3.223 \(\int \frac{\sqrt{a-a \cos (e+f x)}}{\sqrt{-\cos (e+f x)}} \, dx\)

Optimal. Leaf size=38 \[ -\frac{2 \sqrt{a} \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a-a \cos (e+f x)}}\right )}{f} \]

[Out]

(-2*Sqrt[a]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a - a*Cos[e + f*x]]])/f

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Rubi [A]  time = 0.0721399, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {2774, 216} \[ -\frac{2 \sqrt{a} \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a-a \cos (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*Cos[e + f*x]]/Sqrt[-Cos[e + f*x]],x]

[Out]

(-2*Sqrt[a]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a - a*Cos[e + f*x]]])/f

Rule 2774

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{\sqrt{a-a \cos (e+f x)}}{\sqrt{-\cos (e+f x)}} \, dx &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,\frac{a \sin (e+f x)}{\sqrt{a-a \cos (e+f x)}}\right )}{f}\\ &=-\frac{2 \sqrt{a} \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a-a \cos (e+f x)}}\right )}{f}\\ \end{align*}

Mathematica [C]  time = 3.46982, size = 188, normalized size = 4.95 \[ \frac{\sqrt{\cos (e)-i \sin (e)} \sqrt{-\cos (e+f x)} \left (\cot \left (\frac{1}{2} (e+f x)\right )+i\right ) \sqrt{a-a \cos (e+f x)} \left (\tanh ^{-1}\left (\frac{e^{i f x}}{\sqrt{\cos (e)-i \sin (e)} \sqrt{e^{2 i f x} (\cos (e)+i \sin (e))-i \sin (e)+\cos (e)}}\right )+\tanh ^{-1}\left (\frac{\sqrt{e^{2 i f x} (\cos (e)+i \sin (e))-i \sin (e)+\cos (e)}}{\sqrt{\cos (e)-i \sin (e)}}\right )\right )}{\sqrt{2} f \sqrt{\cos (e+f x) (\cos (f x)+i \sin (f x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - a*Cos[e + f*x]]/Sqrt[-Cos[e + f*x]],x]

[Out]

((ArcTanh[E^(I*f*x)/(Sqrt[Cos[e] - I*Sin[e]]*Sqrt[Cos[e] + E^((2*I)*f*x)*(Cos[e] + I*Sin[e]) - I*Sin[e]])] + A
rcTanh[Sqrt[Cos[e] + E^((2*I)*f*x)*(Cos[e] + I*Sin[e]) - I*Sin[e]]/Sqrt[Cos[e] - I*Sin[e]]])*Sqrt[-Cos[e + f*x
]]*Sqrt[a - a*Cos[e + f*x]]*(I + Cot[(e + f*x)/2])*Sqrt[Cos[e] - I*Sin[e]])/(Sqrt[2]*f*Sqrt[Cos[e + f*x]*(Cos[
f*x] + I*Sin[f*x])])

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Maple [B]  time = 0.336, size = 91, normalized size = 2.4 \begin{align*} -{\frac{\sin \left ( fx+e \right ) }{f \left ( -1+\cos \left ( fx+e \right ) \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sqrt{-2\,a \left ( -1+\cos \left ( fx+e \right ) \right ) }\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}} \right ){\frac{1}{\sqrt{-\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-a*cos(f*x+e))^(1/2)/(-cos(f*x+e))^(1/2),x)

[Out]

-1/f*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-2*a*(-1+cos(f*x+e)))^(1/2)*sin(f*x+e)*arctan(1/2*2^(1/2)*(-2*cos(f
*x+e)/(cos(f*x+e)+1))^(1/2))/(-cos(f*x+e))^(1/2)/(-1+cos(f*x+e))

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Maxima [B]  time = 1.84935, size = 567, normalized size = 14.92 \begin{align*} \frac{\sqrt{-a}{\left (\log \left (4 \, \sqrt{\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1} \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )^{2} + 4 \, \sqrt{\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )^{2} + 8 \,{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac{1}{4}} \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + 4\right ) - \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + \sqrt{\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1}{\left (\cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )^{2} + \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )^{2}\right )} + 2 \,{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac{1}{4}}{\left (\cos \left (f x + e\right ) \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + \sin \left (f x + e\right ) \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )}\right )\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cos(f*x+e))^(1/2)/(-cos(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(-a)*(log(4*sqrt(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin
(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + 4*sqrt(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e)
+ 1)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + 8*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 +
2*cos(2*f*x + 2*e) + 1)^(1/4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 4) - log(cos(f*x + e
)^2 + sin(f*x + e)^2 + sqrt(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*(cos(1/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2) + 2*
(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/4)*(cos(f*x + e)*cos(1/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e) + 1)) + sin(f*x + e)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))))/f

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Fricas [A]  time = 2.2619, size = 435, normalized size = 11.45 \begin{align*} \left [\frac{\sqrt{-a} \log \left (\frac{4 \, \sqrt{-a \cos \left (f x + e\right ) + a}{\left (2 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1\right )} \sqrt{-a} \sqrt{-\cos \left (f x + e\right )} -{\left (8 \, a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}\right )}{2 \, f}, \frac{\sqrt{a} \arctan \left (\frac{\sqrt{-a \cos \left (f x + e\right ) + a} \sqrt{-\cos \left (f x + e\right )}{\left (2 \, \cos \left (f x + e\right ) + 1\right )}}{2 \, \sqrt{a} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cos(f*x+e))^(1/2)/(-cos(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/2*sqrt(-a)*log((4*sqrt(-a*cos(f*x + e) + a)*(2*cos(f*x + e)^2 + 3*cos(f*x + e) + 1)*sqrt(-a)*sqrt(-cos(f*x
+ e)) - (8*a*cos(f*x + e)^2 + 8*a*cos(f*x + e) + a)*sin(f*x + e))/sin(f*x + e))/f, sqrt(a)*arctan(1/2*sqrt(-a*
cos(f*x + e) + a)*sqrt(-cos(f*x + e))*(2*cos(f*x + e) + 1)/(sqrt(a)*cos(f*x + e)*sin(f*x + e)))/f]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cos(f*x+e))**(1/2)/(-cos(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(-a*(cos(e + f*x) - 1))/sqrt(-cos(e + f*x)), x)

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Giac [B]  time = 2.23476, size = 188, normalized size = 4.95 \begin{align*} \frac{\sqrt{2}{\left (\frac{a^{2}{\left (\frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a}}{2 \, \sqrt{a}}\right )}{\sqrt{a}} - \frac{\sqrt{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{a}}\right )}{\sqrt{a}}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{{\left | a \right |}} - \frac{\sqrt{2}{\left (a^{2} \arctan \left (\frac{\sqrt{2} \sqrt{-a}}{2 \, \sqrt{a}}\right ) - a^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{a}}\right )\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}{\sqrt{a}{\left | a \right |}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*cos(f*x+e))^(1/2)/(-cos(f*x+e))^(1/2),x, algorithm="giac")

[Out]

sqrt(2)*(a^2*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*f*x + 1/2*e)^2 - a)/sqrt(a))/sqrt(a) - sqrt(2)*arctan(
sqrt(-a)/sqrt(a))/sqrt(a))*sgn(tan(1/2*f*x + 1/2*e))/abs(a) - sqrt(2)*(a^2*arctan(1/2*sqrt(2)*sqrt(-a)/sqrt(a)
) - a^2*arctan(sqrt(-a)/sqrt(a)))*sgn(tan(1/2*f*x + 1/2*e))/(sqrt(a)*abs(a)))/f