\(\int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {-2+3 \cos (c+d x)}} \, dx\) [669]

Optimal result

Integrand size = 27, antiderivative size = 97 \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {-2+3 \cos (c+d x)}} \, dx=-\frac {4 \sqrt {-\cos (c+d x)} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {1}{3},\arcsin \left (\frac {\sqrt {-2+3 \cos (c+d x)}}{\sqrt {\cos (c+d x)}}\right ),\frac {1}{5}\right ) \sqrt {-1+\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{3 \sqrt {5} d} \] Output:

-4/15*(-cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticPi((-2+3*cos 
(d*x+c))^(1/2)/cos(d*x+c)^(1/2),1/3,1/5*5^(1/2))*(-1+sec(d*x+c))^(1/2)*(1+ 
sec(d*x+c))^(1/2)*5^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.46 \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {-2+3 \cos (c+d x)}} \, dx=\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {-2+3 \cos (c+d x)}{1+\cos (c+d x)}} \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {1}{5}\right )-2 \operatorname {EllipticPi}\left (-\frac {1}{5},\arcsin \left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {1}{5}\right )\right )}{\sqrt {5} d \sqrt {-\cos (c+d x)} \sqrt {-2+3 \cos (c+d x)}} \] Input:

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

Output:

(4*Cos[(c + d*x)/2]^2*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(-2 + 3*C 
os[c + d*x])/(1 + Cos[c + d*x])]*(EllipticF[ArcSin[Sqrt[5]*Tan[(c + d*x)/2 
]], 1/5] - 2*EllipticPi[-1/5, ArcSin[Sqrt[5]*Tan[(c + d*x)/2]], 1/5]))/(Sq 
rt[5]*d*Sqrt[-Cos[c + d*x]]*Sqrt[-2 + 3*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.35 (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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3289

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

\(\displaystyle -\frac {4 \sqrt {-\cos (c+d x)} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\sec (c+d x)-1} \sqrt {\sec (c+d x)+1} \operatorname {EllipticPi}\left (\frac {1}{3},\arcsin \left (\frac {\sqrt {3 \cos (c+d x)-2}}{\sqrt {\cos (c+d x)}}\right ),\frac {1}{5}\right )}{3 \sqrt {5} d}\)

Input:

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

Output:

(-4*Sqrt[-Cos[c + d*x]]*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticPi[1/3, Ar 
cSin[Sqrt[-2 + 3*Cos[c + d*x]]/Sqrt[Cos[c + d*x]]], 1/5]*Sqrt[-1 + Sec[c + 
 d*x]]*Sqrt[1 + Sec[c + d*x]])/(3*Sqrt[5]*d)
 

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 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 = 8.86 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {2 \sqrt {-\cos \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {-2+3 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {5}\right )-2 \operatorname {EllipticPi}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {5}\right )\right ) \left (1+\sec \left (d x +c \right )\right )}{d \sqrt {-2+3 \cos \left (d x +c \right )}}\)	\(120\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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