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

Optimal. Leaf size=99 \[ -\frac{4 \cos ^{\frac{3}{2}}(c+d x) \csc (c+d x) \sqrt{\sec (c+d x)-1} \sqrt{\sec (c+d x)+1} \Pi \left (\frac{1}{3};\sin ^{-1}\left (\frac{\sqrt{2-3 \cos (c+d x)}}{\sqrt{-\cos (c+d x)}}\right )|\frac{1}{5}\right )}{3 \sqrt{5} d \sqrt{-\cos (c+d x)}} \]

[Out]

(-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]])

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Rubi [A]  time = 0.116352, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2810, 2809} \[ -\frac{4 \cos ^{\frac{3}{2}}(c+d x) \csc (c+d x) \sqrt{\sec (c+d x)-1} \sqrt{\sec (c+d x)+1} \Pi \left (\frac{1}{3};\sin ^{-1}\left (\frac{\sqrt{2-3 \cos (c+d x)}}{\sqrt{-\cos (c+d x)}}\right )|\frac{1}{5}\right )}{3 \sqrt{5} d \sqrt{-\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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]])

Rule 2810

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*S
in[e + f*x]]/Sqrt[-(b*Sin[e + f*x])], Int[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]

Rule 2809

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(2*b*Tan
[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c - d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticP
i[(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), x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)}}{\sqrt{2-3 \cos (c+d x)}} \, dx &=\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)}}\\ &=-\frac{4 \cos ^{\frac{3}{2}}(c+d x) \csc (c+d x) \Pi \left (\frac{1}{3};\sin ^{-1}\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)}}\\ \end{align*}

Mathematica [A]  time = 1.69056, size = 147, normalized size = 1.48 \[ -\frac{4 \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{-\frac{(2-3 \cos (c+d x))^2}{(\cos (c+d x)+1)^2}} \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \left (F\left (\left .\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )\right |5\right )+2 \Pi \left (-1;\left .-\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )\right |5\right )\right )}{d \sqrt{2-3 \cos (c+d x)} \sqrt{\cos (c+d x)} \sqrt{\frac{2-3 \cos (c+d x)}{\cos (c+d x)+1}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.399, size = 144, normalized size = 1.5 \begin{align*} 2\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}\sqrt{2-3\,\cos \left ( dx+c \right ) }}{d \left ( 3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-5\,\cos \left ( dx+c \right ) +2 \right ) \sqrt{\cos \left ( dx+c \right ) }} \left ({\it EllipticF} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }},\sqrt{5} \right ) -2\,{\it EllipticPi} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }},-1,\sqrt{5} \right ) \right ) \sqrt{{\frac{-2+3\,\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\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)/(2-3*cos(d*x+c))^(1/2),x)

[Out]

2/d*(EllipticF((-1+cos(d*x+c))/sin(d*x+c),5^(1/2))-2*EllipticPi((-1+cos(d*x+c))/sin(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)*sin(d*x+c)^2*(2-3*cos(d*x+c))^(1/2)/(3*co
s(d*x+c)^2-5*cos(d*x+c)+2)/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{-3 \, \cos \left (d x + c\right ) + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-3*cos(d*x + c) + 2)*sqrt(cos(d*x + c))/(3*cos(d*x + c) - 2), 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{2 - 3 \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)/(2-3*cos(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(cos(c + d*x))/sqrt(2 - 3*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{-3 \, \cos \left (d x + c\right ) + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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