3.87 \(\int \frac{\coth ^{-1}(\sqrt{x})}{x^2} \, dx\)

Optimal. Leaf size=25 \[ -\frac{1}{\sqrt{x}}+\tanh ^{-1}\left (\sqrt{x}\right )-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{x} \]

[Out]

-(1/Sqrt[x]) - ArcCoth[Sqrt[x]]/x + ArcTanh[Sqrt[x]]

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Rubi [A]  time = 0.013518, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6098, 51, 63, 206} \[ -\frac{1}{\sqrt{x}}+\tanh ^{-1}\left (\sqrt{x}\right )-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Sqrt[x]]/x^2,x]

[Out]

-(1/Sqrt[x]) - ArcCoth[Sqrt[x]]/x + ArcTanh[Sqrt[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]

Rule 51

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

Rule 63

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 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 \frac{\coth ^{-1}\left (\sqrt{x}\right )}{x^2} \, dx &=-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{x}+\frac{1}{2} \int \frac{1}{(1-x) x^{3/2}} \, dx\\ &=-\frac{1}{\sqrt{x}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{x}+\frac{1}{2} \int \frac{1}{(1-x) \sqrt{x}} \, dx\\ &=-\frac{1}{\sqrt{x}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{x}+\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{\sqrt{x}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{x}+\tanh ^{-1}\left (\sqrt{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0240027, size = 45, normalized size = 1.8 \[ -\frac{1}{\sqrt{x}}-\frac{1}{2} \log \left (1-\sqrt{x}\right )+\frac{1}{2} \log \left (\sqrt{x}+1\right )-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Sqrt[x]]/x^2,x]

[Out]

-(1/Sqrt[x]) - ArcCoth[Sqrt[x]]/x - Log[1 - Sqrt[x]]/2 + Log[1 + Sqrt[x]]/2

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Maple [A]  time = 0.038, size = 32, normalized size = 1.3 \begin{align*} -{\frac{1}{x}{\rm arccoth} \left (\sqrt{x}\right )}-{\frac{1}{\sqrt{x}}}-{\frac{1}{2}\ln \left ( -1+\sqrt{x} \right ) }+{\frac{1}{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^2,x)

[Out]

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

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Maxima [A]  time = 0.969192, size = 42, normalized size = 1.68 \begin{align*} -\frac{\operatorname{arcoth}\left (\sqrt{x}\right )}{x} - \frac{1}{\sqrt{x}} + \frac{1}{2} \, \log \left (\sqrt{x} + 1\right ) - \frac{1}{2} \, \log \left (\sqrt{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.57107, size = 84, normalized size = 3.36 \begin{align*} \frac{{\left (x - 1\right )} \log \left (\frac{x + 2 \, \sqrt{x} + 1}{x - 1}\right ) - 2 \, \sqrt{x}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 3.66259, size = 92, normalized size = 3.68 \begin{align*} \frac{x^{\frac{5}{2}} \operatorname{acoth}{\left (\sqrt{x} \right )}}{x^{\frac{5}{2}} - x^{\frac{3}{2}}} - \frac{2 x^{\frac{3}{2}} \operatorname{acoth}{\left (\sqrt{x} \right )}}{x^{\frac{5}{2}} - x^{\frac{3}{2}}} + \frac{\sqrt{x} \operatorname{acoth}{\left (\sqrt{x} \right )}}{x^{\frac{5}{2}} - x^{\frac{3}{2}}} - \frac{x^{2}}{x^{\frac{5}{2}} - x^{\frac{3}{2}}} + \frac{x}{x^{\frac{5}{2}} - x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**(5/2)*acoth(sqrt(x))/(x**(5/2) - x**(3/2)) - 2*x**(3/2)*acoth(sqrt(x))/(x**(5/2) - x**(3/2)) + sqrt(x)*acot
h(sqrt(x))/(x**(5/2) - x**(3/2)) - x**2/(x**(5/2) - x**(3/2)) + x/(x**(5/2) - x**(3/2))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (\sqrt{x}\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(sqrt(x))/x^2, x)