\(\int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx\) [667]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx=-\frac {3 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \operatorname {EllipticPi}\left (\frac {5}{2},\arcsin \left (\frac {\sqrt {-3-2 \cos (c+d x)}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right ) \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{d \sqrt {-\cos (c+d x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx=\frac {2 i \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {5}}\right ),-5\right )-2 \operatorname {EllipticPi}\left (5,i \text {arcsinh}\left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {5}}\right ),-5\right )\right ) \sqrt {\cos (c+d x) (3+2 \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right )}}{d \sqrt {-3-2 \cos (c+d x)} \sqrt {\cos (c+d x)}} \] Input:

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

Output:

((2*I)*Cos[(c + d*x)/2]^2*(EllipticF[I*ArcSinh[Tan[(c + d*x)/2]/Sqrt[5]], 
-5] - 2*EllipticPi[5, I*ArcSinh[Tan[(c + d*x)/2]/Sqrt[5]], -5])*Sqrt[Cos[c 
 + d*x]*(3 + 2*Cos[c + d*x])*Sec[(c + d*x)/2]^4])/(d*Sqrt[-3 - 2*Cos[c + d 
*x]]*Sqrt[Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3289

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3287

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

Input:

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

Output:

(-3*Cos[c + d*x]^(3/2)*Csc[c + d*x]*EllipticPi[5/2, ArcSin[Sqrt[-3 - 2*Cos 
[c + d*x]]/(Sqrt[5]*Sqrt[-Cos[c + d*x]])], -5]*Sqrt[1 - Sec[c + d*x]]*Sqrt 
[1 + Sec[c + d*x]])/(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 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 = 8.80 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.51

method	result	size
default	\(\frac {i \sqrt {2}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \left (\operatorname {EllipticF}\left (\frac {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}}{5}, i \sqrt {5}\right )-2 \operatorname {EllipticPi}\left (\frac {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}}{5}, 5, i \sqrt {5}\right )\right ) \sqrt {5}}{5 d \sqrt {-3-2 \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}}\)	\(146\)

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

Sympy [F]

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

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

Output:

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

Reduce [F]

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