3.1.87 \(\int \frac {\coth ^{-1}(\sqrt {x})}{x^2} \, dx\) [87]

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

[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 = {6038, 53, 65, 212} \begin {gather*} -\frac {1}{\sqrt {x}}+\tanh ^{-1}\left (\sqrt {x}\right )-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 53

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 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 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*} \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}+\text {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 \begin {gather*} -\frac {1}{\sqrt {x}}-\frac {\coth ^{-1}\left (\sqrt {x}\right )}{x}-\frac {1}{2} \log \left (1-\sqrt {x}\right )+\frac {1}{2} \log \left (1+\sqrt {x}\right ) \end {gather*}

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 = 0.04, size = 32, normalized size = 1.28

method	result	size
derivativedivides	\(-\frac {\mathrm {arccoth}\left (\sqrt {x}\right )}{x}-\frac {1}{\sqrt {x}}-\frac {\ln \left (\sqrt {x}-1\right )}{2}+\frac {\ln \left (\sqrt {x}+1\right )}{2}\)	\(32\)
default	\(-\frac {\mathrm {arccoth}\left (\sqrt {x}\right )}{x}-\frac {1}{\sqrt {x}}-\frac {\ln \left (\sqrt {x}-1\right )}{2}+\frac {\ln \left (\sqrt {x}+1\right )}{2}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 31, normalized size = 1.24 \begin {gather*} -\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 {gather*}

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 = 0.34, size = 30, normalized size = 1.20 \begin {gather*} \frac {{\left (x - 1\right )} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) - 2 \, \sqrt {x}}{2 \, x} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (20) = 40\).
time = 0.59, size = 92, normalized size = 3.68 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (19) = 38\).
time = 0.39, size = 65, normalized size = 2.60 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(x^(1/2))/x^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)

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

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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