3.285 \(\int \frac{1}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}} \, dx\)

Optimal. Leaf size=47 \[ -\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{2} \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d} \]

[Out]

-((Sqrt[2]*ArcTanh[Sin[c + d*x]/(Sqrt[2]*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/d)

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Rubi [A]  time = 0.0482254, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2782, 206} \[ -\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{2} \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*ArcTanh[Sin[c + d*x]/(Sqrt[2]*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/d)

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - 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 206

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

Rubi steps

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

Mathematica [C]  time = 0.138234, size = 110, normalized size = 2.34 \[ \frac{i e^{-i (c+d x)} \left (-1+e^{i (c+d x)}\right ) \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{2} d \sqrt{-(\cos (c+d x)-1) \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.289, size = 84, normalized size = 1.8 \begin{align*} 4\,{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) }{d\sqrt{\cos \left ( dx+c \right ) } \left ( 2-2\,\cos \left ( dx+c \right ) \right ) ^{3/2}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}{\it Artanh} \left ( 1/2\,{\sqrt{2}{\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/d*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))*sin(d*x+c)*arctanh(1/2*2^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2))/cos(d*x+c)^(1/2)/(2-2*cos(d*x+c))^(3/2)

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Maxima [C]  time = 1.91575, size = 335, normalized size = 7.13 \begin{align*} \frac{\sqrt{2} \log \left (\frac{4 \,{\left ({\left | i \, e^{\left (i \, d x + i \, c\right )} - i \right |}^{2} + 2 \, \sqrt{\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}{\left (\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 )^{2} + \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 )^{2}\right )} - 2 \,{\left (\sqrt{2}{\left | i \, e^{\left (i \, d x + i \, c\right )} - i \right |} \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 ) - 2 \, \sqrt{2} \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 )\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}} + 4\right )}}{{\left | i \, e^{\left (i \, d x + i \, c\right )} - i \right |}^{2}}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log(4*(abs(I*e^(I*d*x + I*c) - I)^2 + 2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), c
os(2*d*x + 2*c) + 1))^2) - 2*(sqrt(2)*abs(I*e^(I*d*x + I*c) - I)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1)) - 2*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*
d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4) + 4)/abs(I*e^(I*d*x + I*c) - I)^2)/d

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Fricas [B]  time = 2.13078, size = 231, normalized size = 4.91 \begin{align*} \frac{\sqrt{2} \log \left (-\frac{2 \,{\left (\sqrt{2} \cos \left (d x + c\right ) + \sqrt{2}\right )} \sqrt{-\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )} -{\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log(-(2*(sqrt(2)*cos(d*x + c) + sqrt(2))*sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) - (3*cos(d*x +
c) + 1)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.04332, size = 107, normalized size = 2.28 \begin{align*} \frac{\sqrt{2}{\left (\log \left (\sqrt{2} + 1\right ) - \log \left (\sqrt{2} - 1\right ) - \log \left (\sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 1\right ) + \log \left (-\sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 1\right )\right )}}{2 \, d \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*(log(sqrt(2) + 1) - log(sqrt(2) - 1) - log(sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1) + 1) + log(-sqrt(-tan
(1/2*d*x + 1/2*c)^2 + 1) + 1))/(d*sgn(tan(1/2*d*x + 1/2*c)))