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

Optimal. Leaf size=75 \[ \frac{3 \cot (c+d x) \sqrt{1-\sec (c+d x)} \sqrt{\sec (c+d x)+1} \Pi \left (-\frac{1}{2};\sin ^{-1}\left (\frac{\sqrt{3-2 \cos (c+d x)}}{\sqrt{\cos (c+d x)}}\right )|-\frac{1}{5}\right )}{\sqrt{5} d} \]

[Out]

(3*Cot[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)

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Rubi [A]  time = 0.049012, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2808} \[ \frac{3 \cot (c+d x) \sqrt{1-\sec (c+d x)} \sqrt{\sec (c+d x)+1} \Pi \left (-\frac{1}{2};\sin ^{-1}\left (\frac{\sqrt{3-2 \cos (c+d x)}}{\sqrt{\cos (c+d x)}}\right )|-\frac{1}{5}\right )}{\sqrt{5} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Cot[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)

Rule 2808

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]]*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))])/(d*f*Sqrt[c^2 - d^2]), x] /; F
reeQ[{b, c, d, e, f}, x] && GtQ[c^2 - d^2, 0] && PosQ[(c + d)/b] && GtQ[c^2, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)}}{\sqrt{3-2 \cos (c+d x)}} \, dx &=\frac{3 \cot (c+d x) \Pi \left (-\frac{1}{2};\sin ^{-1}\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}\\ \end{align*}

Mathematica [A]  time = 0.813829, size = 119, normalized size = 1.59 \[ -\frac{4 \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{3-2 \cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \left (F\left (\left .\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )\right |-5\right )+2 \Pi \left (-1;\left .-\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )\right |-5\right )\right )}{d \sqrt{3-2 \cos (c+d x)} \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[c + d*x]]/Sqrt[3 - 2*Cos[c + d*x]],x]

[Out]

(-4*Cos[(c + d*x)/2]^2*Sqrt[(3 - 2*Cos[c + d*x])/(1 + Cos[c + d*x])]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*(El
lipticF[ArcSin[Tan[(c + d*x)/2]], -5] + 2*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], -5]))/(d*Sqrt[3 - 2*Cos[c
+ d*x]]*Sqrt[Cos[c + d*x]])

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Maple [B]  time = 0.4, size = 153, normalized size = 2. \begin{align*}{\frac{\sqrt{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d \left ( 2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-5\,\cos \left ( dx+c \right ) +3 \right ) } \left ({\it EllipticF} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }},i\sqrt{5} \right ) -2\,{\it EllipticPi} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }},-1,i\sqrt{5} \right ) \right ) \sqrt{3-2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{-3+2\,\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*2^(1/2)*(EllipticF((-1+cos(d*x+c))/sin(d*x+c),I*5^(1/2))-2*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,I*5^(1
/2)))*(3-2*cos(d*x+c))^(1/2)*(-2*(-3+2*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin
(d*x+c)^2/(2*cos(d*x+c)^2-5*cos(d*x+c)+3)/cos(d*x+c)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos \left (d x + c\right )}}{\sqrt{-2 \, \cos \left (d x + c\right ) + 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-2 \, \cos \left (d x + c\right ) + 3} \sqrt{\cos \left (d x + c\right )}}{2 \, \cos \left (d x + c\right ) - 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos{\left (c + d x \right )}}}{\sqrt{3 - 2 \cos{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos \left (d x + c\right )}}{\sqrt{-2 \, \cos \left (d x + c\right ) + 3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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