3.7.69 \(\int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {-2+3 \cos (c+d x)}} \, dx\) [669]

Optimal. Leaf size=97 \[ -\frac {4 \sqrt {-\cos (c+d x)} \sqrt {\cos (c+d x)} \csc (c+d x) \Pi \left (\frac {1}{3};\text {ArcSin}\left (\frac {\sqrt {-2+3 \cos (c+d x)}}{\sqrt {\cos (c+d x)}}\right )|\frac {1}{5}\right ) \sqrt {-1+\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{3 \sqrt {5} d} \]

[Out]

-4/15*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))*(-cos(d*x+c))^(1/2)*cos(
d*x+c)^(1/2)*(-1+sec(d*x+c))^(1/2)*(1+sec(d*x+c))^(1/2)/d*5^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 97, 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 = {2889, 2888} \begin {gather*} -\frac {4 \sqrt {-\cos (c+d x)} \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\sec (c+d x)-1} \sqrt {\sec (c+d x)+1} \Pi \left (\frac {1}{3};\text {ArcSin}\left (\frac {\sqrt {3 \cos (c+d x)-2}}{\sqrt {\cos (c+d x)}}\right )|\frac {1}{5}\right )}{3 \sqrt {5} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[-Cos[c + d*x]]*Sqrt[Cos[c + d*x]]*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)

Rule 2888

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))]*El
lipticPi[(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]

Rule 2889

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]

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 \sqrt {-\cos (c+d x)} \sqrt {\cos (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}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 142, normalized size = 1.46 \begin {gather*} \frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {-2+3 \cos (c+d x)}{1+\cos (c+d x)}} \left (F\left (\text {ArcSin}\left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {1}{5}\right )-2 \Pi \left (-\frac {1}{5};\text {ArcSin}\left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {1}{5}\right )\right )}{\sqrt {5} d \sqrt {-\cos (c+d x)} \sqrt {-2+3 \cos (c+d x)}} \end {gather*}

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

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Maple [A]
time = 0.22, size = 142, normalized size = 1.46

method	result	size
default	\(-\frac {2 \left (\EllipticF \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, \sqrt {5}\right )-2 \EllipticPi \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}, -1, \sqrt {5}\right )\right ) \left (\sin ^{2}\left (d x +c \right )\right ) \sqrt {-\cos \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {-2+3 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{d \sqrt {-2+3 \cos \left (d x +c \right )}\, \left (-1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}\)	\(142\)

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,method=_RETURNVERBOSE)

[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)))*sin(d
*x+c)^2*(-cos(d*x+c))^(1/2)/(-2+3*cos(d*x+c))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((-2+3*cos(d*x+c))/(1+co
s(d*x+c)))^(1/2)/(-1+cos(d*x+c))/cos(d*x+c)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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(3*cos(c + d*x) - 2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {-\cos \left (c+d\,x\right )}}{\sqrt {3\,\cos \left (c+d\,x\right )-2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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