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]

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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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])

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 )}{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*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 1.80 \[ -\frac {1}{\sqrt {x}}-\frac {1}{2} \log \left (1-\sqrt {x}\right )+\frac {1}{2} \log \left (\sqrt {x}+1\right )-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{x} \]

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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fricas [A]  time = 0.49, size = 30, normalized size = 1.20 \[ \frac {{\left (x - 1\right )} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) - 2 \, \sqrt {x}}{2 \, x} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (\sqrt {x}\right )}{x^{2}}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 32, normalized size = 1.28 \[ -\frac {\mathrm {arccoth}\left (\sqrt {x}\right )}{x}-\frac {1}{\sqrt {x}}-\frac {\ln \left (-1+\sqrt {x}\right )}{2}+\frac {\ln \left (1+\sqrt {x}\right )}{2} \]

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 = 0.30, size = 31, normalized size = 1.24 \[ -\frac {\operatorname {arcoth}\left (\sqrt {x}\right )}{x} - \frac {1}{\sqrt {x}} + \frac {1}{2} \, \log \left (\sqrt {x} + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {x} - 1\right ) \]

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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mupad [B]  time = 1.27, size = 18, normalized size = 0.72 \[ \mathrm {atanh}\left (\sqrt {x}\right )-\frac {\mathrm {acoth}\left (\sqrt {x}\right )+\sqrt {x}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.12, size = 92, normalized size = 3.68 \[ \frac {x^{\frac {5}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{x^{\frac {5}{2}} - x^{\frac {3}{2}}} - \frac {2 x^{\frac {3}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{x^{\frac {5}{2}} - x^{\frac {3}{2}}} + \frac {\sqrt {x} \operatorname {acoth}{\left (\sqrt {x} \right )}}{x^{\frac {5}{2}} - x^{\frac {3}{2}}} - \frac {x^{2}}{x^{\frac {5}{2}} - x^{\frac {3}{2}}} + \frac {x}{x^{\frac {5}{2}} - x^{\frac {3}{2}}} \]

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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