3.1.84 \(\int x \coth ^{-1}(\sqrt {x}) \, dx\) [84]

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

[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 = {6038, 52, 65, 212} \begin {gather*} \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 ) \end {gather*}

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 52

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 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 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} \text {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 \begin {gather*} \frac {1}{12} \left (6 \sqrt {x}+2 x^{3/2}+6 x^2 \coth ^{-1}\left (\sqrt {x}\right )+3 \log \left (1-\sqrt {x}\right )-3 \log \left (1+\sqrt {x}\right )\right ) \end {gather*}

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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Maple [A]
time = 0.03, size = 37, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {x^{2} \mathrm {arccoth}\left (\sqrt {x}\right )}{2}+\frac {x^{\frac {3}{2}}}{6}+\frac {\sqrt {x}}{2}+\frac {\ln \left (\sqrt {x}-1\right )}{4}-\frac {\ln \left (\sqrt {x}+1\right )}{4}\)	\(37\)
default	\(\frac {x^{2} \mathrm {arccoth}\left (\sqrt {x}\right )}{2}+\frac {x^{\frac {3}{2}}}{6}+\frac {\sqrt {x}}{2}+\frac {\ln \left (\sqrt {x}-1\right )}{4}-\frac {\ln \left (\sqrt {x}+1\right )}{4}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 36, normalized size = 0.86 \begin {gather*} \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 ) \end {gather*}

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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Fricas [A]
time = 0.36, size = 31, normalized size = 0.74 \begin {gather*} \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} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {acoth}{\left (\sqrt {x} \right )}\, dx \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (26) = 52\).
time = 0.39, size = 114, normalized size = 2.71 \begin {gather*} \frac {2 \, {\left (\frac {3 \, {\left (\sqrt {x} + 1\right )}^{2}}{{\left (\sqrt {x} - 1\right )}^{2}} - \frac {3 \, {\left (\sqrt {x} + 1\right )}}{\sqrt {x} - 1} + 2\right )}}{3 \, {\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1\right )}^{3}} + \frac {2 \, {\left (\frac {{\left (\sqrt {x} + 1\right )}^{3}}{{\left (\sqrt {x} - 1\right )}^{3}} + \frac {\sqrt {x} + 1}{\sqrt {x} - 1}\right )} \log \left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1}\right )}{{\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.26, size = 26, normalized size = 0.62 \begin {gather*} \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} \end {gather*}

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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