\(\int \frac {\coth ^{-1}(\sqrt {x})}{\sqrt {x}} \, dx\) [91]

Optimal result

Integrand size = 12, antiderivative size = 20 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=2 \sqrt {x} \coth ^{-1}\left (\sqrt {x}\right )+\log (1-x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6038, 31} \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=\log (1-x)+2 \sqrt {x} \coth ^{-1}\left (\sqrt {x}\right ) \]

[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 6038

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCoth[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rubi steps \begin{align*} \text {integral}& = 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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=2 \sqrt {x} \coth ^{-1}\left (\sqrt {x}\right )+\log (1-x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(2 \,\operatorname {arccoth}\left (\sqrt {x}\right ) \sqrt {x}+\ln \left (x -1\right )\)	\(15\)
default	\(2 \,\operatorname {arccoth}\left (\sqrt {x}\right ) \sqrt {x}+\ln \left (x -1\right )\)	\(15\)

[In]

int(arccoth(x^(1/2))/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=\sqrt {x} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) + \log \left (x - 1\right ) \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (17) = 34\).

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 4.35 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=\frac {2 x^{\frac {3}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{x - 1} - \frac {2 \sqrt {x} \operatorname {acoth}{\left (\sqrt {x} \right )}}{x - 1} + \frac {2 x \log {\left (\sqrt {x} + 1 \right )}}{x - 1} - \frac {2 x \operatorname {acoth}{\left (\sqrt {x} \right )}}{x - 1} - \frac {2 \log {\left (\sqrt {x} + 1 \right )}}{x - 1} + \frac {2 \operatorname {acoth}{\left (\sqrt {x} \right )}}{x - 1} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=2 \, \sqrt {x} \operatorname {arcoth}\left (\sqrt {x}\right ) + \log \left (-x + 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (16) = 32\).

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.50 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=\frac {2 \, \log \left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1}\right )}{\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1} + 2 \, \log \left (\frac {\sqrt {x} + 1}{{\left | \sqrt {x} - 1 \right |}}\right ) - 2 \, \log \left ({\left | \frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.98 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {\coth ^{-1}\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=\ln \left (x-1\right )+2\,\sqrt {x}\,\mathrm {acoth}\left (\sqrt {x}\right ) \]

[In]

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

[Out]

log(x - 1) + 2*x^(1/2)*acoth(x^(1/2))