\(\int \coth ^{-1}(\sqrt {x}) \, dx\) [85]

Optimal result

Integrand size = 6, antiderivative size = 22 \[ \int \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\sqrt {x}+x \coth ^{-1}\left (\sqrt {x}\right )-\text {arctanh}\left (\sqrt {x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6022, 52, 65, 212} \[ \int \coth ^{-1}\left (\sqrt {x}\right ) \, dx=-\text {arctanh}\left (\sqrt {x}\right )+\sqrt {x}+x \coth ^{-1}\left (\sqrt {x}\right ) \]

[In]

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

[Out]

Sqrt[x] + x*ArcCoth[Sqrt[x]] - ArcTanh[Sqrt[x]]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6022

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

Rubi steps \begin{align*} \text {integral}& = x \coth ^{-1}\left (\sqrt {x}\right )-\frac {1}{2} \int \frac {\sqrt {x}}{1-x} \, dx \\ & = \sqrt {x}+x \coth ^{-1}\left (\sqrt {x}\right )-\frac {1}{2} \int \frac {1}{(1-x) \sqrt {x}} \, dx \\ & = \sqrt {x}+x \coth ^{-1}\left (\sqrt {x}\right )-\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {x}\right ) \\ & = \sqrt {x}+x \coth ^{-1}\left (\sqrt {x}\right )-\text {arctanh}\left (\sqrt {x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\sqrt {x}+x \coth ^{-1}\left (\sqrt {x}\right )-\text {arctanh}\left (\sqrt {x}\right ) \]

[In]

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

[Out]

Sqrt[x] + x*ArcCoth[Sqrt[x]] - ArcTanh[Sqrt[x]]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(x \,\operatorname {arccoth}\left (\sqrt {x}\right )+\sqrt {x}+\frac {\ln \left (\sqrt {x}-1\right )}{2}-\frac {\ln \left (\sqrt {x}+1\right )}{2}\)	\(27\)
default	\(x \,\operatorname {arccoth}\left (\sqrt {x}\right )+\sqrt {x}+\frac {\ln \left (\sqrt {x}-1\right )}{2}-\frac {\ln \left (\sqrt {x}+1\right )}{2}\)	\(27\)
parts	\(x \,\operatorname {arccoth}\left (\sqrt {x}\right )+\sqrt {x}+\frac {\ln \left (\sqrt {x}-1\right )}{2}-\frac {\ln \left (\sqrt {x}+1\right )}{2}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{2} \, {\left (x - 1\right )} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) + \sqrt {x} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\int \operatorname {acoth}{\left (\sqrt {x} \right )}\, dx \]

[In]

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

[Out]

Integral(acoth(sqrt(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \coth ^{-1}\left (\sqrt {x}\right ) \, dx=x \operatorname {arcoth}\left (\sqrt {x}\right ) + \sqrt {x} - \frac {1}{2} \, \log \left (\sqrt {x} + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {x} - 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (16) = 32\).

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.95 \[ \int \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {2}{\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1} + \frac {2 \, {\left (\sqrt {x} + 1\right )} \log \left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1}\right )}{{\left (\sqrt {x} - 1\right )} {\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1\right )}^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \coth ^{-1}\left (\sqrt {x}\right ) \, dx=x\,\mathrm {acoth}\left (\sqrt {x}\right )-\mathrm {acoth}\left (\sqrt {x}\right )+\sqrt {x} \]

[In]

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

[Out]

x*acoth(x^(1/2)) - acoth(x^(1/2)) + x^(1/2)