3.392 \(\int \frac{1}{\sqrt{1+\cos (2 x)}} \, dx\)

Optimal. Leaf size=27 \[ \frac{\tanh ^{-1}\left (\frac{\sin (2 x)}{\sqrt{2} \sqrt{\cos (2 x)+1}}\right )}{\sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0125423, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2649, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sin (2 x)}{\sqrt{2} \sqrt{\cos (2 x)+1}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + Cos[2*x]],x]

[Out]

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^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 (2 x)}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{2-x^2} \, dx,x,-\frac{\sin (2 x)}{\sqrt{1+\cos (2 x)}}\right )\\ &=\frac{\tanh ^{-1}\left (\frac{\sin (2 x)}{\sqrt{2} \sqrt{1+\cos (2 x)}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0143964, size = 47, normalized size = 1.74 \[ -\frac{\cos (x) \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{\sqrt{\cos (2 x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 + Cos[2*x]],x]

[Out]

-((Cos[x]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]]))/Sqrt[1 + Cos[2*x]])

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Maple [C]  time = 0.015, size = 9, normalized size = 0.3 \begin{align*}{\frac{\sqrt{2}{\it InverseJacobiAM} \left ( x,1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*2^(1/2)*InverseJacobiAM(x,1)

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Maxima [A]  time = 1.58802, size = 55, normalized size = 2.04 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) - \frac{1}{4} \, \sqrt{2} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) - 1/4*sqrt(2)*log(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1)

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Fricas [B]  time = 1.85242, size = 161, normalized size = 5.96 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (-\frac{\cos \left (2 \, x\right )^{2} - 2 \, \sqrt{2} \sqrt{\cos \left (2 \, x\right ) + 1} \sin \left (2 \, x\right ) - 2 \, \cos \left (2 \, x\right ) - 3}{\cos \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log(-(cos(2*x)^2 - 2*sqrt(2)*sqrt(cos(2*x) + 1)*sin(2*x) - 2*cos(2*x) - 3)/(cos(2*x)^2 + 2*cos(2*x
) + 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\cos{\left (2 x \right )} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cos(2*x))**(1/2),x)

[Out]

Integral(1/sqrt(cos(2*x) + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\cos \left (2 \, x\right ) + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(cos(2*x) + 1), x)