Integrand size = 25, antiderivative size = 99 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}} \, dx=-\frac {4 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \operatorname {EllipticPi}\left (\frac {1}{3},\arcsin \left (\frac {\sqrt {2-3 \cos (c+d x)}}{\sqrt {-\cos (c+d x)}}\right ),\frac {1}{5}\right ) \sqrt {-1+\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{3 \sqrt {5} d \sqrt {-\cos (c+d x)}} \]
-4/15*cos(d*x+c)^(3/2)*csc(d*x+c)*EllipticPi((2-3*cos(d*x+c))^(1/2)/(-cos( d*x+c))^(1/2),1/3,1/5*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)
Leaf count is larger than twice the leaf count of optimal. \(201\) vs. \(2(99)=198\).
Time = 1.62 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.03 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}} \, dx=\frac {4 \sqrt {\cot ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {-\left ((-2+3 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )\right )} \csc (c+d x) \left (3 \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {(2-3 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}\right ),\frac {4}{5}\right )-\operatorname {EllipticPi}\left (\frac {2}{3},\arcsin \left (\frac {1}{2} \sqrt {(2-3 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}\right ),\frac {4}{5}\right )\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{3 \sqrt {5} d \sqrt {2-3 \cos (c+d x)} \sqrt {\cos (c+d x)}} \]
(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], 4/5])*Sin[(c + d* x)/2]^4)/(3*Sqrt[5]*d*Sqrt[2 - 3*Cos[c + d*x]]*Sqrt[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, 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.
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {2-3 \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3289 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}}dx}{\sqrt {-\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {\sqrt {-\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {2-3 \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {-\cos (c+d x)}}\) |
\(\Big \downarrow \) 3288 |
\(\displaystyle -\frac {4 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \sqrt {\sec (c+d x)-1} \sqrt {\sec (c+d x)+1} \operatorname {EllipticPi}\left (\frac {1}{3},\arcsin \left (\frac {\sqrt {2-3 \cos (c+d x)}}{\sqrt {-\cos (c+d x)}}\right ),\frac {1}{5}\right )}{3 \sqrt {5} d \sqrt {-\cos (c+d x)}}\) |
(-4*Cos[c + d*x]^(3/2)*Csc[c + d*x]*EllipticPi[1/3, ArcSin[Sqrt[2 - 3*Cos[ c + d*x]]/Sqrt[-Cos[c + d*x]]], 1/5]*Sqrt[-1 + Sec[c + d*x]]*Sqrt[1 + Sec[ c + d*x]])/(3*Sqrt[5]*d*Sqrt[-Cos[c + d*x]])
3.7.62.3.1 Defintions of rubi rules used
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*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]
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 = 6.71 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.31
method | result | size |
default | \(-\frac {2 \left (F\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {5}\right )-2 \Pi \left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {5}\right )\right ) \sqrt {\frac {-2+3 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2-3 \cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right )}\, \left (-2+3 \cos \left (d x +c \right )\right )}\) | \(130\) |
-2/d*(EllipticF(cot(d*x+c)-csc(d*x+c),5^(1/2))-2*EllipticPi(cot(d*x+c)-csc (d*x+c),-1,5^(1/2)))*((-2+3*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/ (1+cos(d*x+c)))^(1/2)*(2-3*cos(d*x+c))^(1/2)*(1+cos(d*x+c))/cos(d*x+c)^(1/ 2)/(-2+3*cos(d*x+c))
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {-3 \, \cos \left (d x + c\right ) + 2}} \,d x } \]
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\sqrt {2 - 3 \cos {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {-3 \, \cos \left (d x + c\right ) + 2}} \,d x } \]
\[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {-3 \, \cos \left (d x + c\right ) + 2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {2-3 \cos (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}}{\sqrt {2-3\,\cos \left (c+d\,x\right )}} \,d x \]