\(\int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx\) [657]

Optimal result

Integrand size = 27, antiderivative size = 82 \[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\frac {2 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 \cos (c+d x)}}{\sqrt {\cos (c+d x)}}\right ),-\frac {1}{5}\right ) \sqrt {-\tan ^2(c+d x)}}{\sqrt {5} d \sqrt {-\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2896, 2894} \[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {-\tan ^2(c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {3-2 \cos (c+d x)}}{\sqrt {\cos (c+d x)}}\right ),-\frac {1}{5}\right )}{\sqrt {5} d \sqrt {-\cos (c+d x)}} \]

[In]

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

[Out]

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

Rule 2894

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

Rule 2896

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {\cos (c+d x)}} \, dx}{\sqrt {-\cos (c+d x)}} \\ & = \frac {2 \cos ^{\frac {3}{2}}(c+d x) \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \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 \sqrt {-\cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.78 \[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\frac {4 \sqrt {\cot ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {(3-2 \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {-\cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {\cos (c+d x)}{-1+\cos (c+d x)}}}{\sqrt {3}}\right ),6\right ) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{d \sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \]

[In]

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

[Out]

(4*Sqrt[Cot[(c + d*x)/2]^2]*Sqrt[(3 - 2*Cos[c + d*x])*Csc[(c + d*x)/2]^2]*Sqrt[-(Cos[c + d*x]*Csc[(c + d*x)/2]
^2)]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[Cos[c + d*x]/(-1 + Cos[c + d*x])]/Sqrt[3]], 6]*Sin[(c + d*x)/2]^4)/(d*
Sqrt[3 - 2*Cos[c + d*x]]*Sqrt[-Cos[c + d*x]])

Maple [A] (verified)

Time = 6.40 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.29

method	result	size
default	\(-\frac {2 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {5}, \frac {i \sqrt {5}}{5}\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {3-2 \cos \left (d x +c \right )}\, \sqrt {5}}{5 d \sqrt {-\frac {2 \left (-3+2 \cos \left (d x +c \right )\right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\cos \left (d x +c \right )}}\)	\(106\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {- \cos {\left (c + d x \right )}} \sqrt {3 - 2 \cos {\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {-\cos \left (d x + c\right )} \sqrt {-2 \, \cos \left (d x + c\right ) + 3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {-\cos \left (c+d\,x\right )}\,\sqrt {3-2\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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