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

Optimal. Leaf size=20 \[ \log (1-x)+2 \sqrt {x} \coth ^{-1}\left (\sqrt {x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6098, 31} \[ \log (1-x)+2 \sqrt {x} \coth ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[x]*ArcCoth[Sqrt[x]] + Log[1 - x]

Rule 31

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

Rule 6098

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
th[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx &=2 \sqrt {x} \coth ^{-1}\left (\sqrt {x}\right )-\int \frac {1}{1-x} \, dx\\ &=2 \sqrt {x} \coth ^{-1}\left (\sqrt {x}\right )+\log (1-x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \[ \log (1-x)+2 \sqrt {x} \coth ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[x]*ArcCoth[Sqrt[x]] + Log[1 - x]

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fricas [A]  time = 0.65, size = 24, normalized size = 1.20 \[ \sqrt {x} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) + \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (\sqrt {x}\right )}{\sqrt {x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(sqrt(x))/sqrt(x), x)

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maple [A]  time = 0.05, size = 15, normalized size = 0.75 \[ 2 \,\mathrm {arccoth}\left (\sqrt {x}\right ) \sqrt {x}+\ln \left (-1+x \right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 16, normalized size = 0.80 \[ 2 \, \sqrt {x} \operatorname {arcoth}\left (\sqrt {x}\right ) + \log \left (-x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.29, size = 14, normalized size = 0.70 \[ \ln \left (x-1\right )+2\,\sqrt {x}\,\mathrm {acoth}\left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 1) + 2*x^(1/2)*acoth(x^(1/2))

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sympy [B]  time = 0.54, size = 87, normalized size = 4.35 \[ \frac {2 x^{\frac {3}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{x - 1} - \frac {2 \sqrt {x} \operatorname {acoth}{\left (\sqrt {x} \right )}}{x - 1} + \frac {2 x \log {\left (\sqrt {x} + 1 \right )}}{x - 1} - \frac {2 x \operatorname {acoth}{\left (\sqrt {x} \right )}}{x - 1} - \frac {2 \log {\left (\sqrt {x} + 1 \right )}}{x - 1} + \frac {2 \operatorname {acoth}{\left (\sqrt {x} \right )}}{x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**(3/2)*acoth(sqrt(x))/(x - 1) - 2*sqrt(x)*acoth(sqrt(x))/(x - 1) + 2*x*log(sqrt(x) + 1)/(x - 1) - 2*x*acot
h(sqrt(x))/(x - 1) - 2*log(sqrt(x) + 1)/(x - 1) + 2*acoth(sqrt(x))/(x - 1)

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