Integrand size = 27, antiderivative size = 60 \[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=-\frac {2 \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {-3-2 \cos (c+d x)}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right ) \sqrt {-\tan ^2(c+d x)}}{d} \] Output:
-2*cot(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
Leaf count is larger than twice the leaf count of optimal. \(155\) vs. \(2(60)=120\).
Time = 0.38 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.58 \[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\frac {4 \sqrt {\cot ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {-\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {(3+2 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {5}{3}} \sqrt {\frac {\cos (c+d x)}{-1+\cos (c+d x)}}\right ),\frac {6}{5}\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{\sqrt {5} d \sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \] Input:
Integrate[1/(Sqrt[-3 - 2*Cos[c + d*x]]*Sqrt[-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[ (3 + 2*Cos[c + d*x])*Csc[(c + d*x)/2]^2]*Csc[c + d*x]*EllipticF[ArcSin[Sqr t[5/3]*Sqrt[Cos[c + d*x]/(-1 + Cos[c + d*x])]], 6/5]*Sin[(c + d*x)/2]^4)/( Sqrt[5]*d*Sqrt[-3 - 2*Cos[c + d*x]]*Sqrt[-Cos[c + d*x]])
Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {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.
\(\displaystyle \int \frac {1}{\sqrt {-2 \cos (c+d x)-3} \sqrt {-\cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {-2 \sin \left (c+d x+\frac {\pi }{2}\right )-3} \sqrt {-\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3294 |
\(\displaystyle -\frac {2 \sqrt {-\tan ^2(c+d x)} \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {-2 \cos (c+d x)-3}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right )}{d}\) |
Input:
Int[1/(Sqrt[-3 - 2*Cos[c + d*x]]*Sqrt[-Cos[c + d*x]]),x]
Output:
(-2*Cot[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
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]
Time = 8.72 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.77
method | result | size |
default | \(-\frac {\sqrt {2}\, \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \frac {i \sqrt {5}}{5}\right )}{5 d \sqrt {-3-2 \cos \left (d x +c \right )}\, \sqrt {-\cos \left (d x +c \right )}}\) | \(106\) |
Input:
int(1/(-3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/5/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(cot(d*x+c)-csc(d*x+c),1/5*I*5^(1/2))
\[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) - 3}} \,d x } \] Input:
integrate(1/(-3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2),x, algorithm="fric as")
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)
\[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {- \cos {\left (c + d x \right )}} \sqrt {- 2 \cos {\left (c + d x \right )} - 3}}\, dx \] Input:
integrate(1/(-3-2*cos(d*x+c))**(1/2)/(-cos(d*x+c))**(1/2),x)
Output:
Integral(1/(sqrt(-cos(c + d*x))*sqrt(-2*cos(c + d*x) - 3)), x)
\[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) - 3}} \,d x } \] Input:
integrate(1/(-3-2*cos(d*x+c))^(1/2)/(-cos(d*x+c))^(1/2),x, algorithm="maxi ma")
Output:
integrate(1/(sqrt(-cos(d*x + c))*sqrt(-2*cos(d*x + c) - 3)), x)
\[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) - 3}} \,d x } \] Input:
integrate(1/(-3-2*cos(d*x+c))^(1/2)/(-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)
Timed out. \[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {-\cos \left (c+d\,x\right )}\,\sqrt {-2\,\cos \left (c+d\,x\right )-3}} \,d x \] 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)
\[ \int \frac {1}{\sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {-2 \cos \left (d x +c \right )-3}\, \sqrt {\cos \left (d x +c \right )}}{2 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )}d x \right ) i \] Input:
int(1/(-3-2*cos(d*x+c))^(1/2)/(-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