Integrand size = 25, antiderivative size = 99 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3+2 \cos (c+d x)}} \, dx=\frac {3 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \operatorname {EllipticPi}\left (-\frac {1}{2},\arcsin \left (\frac {\sqrt {-3+2 \cos (c+d x)}}{\sqrt {-\cos (c+d x)}}\right ),-\frac {1}{5}\right ) \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{\sqrt {5} d \sqrt {-\cos (c+d x)}} \]
3/5*cos(d*x+c)^(3/2)*csc(d*x+c)*EllipticPi((-3+2*cos(d*x+c))^(1/2)/(-cos(d *x+c))^(1/2),-1/2,1/5*I*5^(1/2))*(1-sec(d*x+c))^(1/2)*(1+sec(d*x+c))^(1/2) /d*5^(1/2)/(-cos(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 0.82 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3+2 \cos (c+d x)}} \, dx=-\frac {2 i \sqrt {-3+2 \cos (c+d x)} \sqrt {\frac {\cos (c+d x)}{5+5 \cos (c+d x)}} \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right ),-\frac {1}{5}\right )-2 \operatorname {EllipticPi}\left (\frac {1}{5},i \text {arcsinh}\left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right ),-\frac {1}{5}\right )\right )}{d \sqrt {\cos (c+d x)} \sqrt {\frac {3-2 \cos (c+d x)}{1+\cos (c+d x)}}} \]
((-2*I)*Sqrt[-3 + 2*Cos[c + d*x]]*Sqrt[Cos[c + d*x]/(5 + 5*Cos[c + d*x])]* (EllipticF[I*ArcSinh[Sqrt[5]*Tan[(c + d*x)/2]], -1/5] - 2*EllipticPi[1/5, I*ArcSinh[Sqrt[5]*Tan[(c + d*x)/2]], -1/5]))/(d*Sqrt[Cos[c + d*x]]*Sqrt[(3 - 2*Cos[c + d*x])/(1 + Cos[c + d*x])])
Time = 0.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 |
\(\displaystyle \frac {3 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \operatorname {EllipticPi}\left (-\frac {1}{2},\arcsin \left (\frac {\sqrt {2 \cos (c+d x)-3}}{\sqrt {-\cos (c+d x)}}\right ),-\frac {1}{5}\right )}{\sqrt {5} d \sqrt {-\cos (c+d x)}}\) |
(3*Cos[c + d*x]^(3/2)*Csc[c + d*x]*EllipticPi[-1/2, ArcSin[Sqrt[-3 + 2*Cos [c + d*x]]/Sqrt[-Cos[c + d*x]]], -1/5]*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec [c + d*x]])/(Sqrt[5]*d*Sqrt[-Cos[c + d*x]])
3.7.66.3.1 Defintions of rubi rules used
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]
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]
Time = 7.06 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.45
method | result | size |
default | \(\frac {i \left (F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}, \frac {i \sqrt {5}}{5}\right )-2 \Pi \left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}, \frac {1}{5}, \frac {i \sqrt {5}}{5}\right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right ) \sqrt {5}}{5 d \sqrt {-3+2 \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}}\) | \(144\) |
1/5*I/d*(EllipticF(I*(csc(d*x+c)-cot(d*x+c))*5^(1/2),1/5*I*5^(1/2))-2*Elli pticPi(I*(csc(d*x+c)-cot(d*x+c))*5^(1/2),1/5,1/5*I*5^(1/2)))*2^(1/2)*(cos( d*x+c)/(1+cos(d*x+c)))^(1/2)*(-2*(-3+2*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)/( -3+2*cos(d*x+c))^(1/2)*(1+cos(d*x+c))/cos(d*x+c)^(1/2)*5^(1/2)
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3+2 \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {2 \, \cos \left (d x + c\right ) - 3}} \,d x } \]
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3+2 \cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\sqrt {2 \cos {\left (c + d x \right )} - 3}}\, dx \]
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3+2 \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {2 \, \cos \left (d x + c\right ) - 3}} \,d x } \]
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3+2 \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {2 \, \cos \left (d x + c\right ) - 3}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3+2 \cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}}{\sqrt {2\,\cos \left (c+d\,x\right )-3}} \,d x \]