\right )-2 \operatorname {EllipticPi}\left (\frac {1}{5},i \text {arcsinh}\left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right ),-\frac {1}{5}\right )\right ) \sqrt {1+5 \tan ^2\left (\frac {1}{2} (c+d x)\right )}}{d \sqrt {30-20 \cos (c+d x)} \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}} \] Input:

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

Output:

((2*I)*Sqrt[-Cos[c + d*x]]*(EllipticF[I*ArcSinh[Sqrt[5]*Tan[(c + d*x)/2]], 
 -1/5] - 2*EllipticPi[1/5, I*ArcSinh[Sqrt[5]*Tan[(c + d*x)/2]], -1/5])*Sqr 
t[1 + 5*Tan[(c + d*x)/2]^2])/(d*Sqrt[30 - 20*Cos[c + d*x]]*Sqrt[Cos[c + d* 
x]/(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 97, 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, 3289, 3042, 3287}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3289

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3287

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

Input:

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

Output:

(3*Sqrt[-Cos[c + d*x]]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticPi[-1/2, Ar 
cSin[Sqrt[3 - 2*Cos[c + d*x]]/Sqrt[Cos[c + d*x]]], -1/5]*Sqrt[1 - Sec[c + 
d*x]]*Sqrt[1 + Sec[c + d*x]])/(Sqrt[5]*d)
 

Defintions of rubi rules used

rule 3042

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

rule 3287

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[2*c*Rt[b*(c + d), 2]*Tan[e + f*x]*Sqrt[1 + Csc[e 
 + f*x]]*(Sqrt[1 - Csc[e + f*x]]/(d*f*Sqrt[c^2 - d^2]))*EllipticPi[(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] && GtQ[c^2 - d^2, 0] && 
PosQ[(c + d)/b] && GtQ[c^2, 0]
 

rule 3289

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[Sqrt[b*Sin[e + f*x]]/Sqrt[(-b)*Sin[e + f*x]]   I 
nt[Sqrt[(-b)*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] /; FreeQ[{b, c, 
 d, e, f}, x] && NeQ[c^2 - d^2, 0] && NegQ[(c + d)/b]
 

Maple [A] (verified)

Time = 9.79 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.53

method	result	size
default	\(\frac {i \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )+1}}\, \sqrt {-\cos \left (d x +c \right )}\, \left (2 \operatorname {EllipticPi}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {5}, \frac {1}{5}, \frac {i \sqrt {5}}{5}\right )-\operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {5}, \frac {i \sqrt {5}}{5}\right )\right ) \left (1+\sec \left (d x +c \right )\right ) \sqrt {5}}{5 d \sqrt {3-2 \cos \left (d x +c \right )}}\)	\(148\)

Input:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((-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) - 
3),x)*i