\(\int \frac {1}{\sqrt {-\cos (c+d x)} \sqrt {3+2 \cos (c+d x)}} \, dx\) [656]

Optimal result

Integrand size = 27, antiderivative size = 80 \[ \int \frac {1}{\sqrt {-\cos (c+d x)} \sqrt {3+2 \cos (c+d x)}} \, dx=\frac {2 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3+2 \cos (c+d x)}}{\sqrt {5} \sqrt {\cos (c+d x)}}\right ),-5\right ) \sqrt {-\tan ^2(c+d x)}}{d \sqrt {-\cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.92 \[ \int \frac {1}{\sqrt {-\cos (c+d x)} \sqrt {3+2 \cos (c+d x)}} \, dx=-\frac {4 \sqrt {-\cot ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {-\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {(3+2 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {(3+2 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}}{\sqrt {6}}\right ),6\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{d \sqrt {-\cos (c+d x)} \sqrt {3+2 \cos (c+d x)}} \] Input:

Integrate[1/(Sqrt[-Cos[c + d*x]]*Sqrt[3 + 2*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)]*Sqr 
t[(3 + 2*Cos[c + d*x])*Csc[(c + d*x)/2]^2]*Csc[c + d*x]*EllipticF[ArcSin[S 
qrt[(3 + 2*Cos[c + d*x])*Csc[(c + d*x)/2]^2]/Sqrt[6]], 6]*Sin[(c + d*x)/2] 
^4)/(d*Sqrt[-Cos[c + d*x]]*Sqrt[3 + 2*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 80, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3296

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3294

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

Input:

Int[1/(Sqrt[-Cos[c + d*x]]*Sqrt[3 + 2*Cos[c + d*x]]),x]
 

Output:

(2*Cos[c + d*x]^(3/2)*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[3 + 2*Cos[c + d*x 
]]/(Sqrt[5]*Sqrt[Cos[c + d*x]])], -5]*Sqrt[-Tan[c + d*x]^2])/(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 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 = 9.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.45

method	result	size
default	\(-\frac {i \left (\cos \left (d x +c \right )+1\right ) \operatorname {EllipticF}\left (\frac {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}}{5}, i \sqrt {5}\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {5}}{5 d \sqrt {3+2 \cos \left (d x +c \right )}\, \sqrt {-\cos \left (d x +c \right )}}\)	\(116\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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