3.265 \(\int \frac{\sqrt{a-a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\)

Optimal. Leaf size=48 \[ -\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{d} \]

[Out]

(-2*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])])/d

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Rubi [A]  time = 0.0661367, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2775, 207} \[ -\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])])/d

Rule 2775

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

Rule 207

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

Rubi steps

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

Mathematica [C]  time = 0.514349, size = 278, normalized size = 5.79 \[ \frac{2 e^{i d x} \left (\cos \left (\frac{c}{2}\right )+i \sin \left (\frac{c}{2}\right )\right ) \sqrt{\cos (c)-i \sin (c)} \sqrt{a-a \cos (c+d x)} \sqrt{e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )} \left (\tanh ^{-1}\left (\frac{e^{i d x}}{\sqrt{\cos (c)-i \sin (c)} \sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}\right )+\tanh ^{-1}\left (\frac{\sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}{\sqrt{\cos (c)-i \sin (c)}}\right )\right )}{d \left (i \cos \left (\frac{c}{2}\right ) \left (-1+e^{i d x}\right )-\sin \left (\frac{c}{2}\right ) \left (1+e^{i d x}\right )\right ) \sqrt{2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(I*d*x)*(ArcTanh[E^(I*d*x)/(Sqrt[Cos[c] - I*Sin[c]]*Sqrt[Cos[c] + E^((2*I)*d*x)*(Cos[c] + I*Sin[c]) - I*S
in[c]])] + ArcTanh[Sqrt[Cos[c] + E^((2*I)*d*x)*(Cos[c] + I*Sin[c]) - I*Sin[c]]/Sqrt[Cos[c] - I*Sin[c]]])*Sqrt[
a - a*Cos[c + d*x]]*(Cos[c/2] + I*Sin[c/2])*Sqrt[Cos[c] - I*Sin[c]]*Sqrt[((1 + E^((2*I)*d*x))*Cos[c] + I*(-1 +
E^((2*I)*d*x))*Sin[c])/E^(I*d*x)])/(d*(I*(-1 + E^(I*d*x))*Cos[c/2] - (1 + E^(I*d*x))*Sin[c/2])*Sqrt[2*(1 + E^
((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c]])

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Maple [B]  time = 0.252, size = 84, normalized size = 1.8 \begin{align*}{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{d \left ( -1+\cos \left ( dx+c \right ) \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\sqrt{-2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) }{\it Artanh} \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \right ){\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-2*a*(-1+cos(d*x+c)))^(1/2)*sin(d*x+c)*arctanh((cos(d*x+c)/(1+c
os(d*x+c)))^(1/2))/cos(d*x+c)^(1/2)/(-1+cos(d*x+c))

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Maxima [B]  time = 1.79409, size = 200, normalized size = 4.17 \begin{align*} \frac{\sqrt{-a} \arctan \left ({\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac{1}{4}} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \sin \left (d x + c\right ),{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac{1}{4}} \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \cos \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c))/d

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Fricas [A]  time = 2.28595, size = 417, normalized size = 8.69 \begin{align*} \left [\frac{\sqrt{a} \log \left (\frac{4 \, \sqrt{-a \cos \left (d x + c\right ) + a}{\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{a} \sqrt{\cos \left (d x + c\right )} -{\left (8 \, a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right )}{2 \, d}, \frac{\sqrt{-a} \arctan \left (\frac{\sqrt{-a \cos \left (d x + c\right ) + a} \sqrt{-a}{\left (2 \, \cos \left (d x + c\right ) + 1\right )}}{2 \, a \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right )}{d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(a)*log((4*sqrt(-a*cos(d*x + c) + a)*(2*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*sqrt(cos(d*x + c
)) - (8*a*cos(d*x + c)^2 + 8*a*cos(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))/d, sqrt(-a)*arctan(1/2*sqrt(-a*co
s(d*x + c) + a)*sqrt(-a)*(2*cos(d*x + c) + 1)/(a*sqrt(cos(d*x + c))*sin(d*x + c)))/d]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.28903, size = 193, normalized size = 4.02 \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} \, d x + \frac{1}{2} \, c\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} \, d x + \frac{1}{2} \, c\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} \, d x + \frac{1}{2} \, c\right )\right )}{\sqrt{-a}{\left | a \right |}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*(a^2*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) - sqrt(2)*arct
an(sqrt(a)/sqrt(-a))/sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c))/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*d*x + 1/2*c))/(sqrt(-a)*abs(a)))/d