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

Optimal. Leaf size=37 \[ -\frac{2 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.0433369, antiderivative size = 37, 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 = {2775, 207} \[ -\frac{2 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[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{1-\cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d}\\ \end{align*}

Mathematica [C]  time = 0.488938, size = 277, normalized size = 7.49 \[ \frac{2 e^{i d x} \left (\cos \left (\frac{c}{2}\right )+i \sin \left (\frac{c}{2}\right )\right ) \sqrt{\cos (c)-i \sin (c)} \sqrt{1-\cos (c+d x)} \sqrt{e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )} \left (\tanh ^{-1}\left (\frac{e^{i d x}}{\sqrt{\cos (c)-i \sin (c)} \sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}\right )+\tanh ^{-1}\left (\frac{\sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}{\sqrt{\cos (c)-i \sin (c)}}\right )\right )}{d \left (i \cos \left (\frac{c}{2}\right ) \left (-1+e^{i d x}\right )-\sin \left (\frac{c}{2}\right ) \left (1+e^{i d x}\right )\right ) \sqrt{2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - 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[
1 - 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 = 0.201, size = 83, normalized size = 2.2 \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{2-2\,\cos \left ( dx+c \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((1-cos(d*x+c))^(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)*(2-2*cos(d*x+c))^(1/2)*sin(d*x+c)*arctanh((cos(d*x+c)/(1+cos(d*x
+c)))^(1/2))/cos(d*x+c)^(1/2)/(-1+cos(d*x+c))

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Maxima [B]  time = 1.7449, size = 298, normalized size = 8.05 \begin{align*} \frac{2 \, \operatorname{arsinh}\left (1\right ) + \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{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 (\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}}{\left (\cos \left (d x + c\right ) \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 ) + \sin \left (d x + c\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 )\right )}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*arcsinh(1) + log(cos(d*x + c)^2 + sin(d*x + c)^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), cos(2*d*x + 2*c) + 1))^2) + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(c
os(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)))))/d

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Fricas [A]  time = 2.12918, size = 167, normalized size = 4.51 \begin{align*} \frac{\log \left (-\frac{2 \,{\left (\cos \left (d x + c\right ) + 1\right )} \sqrt{-\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )} -{\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Giac [B]  time = 2.07705, size = 149, normalized size = 4.03 \begin{align*} -\frac{\sqrt{2}{\left ({\left (\sqrt{2} \log \left (\sqrt{2} + \sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}\right ) - \sqrt{2} \log \left (\sqrt{2} - \sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}\right )\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right ) -{\left (\sqrt{2} \log \left (\sqrt{2} + 1\right ) - \sqrt{2} \log \left (\sqrt{2} - 1\right )\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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