Integrand size = 25, antiderivative size = 97 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx=-\frac {3 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \operatorname {EllipticPi}\left (\frac {5}{2},\arcsin \left (\frac {\sqrt {-3-2 \cos (c+d x)}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right ) \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{d \sqrt {-\cos (c+d x)}} \]
-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)
Result contains complex when optimal does not.
Time = 0.66 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx=\frac {2 i \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {5}}\right ),-5\right )-2 \operatorname {EllipticPi}\left (5,i \text {arcsinh}\left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {5}}\right ),-5\right )\right ) \sqrt {\cos (c+d x) (3+2 \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right )}}{d \sqrt {-3-2 \cos (c+d x)} \sqrt {\cos (c+d x)}} \]
((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]])
Time = 0.35 (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 |
\(\displaystyle -\frac {3 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \operatorname {EllipticPi}\left (\frac {5}{2},\arcsin \left (\frac {\sqrt {-2 \cos (c+d x)-3}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right )}{d \sqrt {-\cos (c+d x)}}\) |
(-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]])
3.7.67.3.1 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 = 6.82 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.63
method | result | size |
default | \(-\frac {i \sqrt {10}\, \sqrt {2}\, \left (F\left (\frac {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}}{5}, i \sqrt {5}\right )-2 \Pi \left (\frac {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}}{5}, 5, i \sqrt {5}\right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-3-2 \cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right ) \sqrt {5}}{5 d \sqrt {\cos \left (d x +c \right )}\, \left (3+2 \cos \left (d x +c \right )\right )}\) | \(158\) |
-1/5*I/d*10^(1/2)*2^(1/2)*(EllipticF(1/5*I*(csc(d*x+c)-cot(d*x+c))*5^(1/2) ,I*5^(1/2))-2*EllipticPi(1/5*I*(csc(d*x+c)-cot(d*x+c))*5^(1/2),5,I*5^(1/2) ))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((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)/(3+2*cos(d*x+c) )*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. \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}}{\sqrt {-2\,\cos \left (c+d\,x\right )-3}} \,d x \]