3.88 \(\int \frac{\coth ^{-1}(\sqrt{x})}{x^3} \, dx\)

Optimal. Leaf size=42 \[ -\frac{1}{6 x^{3/2}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{2 x^2}-\frac{1}{2 \sqrt{x}}+\frac{1}{2} \tanh ^{-1}\left (\sqrt{x}\right ) \]

[Out]

-1/(6*x^(3/2)) - 1/(2*Sqrt[x]) - ArcCoth[Sqrt[x]]/(2*x^2) + ArcTanh[Sqrt[x]]/2

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Rubi [A]  time = 0.0147932, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6098, 51, 63, 206} \[ -\frac{1}{6 x^{3/2}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{2 x^2}-\frac{1}{2 \sqrt{x}}+\frac{1}{2} \tanh ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[Sqrt[x]]/x^3,x]

[Out]

-1/(6*x^(3/2)) - 1/(2*Sqrt[x]) - ArcCoth[Sqrt[x]]/(2*x^2) + ArcTanh[Sqrt[x]]/2

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^3} \, dx &=-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{1}{4} \int \frac{1}{(1-x) x^{5/2}} \, dx\\ &=-\frac{1}{6 x^{3/2}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{1}{4} \int \frac{1}{(1-x) x^{3/2}} \, dx\\ &=-\frac{1}{6 x^{3/2}}-\frac{1}{2 \sqrt{x}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{1}{4} \int \frac{1}{(1-x) \sqrt{x}} \, dx\\ &=-\frac{1}{6 x^{3/2}}-\frac{1}{2 \sqrt{x}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{6 x^{3/2}}-\frac{1}{2 \sqrt{x}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{2 x^2}+\frac{1}{2} \tanh ^{-1}\left (\sqrt{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0191363, size = 58, normalized size = 1.38 \[ -\frac{1}{6 x^{3/2}}-\frac{\coth ^{-1}\left (\sqrt{x}\right )}{2 x^2}-\frac{1}{2 \sqrt{x}}-\frac{1}{4} \log \left (1-\sqrt{x}\right )+\frac{1}{4} \log \left (\sqrt{x}+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[Sqrt[x]]/x^3,x]

[Out]

-1/(6*x^(3/2)) - 1/(2*Sqrt[x]) - ArcCoth[Sqrt[x]]/(2*x^2) - Log[1 - Sqrt[x]]/4 + Log[1 + Sqrt[x]]/4

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Maple [A]  time = 0.035, size = 37, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{x}^{2}}{\rm arccoth} \left (\sqrt{x}\right )}-{\frac{1}{4}\ln \left ( -1+\sqrt{x} \right ) }-{\frac{1}{6}{x}^{-{\frac{3}{2}}}}-{\frac{1}{2}{\frac{1}{\sqrt{x}}}}+{\frac{1}{4}\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^3,x)

[Out]

-1/2*arccoth(x^(1/2))/x^2-1/4*ln(-1+x^(1/2))-1/6/x^(3/2)-1/2/x^(1/2)+1/4*ln(1+x^(1/2))

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Maxima [A]  time = 0.985373, size = 49, normalized size = 1.17 \begin{align*} -\frac{3 \, x + 1}{6 \, x^{\frac{3}{2}}} - \frac{\operatorname{arcoth}\left (\sqrt{x}\right )}{2 \, x^{2}} + \frac{1}{4} \, \log \left (\sqrt{x} + 1\right ) - \frac{1}{4} \, \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^3,x, algorithm="maxima")

[Out]

-1/6*(3*x + 1)/x^(3/2) - 1/2*arccoth(sqrt(x))/x^2 + 1/4*log(sqrt(x) + 1) - 1/4*log(sqrt(x) - 1)

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Fricas [A]  time = 1.59093, size = 107, normalized size = 2.55 \begin{align*} \frac{3 \,{\left (x^{2} - 1\right )} \log \left (\frac{x + 2 \, \sqrt{x} + 1}{x - 1}\right ) - 2 \,{\left (3 \, x + 1\right )} \sqrt{x}}{12 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(x^(1/2))/x^3,x, algorithm="fricas")

[Out]

1/12*(3*(x^2 - 1)*log((x + 2*sqrt(x) + 1)/(x - 1)) - 2*(3*x + 1)*sqrt(x))/x^2

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Sympy [B]  time = 9.8067, size = 160, normalized size = 3.81 \begin{align*} \frac{3 x^{\frac{7}{2}} \operatorname{acoth}{\left (\sqrt{x} \right )}}{6 x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} - \frac{3 x^{\frac{5}{2}} \operatorname{acoth}{\left (\sqrt{x} \right )}}{6 x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} - \frac{3 x^{\frac{3}{2}} \operatorname{acoth}{\left (\sqrt{x} \right )}}{6 x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} + \frac{3 \sqrt{x} \operatorname{acoth}{\left (\sqrt{x} \right )}}{6 x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} - \frac{3 x^{3}}{6 x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} + \frac{2 x^{2}}{6 x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} + \frac{x}{6 x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(x**(1/2))/x**3,x)

[Out]

3*x**(7/2)*acoth(sqrt(x))/(6*x**(7/2) - 6*x**(5/2)) - 3*x**(5/2)*acoth(sqrt(x))/(6*x**(7/2) - 6*x**(5/2)) - 3*
x**(3/2)*acoth(sqrt(x))/(6*x**(7/2) - 6*x**(5/2)) + 3*sqrt(x)*acoth(sqrt(x))/(6*x**(7/2) - 6*x**(5/2)) - 3*x**
3/(6*x**(7/2) - 6*x**(5/2)) + 2*x**2/(6*x**(7/2) - 6*x**(5/2)) + x/(6*x**(7/2) - 6*x**(5/2))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (\sqrt{x}\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(x^(1/2))/x^3,x, algorithm="giac")

[Out]

integrate(arccoth(sqrt(x))/x^3, x)