3.89 \(\int x^{3/2} \coth ^{-1}(\sqrt {x}) \, dx\)

Optimal. Leaf size=38 \[ \frac {2}{5} x^{5/2} \coth ^{-1}\left (\sqrt {x}\right )+\frac {x^2}{10}+\frac {x}{5}+\frac {1}{5} \log (1-x) \]

[Out]

1/5*x+1/10*x^2+2/5*x^(5/2)*arccoth(x^(1/2))+1/5*ln(1-x)

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6098, 43} \[ \frac {x^2}{10}+\frac {2}{5} x^{5/2} \coth ^{-1}\left (\sqrt {x}\right )+\frac {x}{5}+\frac {1}{5} \log (1-x) \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*ArcCoth[Sqrt[x]],x]

[Out]

x/5 + x^2/10 + (2*x^(5/2)*ArcCoth[Sqrt[x]])/5 + Log[1 - x]/5

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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^{3/2} \coth ^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {2}{5} x^{5/2} \coth ^{-1}\left (\sqrt {x}\right )-\frac {1}{5} \int \frac {x^2}{1-x} \, dx\\ &=\frac {2}{5} x^{5/2} \coth ^{-1}\left (\sqrt {x}\right )-\frac {1}{5} \int \left (-1+\frac {1}{1-x}-x\right ) \, dx\\ &=\frac {x}{5}+\frac {x^2}{10}+\frac {2}{5} x^{5/2} \coth ^{-1}\left (\sqrt {x}\right )+\frac {1}{5} \log (1-x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.82 \[ \frac {1}{10} \left (4 x^{5/2} \coth ^{-1}\left (\sqrt {x}\right )+(x+2) x+2 \log (1-x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*ArcCoth[Sqrt[x]],x]

[Out]

(x*(2 + x) + 4*x^(5/2)*ArcCoth[Sqrt[x]] + 2*Log[1 - x])/10

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fricas [A]  time = 1.27, size = 35, normalized size = 0.92 \[ \frac {1}{5} \, x^{\frac {5}{2}} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) + \frac {1}{10} \, x^{2} + \frac {1}{5} \, x + \frac {1}{5} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*arccoth(x^(1/2)),x, algorithm="fricas")

[Out]

1/5*x^(5/2)*log((x + 2*sqrt(x) + 1)/(x - 1)) + 1/10*x^2 + 1/5*x + 1/5*log(x - 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \operatorname {arcoth}\left (\sqrt {x}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*arccoth(x^(1/2)),x, algorithm="giac")

[Out]

integrate(x^(3/2)*arccoth(sqrt(x)), x)

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maple [A]  time = 0.05, size = 35, normalized size = 0.92 \[ \frac {2 x^{\frac {5}{2}} \mathrm {arccoth}\left (\sqrt {x}\right )}{5}+\frac {x^{2}}{10}+\frac {x}{5}+\frac {\ln \left (-1+\sqrt {x}\right )}{5}+\frac {\ln \left (1+\sqrt {x}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*arccoth(x^(1/2)),x)

[Out]

2/5*x^(5/2)*arccoth(x^(1/2))+1/10*x^2+1/5*x+1/5*ln(-1+x^(1/2))+1/5*ln(1+x^(1/2))

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maxima [A]  time = 0.31, size = 24, normalized size = 0.63 \[ \frac {2}{5} \, x^{\frac {5}{2}} \operatorname {arcoth}\left (\sqrt {x}\right ) + \frac {1}{10} \, x^{2} + \frac {1}{5} \, x + \frac {1}{5} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*arccoth(x^(1/2)),x, algorithm="maxima")

[Out]

2/5*x^(5/2)*arccoth(sqrt(x)) + 1/10*x^2 + 1/5*x + 1/5*log(x - 1)

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mupad [B]  time = 1.26, size = 24, normalized size = 0.63 \[ \frac {x}{5}+\frac {\ln \left (x-1\right )}{5}+\frac {2\,x^{5/2}\,\mathrm {acoth}\left (\sqrt {x}\right )}{5}+\frac {x^2}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*acoth(x^(1/2)),x)

[Out]

x/5 + log(x - 1)/5 + (2*x^(5/2)*acoth(x^(1/2)))/5 + x^2/10

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sympy [B]  time = 5.35, size = 121, normalized size = 3.18 \[ \frac {4 x^{\frac {7}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{10 x - 10} - \frac {4 x^{\frac {5}{2}} \operatorname {acoth}{\left (\sqrt {x} \right )}}{10 x - 10} + \frac {x^{3}}{10 x - 10} + \frac {x^{2}}{10 x - 10} + \frac {4 x \log {\left (\sqrt {x} + 1 \right )}}{10 x - 10} - \frac {4 x \operatorname {acoth}{\left (\sqrt {x} \right )}}{10 x - 10} - \frac {4 \log {\left (\sqrt {x} + 1 \right )}}{10 x - 10} + \frac {4 \operatorname {acoth}{\left (\sqrt {x} \right )}}{10 x - 10} - \frac {2}{10 x - 10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*acoth(x**(1/2)),x)

[Out]

4*x**(7/2)*acoth(sqrt(x))/(10*x - 10) - 4*x**(5/2)*acoth(sqrt(x))/(10*x - 10) + x**3/(10*x - 10) + x**2/(10*x
- 10) + 4*x*log(sqrt(x) + 1)/(10*x - 10) - 4*x*acoth(sqrt(x))/(10*x - 10) - 4*log(sqrt(x) + 1)/(10*x - 10) + 4
*acoth(sqrt(x))/(10*x - 10) - 2/(10*x - 10)

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