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

Optimal. Leaf size=47 \[ \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\sin ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )\right |5\right )}{d \sqrt {-\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2814, 2813} \[ \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\sin ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )\right |5\right )}{d \sqrt {-\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])], 5])/(d*Sqrt[-Cos[c + d*x]])

Rule 2813

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

Rule 2814

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

Rubi steps

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

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Mathematica [B]  time = 0.40, size = 158, normalized size = 3.36 \[ \frac {4 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cot ^2\left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) \sqrt {\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {-\left ((3 \cos (c+d x)-2) \csc ^2\left (\frac {1}{2} (c+d x)\right )\right )} F\left (\sin ^{-1}\left (\frac {1}{2} \sqrt {-\left ((3 \cos (c+d x)-2) \csc ^2\left (\frac {1}{2} (c+d x)\right )\right )}\right )|\frac {4}{5}\right )}{\sqrt {5} d \sqrt {-\cos (c+d x)} \sqrt {3 \cos (c+d x)-2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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]*EllipticF[ArcSin[Sqrt[-((-2 + 3*Cos[c + d*x])*Csc[(c + d*x)/2]^2)]/2], 4/5]*Sin[(c + d*x)/2
]^4)/(Sqrt[5]*d*Sqrt[-Cos[c + d*x]]*Sqrt[-2 + 3*Cos[c + d*x]])

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-\cos \left (d x + c\right )} \sqrt {3 \, \cos \left (d x + c\right ) - 2}}{3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {3 \, \cos \left (d x + c\right ) - 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.19, size = 109, normalized size = 2.32 \[ \frac {2 \EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {5}\right ) \sqrt {\frac {-2+3 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\sin ^{2}\left (d x +c \right )\right )}{d \sqrt {-2+3 \cos \left (d x +c \right )}\, \left (-1+\cos \left (d x +c \right )\right ) \sqrt {-\cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {3 \, \cos \left (d x + c\right ) - 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {-\cos \left (c+d\,x\right )}\,\sqrt {3\,\cos \left (c+d\,x\right )-2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- \cos {\left (c + d x \right )}} \sqrt {3 \cos {\left (c + d x \right )} - 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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