3.650 \(\int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{-3+2 \cos (c+d x)}} \, dx\)

Optimal. Leaf size=84 \[ -\frac{2 \sqrt{-\cos (c+d x)} \sqrt{\cos (c+d x)} \sqrt{-\tan ^2(c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{2 \cos (c+d x)-3}}{\sqrt{-\cos (c+d x)}}\right )|-\frac{1}{5}\right )}{\sqrt{5} d} \]

[Out]

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

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Rubi [A]  time = 0.123599, antiderivative size = 84, 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 = {2817, 2815} \[ -\frac{2 \sqrt{-\cos (c+d x)} \sqrt{\cos (c+d x)} \sqrt{-\tan ^2(c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{2 \cos (c+d x)-3}}{\sqrt{-\cos (c+d x)}}\right )|-\frac{1}{5}\right )}{\sqrt{5} d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cos[c + d*x]]*Sqrt[-3 + 2*Cos[c + d*x]]),x]

[Out]

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

Rule 2817

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt
[-(d*Sin[e + f*x])]/Sqrt[d*Sin[e + f*x]], Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[-(d*Sin[e + f*x])]), x], x] /;
FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && NegQ[(a + b)/d]

Rule 2815

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Sqrt[a^2]*Sqrt[-Cot[e + f*x]^2]*Rt[(a + b)/d, 2]*EllipticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x
]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f*Sqrt[a^2 - b^2]*Cot[e + f*x]), x] /; FreeQ[{a, b, d, e, f}, x
] && GtQ[a^2 - b^2, 0] && PosQ[(a + b)/d] && GtQ[a^2, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{-3+2 \cos (c+d x)}} \, dx &=\frac{\sqrt{-\cos (c+d x)} \int \frac{1}{\sqrt{-\cos (c+d x)} \sqrt{-3+2 \cos (c+d x)}} \, dx}{\sqrt{\cos (c+d x)}}\\ &=-\frac{2 \sqrt{-\cos (c+d x)} \sqrt{\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{-3+2 \cos (c+d x)}}{\sqrt{-\cos (c+d x)}}\right )|-\frac{1}{5}\right ) \sqrt{-\tan ^2(c+d x)}}{\sqrt{5} d}\\ \end{align*}

Mathematica [A]  time = 1.22856, size = 144, normalized size = 1.71 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\frac{2 \cos (c+d x)-3}{\cos (c+d x)-1}} \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{-\cot ^2\left (\frac{1}{2} (c+d x)\right )} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{2 \cos (c+d x)-3}{\cos (c+d x)-1}}}{\sqrt{3}}\right )|\frac{6}{5}\right )}{\sqrt{5} d \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)-1}} \sqrt{2 \cos (c+d x)-3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Cos[c + d*x]]*Sqrt[-3 + 2*Cos[c + d*x]]),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*Sqrt[(-3 + 2*Cos[c + d*x])/(-1 + Cos[c + d*x])]*Sqrt[-Cot[(c + d*x)/2]^2]*EllipticF[ArcS
in[Sqrt[(-3 + 2*Cos[c + d*x])/(-1 + Cos[c + d*x])]/Sqrt[3]], 6/5]*Tan[(c + d*x)/2])/(Sqrt[5]*d*Sqrt[Cos[c + d*
x]/(-1 + Cos[c + d*x])]*Sqrt[-3 + 2*Cos[c + d*x]])

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Maple [A]  time = 0.477, size = 123, normalized size = 1.5 \begin{align*}{\frac{{\frac{i}{5}}\sqrt{5}\sqrt{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{-2\,{\frac{-3+2\,\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) \sqrt{5}}{\sin \left ( dx+c \right ) }},{\frac{i}{5}}\sqrt{5} \right ){\frac{1}{\sqrt{-3+2\,\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*I/d*5^(1/2)*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)/(-3+2*cos(d*x+c))^(1/2)*sin(d*x+c)^4*(-2*(-3+2*cos(d
*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))*5^(1/2)/sin(d*x+c),1/5*I*5^(1/2))/cos(d*x+c)^(3/2)/(-
1+cos(d*x+c))^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 \, \cos \left (d x + c\right ) - 3} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(2*cos(d*x + c) - 3)*sqrt(cos(d*x + c))), 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 )^{2} - 3 \, \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/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)^2 - 3*cos(d*x + c)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 \cos{\left (c + d x \right )} - 3} \sqrt{\cos{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{2 \, \cos \left (d x + c\right ) - 3} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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