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

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

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2810, 2808} \[ \frac {3 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \Pi \left (-\frac {1}{2};\sin ^{-1}\left (\frac {\sqrt {2 \cos (c+d x)-3}}{\sqrt {-\cos (c+d x)}}\right )|-\frac {1}{5}\right )}{\sqrt {5} d \sqrt {-\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Cos[c + d*x]^(3/2)*Csc[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*Sqrt[-Cos[c + d*x]])

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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.98, size = 135, normalized size = 1.36 \[ -\frac {2 i \sqrt {2 \cos (c+d x)-3} \sqrt {\frac {\cos (c+d x)}{5 \cos (c+d x)+5}} \left (F\left (i \sinh ^{-1}\left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right )|-\frac {1}{5}\right )-2 \Pi \left (\frac {1}{5};i \sinh ^{-1}\left (\sqrt {5} \tan \left (\frac {1}{2} (c+d x)\right )\right )|-\frac {1}{5}\right )\right )}{d \sqrt {\cos (c+d x)} \sqrt {\frac {3-2 \cos (c+d x)}{\cos (c+d x)+1}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[-3 + 2*Cos[c + d*x]]*Sqrt[Cos[c + d*x]/(5 + 5*Cos[c + d*x])]*(EllipticF[I*ArcSinh[Sqrt[5]*Tan[(c
+ d*x)/2]], -1/5] - 2*EllipticPi[1/5, I*ArcSinh[Sqrt[5]*Tan[(c + d*x)/2]], -1/5]))/(d*Sqrt[Cos[c + d*x]]*Sqrt[
(3 - 2*Cos[c + d*x])/(1 + Cos[c + d*x])])

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {2 \, \cos \left (d x + c\right ) - 3}}, x\right ) \]

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

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

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)

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maple [A]  time = 0.18, size = 158, normalized size = 1.60 \[ -\frac {i \sqrt {2}\, \left (2 \EllipticPi \left (\frac {i \left (-1+\cos \left (d x +c \right )\right ) \sqrt {5}}{\sin \left (d x +c \right )}, \frac {1}{5}, \frac {i \sqrt {5}}{5}\right )-\EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right ) \sqrt {5}}{\sin \left (d x +c \right )}, \frac {i \sqrt {5}}{5}\right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{1+\cos \left (d x +c \right )}}\, \left (\sin ^{2}\left (d x +c \right )\right ) \sqrt {5}}{5 d \sqrt {-3+2 \cos \left (d x +c \right )}\, \left (-1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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(2*cos(c + d*x) - 3), x)

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