3.86 \(\int \frac{\coth ^{-1}(\sqrt{x})}{x} \, dx\)

Optimal. Leaf size=19 \[ \text{PolyLog}\left (2,-\frac{1}{\sqrt{x}}\right )-\text{PolyLog}\left (2,\frac{1}{\sqrt{x}}\right ) \]

[Out]

PolyLog[2, -(1/Sqrt[x])] - PolyLog[2, 1/Sqrt[x]]

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Rubi [A]  time = 0.0197034, antiderivative size = 19, 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 = {6096, 5913} \[ \text{PolyLog}\left (2,-\frac{1}{\sqrt{x}}\right )-\text{PolyLog}\left (2,\frac{1}{\sqrt{x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Sqrt[x]]/x,x]

[Out]

PolyLog[2, -(1/Sqrt[x])] - PolyLog[2, 1/Sqrt[x]]

Rule 6096

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcCoth[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rule 5913

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[(b*PolyLog[2, -(c*x)^(-1)
])/2, x] - Simp[(b*PolyLog[2, 1/(c*x)])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}\left (\sqrt{x}\right )}{x} \, dx &=2 \operatorname{Subst}\left (\int \frac{\coth ^{-1}(x)}{x} \, dx,x,\sqrt{x}\right )\\ &=\text{Li}_2\left (-\frac{1}{\sqrt{x}}\right )-\text{Li}_2\left (\frac{1}{\sqrt{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0101691, size = 19, normalized size = 1. \[ \text{PolyLog}\left (2,-\frac{1}{\sqrt{x}}\right )-\text{PolyLog}\left (2,\frac{1}{\sqrt{x}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Sqrt[x]]/x,x]

[Out]

PolyLog[2, -(1/Sqrt[x])] - PolyLog[2, 1/Sqrt[x]]

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Maple [B]  time = 0.054, size = 33, normalized size = 1.7 \begin{align*} \ln \left ( x \right ){\rm arccoth} \left (\sqrt{x}\right )-{\it dilog} \left ( \sqrt{x} \right ) -{\it dilog} \left ( 1+\sqrt{x} \right ) -{\frac{\ln \left ( x \right ) }{2}\ln \left ( 1+\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(x^(1/2))/x,x)

[Out]

ln(x)*arccoth(x^(1/2))-dilog(x^(1/2))-dilog(1+x^(1/2))-1/2*ln(x)*ln(1+x^(1/2))

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Maxima [B]  time = 0.959881, size = 89, normalized size = 4.68 \begin{align*} -\frac{1}{2} \,{\left (\log \left (\sqrt{x} + 1\right ) - \log \left (\sqrt{x} - 1\right )\right )} \log \left (x\right ) + \operatorname{arcoth}\left (\sqrt{x}\right ) \log \left (x\right ) + \log \left (-\sqrt{x}\right ) \log \left (\sqrt{x} + 1\right ) - \frac{1}{2} \, \log \left (x\right ) \log \left (\sqrt{x} - 1\right ) +{\rm Li}_2\left (\sqrt{x} + 1\right ) -{\rm Li}_2\left (-\sqrt{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(log(sqrt(x) + 1) - log(sqrt(x) - 1))*log(x) + arccoth(sqrt(x))*log(x) + log(-sqrt(x))*log(sqrt(x) + 1) -
1/2*log(x)*log(sqrt(x) - 1) + dilog(sqrt(x) + 1) - dilog(-sqrt(x) + 1)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcoth}\left (\sqrt{x}\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccoth(sqrt(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(x**(1/2))/x,x)

[Out]

Integral(acoth(sqrt(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(sqrt(x))/x, x)