3.84 \(\int x \coth ^{-1}(\sqrt {x}) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6098, 50, 63, 206} \[ \frac {x^{3/2}}{6}+\frac {1}{2} x^2 \coth ^{-1}\left (\sqrt {x}\right )+\frac {\sqrt {x}}{2}-\frac {1}{2} \tanh ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

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

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Mathematica [A]  time = 0.01, size = 52, normalized size = 1.24 \[ \frac {1}{12} \left (2 x^{3/2}+6 x^2 \coth ^{-1}\left (\sqrt {x}\right )+6 \sqrt {x}+3 \log \left (1-\sqrt {x}\right )-3 \log \left (\sqrt {x}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[x] + 2*x^(3/2) + 6*x^2*ArcCoth[Sqrt[x]] + 3*Log[1 - Sqrt[x]] - 3*Log[1 + Sqrt[x]])/12

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fricas [A]  time = 0.57, size = 31, normalized size = 0.74 \[ \frac {1}{4} \, {\left (x^{2} - 1\right )} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) + \frac {1}{6} \, {\left (x + 3\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccoth(sqrt(x)), x)

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maple [A]  time = 0.05, size = 37, normalized size = 0.88 \[ \frac {x^{2} \mathrm {arccoth}\left (\sqrt {x}\right )}{2}+\frac {x^{\frac {3}{2}}}{6}+\frac {\sqrt {x}}{2}+\frac {\ln \left (-1+\sqrt {x}\right )}{4}-\frac {\ln \left (1+\sqrt {x}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(x^(1/2)),x)

[Out]

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

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maxima [A]  time = 0.31, size = 36, normalized size = 0.86 \[ \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (\sqrt {x}\right ) + \frac {1}{6} \, x^{\frac {3}{2}} + \frac {1}{2} \, \sqrt {x} - \frac {1}{4} \, \log \left (\sqrt {x} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.26, size = 26, normalized size = 0.62 \[ \frac {x^2\,\mathrm {acoth}\left (\sqrt {x}\right )}{2}-\frac {\mathrm {acoth}\left (\sqrt {x}\right )}{2}+\frac {\sqrt {x}}{2}+\frac {x^{3/2}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*acoth(x^(1/2)))/2 - acoth(x^(1/2))/2 + x^(1/2)/2 + x^(3/2)/6

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {acoth}{\left (\sqrt {x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*acoth(sqrt(x)), x)

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