3.83 \(\int x^2 \coth ^{-1}(\sqrt{x}) \, dx\)

Optimal. Leaf size=51 \[ \frac{x^{5/2}}{15}+\frac{x^{3/2}}{9}+\frac{1}{3} x^3 \coth ^{-1}\left (\sqrt{x}\right )+\frac{\sqrt{x}}{3}-\frac{1}{3} \tanh ^{-1}\left (\sqrt{x}\right ) \]

[Out]

Sqrt[x]/3 + x^(3/2)/9 + x^(5/2)/15 + (x^3*ArcCoth[Sqrt[x]])/3 - ArcTanh[Sqrt[x]]/3

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Rubi [A]  time = 0.0147834, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6098, 50, 63, 206} \[ \frac{x^{5/2}}{15}+\frac{x^{3/2}}{9}+\frac{1}{3} x^3 \coth ^{-1}\left (\sqrt{x}\right )+\frac{\sqrt{x}}{3}-\frac{1}{3} \tanh ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[x]/3 + x^(3/2)/9 + x^(5/2)/15 + (x^3*ArcCoth[Sqrt[x]])/3 - ArcTanh[Sqrt[x]]/3

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 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])

Rubi steps

\begin{align*} \int x^2 \coth ^{-1}\left (\sqrt{x}\right ) \, dx &=\frac{1}{3} x^3 \coth ^{-1}\left (\sqrt{x}\right )-\frac{1}{6} \int \frac{x^{5/2}}{1-x} \, dx\\ &=\frac{x^{5/2}}{15}+\frac{1}{3} x^3 \coth ^{-1}\left (\sqrt{x}\right )-\frac{1}{6} \int \frac{x^{3/2}}{1-x} \, dx\\ &=\frac{x^{3/2}}{9}+\frac{x^{5/2}}{15}+\frac{1}{3} x^3 \coth ^{-1}\left (\sqrt{x}\right )-\frac{1}{6} \int \frac{\sqrt{x}}{1-x} \, dx\\ &=\frac{\sqrt{x}}{3}+\frac{x^{3/2}}{9}+\frac{x^{5/2}}{15}+\frac{1}{3} x^3 \coth ^{-1}\left (\sqrt{x}\right )-\frac{1}{6} \int \frac{1}{(1-x) \sqrt{x}} \, dx\\ &=\frac{\sqrt{x}}{3}+\frac{x^{3/2}}{9}+\frac{x^{5/2}}{15}+\frac{1}{3} x^3 \coth ^{-1}\left (\sqrt{x}\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{x}\right )\\ &=\frac{\sqrt{x}}{3}+\frac{x^{3/2}}{9}+\frac{x^{5/2}}{15}+\frac{1}{3} x^3 \coth ^{-1}\left (\sqrt{x}\right )-\frac{1}{3} \tanh ^{-1}\left (\sqrt{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0175619, size = 59, normalized size = 1.16 \[ \frac{1}{90} \left (6 x^{5/2}+10 x^{3/2}+30 x^3 \coth ^{-1}\left (\sqrt{x}\right )+30 \sqrt{x}+15 \log \left (1-\sqrt{x}\right )-15 \log \left (\sqrt{x}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCoth[Sqrt[x]],x]

[Out]

(30*Sqrt[x] + 10*x^(3/2) + 6*x^(5/2) + 30*x^3*ArcCoth[Sqrt[x]] + 15*Log[1 - Sqrt[x]] - 15*Log[1 + Sqrt[x]])/90

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Maple [A]  time = 0.036, size = 42, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{3}{\rm arccoth} \left (\sqrt{x}\right )}+{\frac{1}{15}{x}^{{\frac{5}{2}}}}+{\frac{1}{9}{x}^{{\frac{3}{2}}}}+{\frac{1}{3}\sqrt{x}}+{\frac{1}{6}\ln \left ( -1+\sqrt{x} \right ) }-{\frac{1}{6}\ln \left ( 1+\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccoth(x^(1/2))+1/15*x^(5/2)+1/9*x^(3/2)+1/3*x^(1/2)+1/6*ln(-1+x^(1/2))-1/6*ln(1+x^(1/2))

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Maxima [A]  time = 0.961345, size = 55, normalized size = 1.08 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{arcoth}\left (\sqrt{x}\right ) + \frac{1}{15} \, x^{\frac{5}{2}} + \frac{1}{9} \, x^{\frac{3}{2}} + \frac{1}{3} \, \sqrt{x} - \frac{1}{6} \, \log \left (\sqrt{x} + 1\right ) + \frac{1}{6} \, \log \left (\sqrt{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccoth(sqrt(x)) + 1/15*x^(5/2) + 1/9*x^(3/2) + 1/3*sqrt(x) - 1/6*log(sqrt(x) + 1) + 1/6*log(sqrt(x) -
1)

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Fricas [A]  time = 1.60583, size = 111, normalized size = 2.18 \begin{align*} \frac{1}{6} \,{\left (x^{3} - 1\right )} \log \left (\frac{x + 2 \, \sqrt{x} + 1}{x - 1}\right ) + \frac{1}{45} \,{\left (3 \, x^{2} + 5 \, x + 15\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(x^3 - 1)*log((x + 2*sqrt(x) + 1)/(x - 1)) + 1/45*(3*x^2 + 5*x + 15)*sqrt(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{acoth}{\left (\sqrt{x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*acoth(sqrt(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arcoth}\left (\sqrt{x}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccoth(sqrt(x)), x)