3.73 \(\int \sqrt{1+\coth (x)} \, dx\)

Optimal. Leaf size=21 \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\coth (x)+1}}{\sqrt{2}}\right ) \]

[Out]

Sqrt[2]*ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]]

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Rubi [A]  time = 0.0120375, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3480, 206} \[ \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\coth (x)+1}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + Coth[x]],x]

[Out]

Sqrt[2]*ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[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 \sqrt{1+\coth (x)} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{2-x^2} \, dx,x,\sqrt{1+\coth (x)}\right )\\ &=\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1+\coth (x)}}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0741134, size = 45, normalized size = 2.14 \[ \frac{(1+i) (\coth (x)+1)^{3/2} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{i (\coth (x)+1)}\right )}{(i (\coth (x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + Coth[x]],x]

[Out]

((1 + I)*ArcTan[(1/2 + I/2)*Sqrt[I*(1 + Coth[x])]]*(1 + Coth[x])^(3/2))/(I*(1 + Coth[x]))^(3/2)

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Maple [A]  time = 0.036, size = 17, normalized size = 0.8 \begin{align*}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+{\rm coth} \left (x\right )}} \right ) \sqrt{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+coth(x))^(1/2),x)

[Out]

arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(coth(x) + 1), x)

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Fricas [B]  time = 2.31598, size = 181, normalized size = 8.62 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (2 \, \sqrt{2} \sqrt{\frac{\sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} + 2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log(2*sqrt(2)*sqrt(sinh(x)/(cosh(x) - sinh(x)))*(cosh(x) + sinh(x)) + 2*cosh(x)^2 + 4*cosh(x)*sinh
(x) + 2*sinh(x)^2 - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(coth(x) + 1), x)

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Giac [B]  time = 1.17684, size = 50, normalized size = 2.38 \begin{align*} -\frac{1}{2} \, \sqrt{2} \log \left ({\left | 2 \, \sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 2 \, e^{\left (2 \, x\right )} + 1 \right |}\right ) \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*log(abs(2*sqrt(e^(4*x) - e^(2*x)) - 2*e^(2*x) + 1))*sgn(e^(2*x) - 1)