3.1.83 \(\int x^2 \coth ^{-1}(\sqrt {x}) \, dx\) [83]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6038, 52, 65, 212} \begin {gather*} \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 ) \end {gather*}

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 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^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} \text {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*}

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Mathematica [A]
time = 0.02, size = 59, normalized size = 1.16 \begin {gather*} \frac {1}{90} \left (30 \sqrt {x}+10 x^{3/2}+6 x^{5/2}+30 x^3 \coth ^{-1}\left (\sqrt {x}\right )+15 \log \left (1-\sqrt {x}\right )-15 \log \left (1+\sqrt {x}\right )\right ) \end {gather*}

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 = 0.03, size = 42, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {x^{3} \mathrm {arccoth}\left (\sqrt {x}\right )}{3}+\frac {x^{\frac {5}{2}}}{15}+\frac {x^{\frac {3}{2}}}{9}+\frac {\sqrt {x}}{3}+\frac {\ln \left (\sqrt {x}-1\right )}{6}-\frac {\ln \left (\sqrt {x}+1\right )}{6}\)	\(42\)
default	\(\frac {x^{3} \mathrm {arccoth}\left (\sqrt {x}\right )}{3}+\frac {x^{\frac {5}{2}}}{15}+\frac {x^{\frac {3}{2}}}{9}+\frac {\sqrt {x}}{3}+\frac {\ln \left (\sqrt {x}-1\right )}{6}-\frac {\ln \left (\sqrt {x}+1\right )}{6}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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(x^(1/2)-1)-1/6*ln(x^(1/2)+1)

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Maxima [A]
time = 0.25, size = 41, normalized size = 0.80 \begin {gather*} \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 {gather*}

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 = 0.36, size = 38, normalized size = 0.75 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {acoth}{\left (\sqrt {x} \right )}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (31) = 62\).
time = 0.39, size = 164, normalized size = 3.22 \begin {gather*} \frac {2 \, {\left (\frac {45 \, {\left (\sqrt {x} + 1\right )}^{4}}{{\left (\sqrt {x} - 1\right )}^{4}} - \frac {90 \, {\left (\sqrt {x} + 1\right )}^{3}}{{\left (\sqrt {x} - 1\right )}^{3}} + \frac {140 \, {\left (\sqrt {x} + 1\right )}^{2}}{{\left (\sqrt {x} - 1\right )}^{2}} - \frac {70 \, {\left (\sqrt {x} + 1\right )}}{\sqrt {x} - 1} + 23\right )}}{45 \, {\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1\right )}^{5}} + \frac {2 \, {\left (\frac {3 \, {\left (\sqrt {x} + 1\right )}^{5}}{{\left (\sqrt {x} - 1\right )}^{5}} + \frac {10 \, {\left (\sqrt {x} + 1\right )}^{3}}{{\left (\sqrt {x} - 1\right )}^{3}} + \frac {3 \, {\left (\sqrt {x} + 1\right )}}{\sqrt {x} - 1}\right )} \log \left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1}\right )}{3 \, {\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1\right )}^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(45*(sqrt(x) + 1)^4/(sqrt(x) - 1)^4 - 90*(sqrt(x) + 1)^3/(sqrt(x) - 1)^3 + 140*(sqrt(x) + 1)^2/(sqrt(x) -
1)^2 - 70*(sqrt(x) + 1)/(sqrt(x) - 1) + 23)/((sqrt(x) + 1)/(sqrt(x) - 1) - 1)^5 + 2/3*(3*(sqrt(x) + 1)^5/(sqr
t(x) - 1)^5 + 10*(sqrt(x) + 1)^3/(sqrt(x) - 1)^3 + 3*(sqrt(x) + 1)/(sqrt(x) - 1))*log((sqrt(x) + 1)/(sqrt(x) -
1))/((sqrt(x) + 1)/(sqrt(x) - 1) - 1)^6

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Mupad [B]
time = 1.30, size = 31, normalized size = 0.61 \begin {gather*} \frac {x^3\,\mathrm {acoth}\left (\sqrt {x}\right )}{3}-\frac {\mathrm {acoth}\left (\sqrt {x}\right )}{3}+\frac {\sqrt {x}}{3}+\frac {x^{3/2}}{9}+\frac {x^{5/2}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3*acoth(x^(1/2)))/3 - acoth(x^(1/2))/3 + x^(1/2)/3 + x^(3/2)/9 + x^(5/2)/15

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