\(\int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {2+3 \cos (c+d x)}} \, dx\) [668]

Optimal result

Integrand size = 27, antiderivative size = 99 \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {2+3 \cos (c+d x)}} \, dx=-\frac {4 \sqrt {-\cos (c+d x)} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {5}{3},\arcsin \left (\frac {\sqrt {2+3 \cos (c+d x)}}{\sqrt {5} \sqrt {\cos (c+d x)}}\right ),5\right ) \sqrt {-1-\sec (c+d x)} \sqrt {1-\sec (c+d x)}}{3 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.96 \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {2+3 \cos (c+d x)}} \, dx=\frac {4 \sqrt {\cot ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {-\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {(2+3 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) \left (3 \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {(2+3 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}\right ),-4\right )-5 \operatorname {EllipticPi}\left (-\frac {2}{3},\arcsin \left (\frac {1}{2} \sqrt {(2+3 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}\right ),-4\right )\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{3 d \sqrt {-\cos (c+d x)} \sqrt {2+3 \cos (c+d x)}} \] Input:

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

Output:

(4*Sqrt[Cot[(c + d*x)/2]^2]*Sqrt[-(Cos[c + d*x]*Csc[(c + d*x)/2]^2)]*Sqrt[ 
(2 + 3*Cos[c + d*x])*Csc[(c + d*x)/2]^2]*Csc[c + d*x]*(3*EllipticF[ArcSin[ 
Sqrt[(2 + 3*Cos[c + d*x])*Csc[(c + d*x)/2]^2]/2], -4] - 5*EllipticPi[-2/3, 
 ArcSin[Sqrt[(2 + 3*Cos[c + d*x])*Csc[(c + d*x)/2]^2]/2], -4])*Sin[(c + d* 
x)/2]^4)/(3*d*Sqrt[-Cos[c + d*x]]*Sqrt[2 + 3*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 99, 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

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

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[5/3, Ar 
cSin[Sqrt[2 + 3*Cos[c + d*x]]/(Sqrt[5]*Sqrt[Cos[c + d*x]])], 5]*Sqrt[-1 - 
Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]])/(3*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 = 9.17 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.40

method	result	size
default	\(-\frac {\sqrt {-\cos \left (d x +c \right )}\, \sqrt {10}\, \sqrt {\frac {2+3 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\operatorname {EllipticF}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}}{5}, \sqrt {5}\right )-2 \operatorname {EllipticPi}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}}{5}, -5, \sqrt {5}\right )\right ) \left (1+\sec \left (d x +c \right )\right ) \sqrt {5}}{5 d \sqrt {2+3 \cos \left (d x +c \right )}}\)	\(139\)

Input:

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

Output:

-1/5/d*(-cos(d*x+c))^(1/2)/(2+3*cos(d*x+c))^(1/2)*10^(1/2)*((2+3*cos(d*x+c 
))/(cos(d*x+c)+1))^(1/2)*2^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(Ellipt 
icF(1/5*(csc(d*x+c)-cot(d*x+c))*5^(1/2),5^(1/2))-2*EllipticPi(1/5*(csc(d*x 
+c)-cot(d*x+c))*5^(1/2),-5,5^(1/2)))*(1+sec(d*x+c))*5^(1/2)
 

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