3.86 \(\int x \cot ^{-1}(\sqrt {x}) \, dx\)

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

[Out]

1/6*x^(3/2)+1/2*x^2*arccot(x^(1/2))+1/2*arctan(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 = {5034, 50, 63, 203} \[ \frac {x^{3/2}}{6}+\frac {1}{2} x^2 \cot ^{-1}\left (\sqrt {x}\right )-\frac {\sqrt {x}}{2}+\frac {1}{2} \tan ^{-1}\left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCot[Sqrt[x]],x]

[Out]

-Sqrt[x]/2 + x^(3/2)/6 + (x^2*ArcCot[Sqrt[x]])/2 + ArcTan[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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 5034

Int[((a_.) + ArcCot[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCot
[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 \cot ^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{2} x^2 \cot ^{-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 \cot ^{-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 \cot ^{-1}\left (\sqrt {x}\right )+\frac {1}{4} \int \frac {1}{\sqrt {x} (1+x)} \, dx\\ &=-\frac {\sqrt {x}}{2}+\frac {x^{3/2}}{6}+\frac {1}{2} x^2 \cot ^{-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 \cot ^{-1}\left (\sqrt {x}\right )+\frac {1}{2} \tan ^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.79 \[ \frac {1}{6} \left (3 x^2 \cot ^{-1}\left (\sqrt {x}\right )+(x-3) \sqrt {x}+3 \tan ^{-1}\left (\sqrt {x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCot[Sqrt[x]],x]

[Out]

((-3 + x)*Sqrt[x] + 3*x^2*ArcCot[Sqrt[x]] + 3*ArcTan[Sqrt[x]])/6

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fricas [A]  time = 0.63, size = 20, normalized size = 0.48 \[ \frac {1}{2} \, {\left (x^{2} - 1\right )} \operatorname {arccot}\left (\sqrt {x}\right ) + \frac {1}{6} \, {\left (x - 3\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 28, normalized size = 0.67 \[ \frac {1}{2} \, x^{2} \arctan \left (\frac {1}{\sqrt {x}}\right ) - \frac {1}{6} \, x^{\frac {3}{2}} {\left (\frac {3}{x} - 1\right )} - \frac {1}{2} \, \arctan \left (\frac {1}{\sqrt {x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(1/sqrt(x)) - 1/6*x^(3/2)*(3/x - 1) - 1/2*arctan(1/sqrt(x))

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maple [A]  time = 0.04, size = 27, normalized size = 0.64 \[ \frac {x^{\frac {3}{2}}}{6}+\frac {x^{2} \mathrm {arccot}\left (\sqrt {x}\right )}{2}+\frac {\arctan \left (\sqrt {x}\right )}{2}-\frac {\sqrt {x}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 26, normalized size = 0.62 \[ \frac {1}{2} \, x^{2} \operatorname {arccot}\left (\sqrt {x}\right ) + \frac {1}{6} \, x^{\frac {3}{2}} - \frac {1}{2} \, \sqrt {x} + \frac {1}{2} \, \arctan \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arccot(sqrt(x)) + 1/6*x^(3/2) - 1/2*sqrt(x) + 1/2*arctan(sqrt(x))

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mupad [B]  time = 0.66, size = 26, normalized size = 0.62 \[ \frac {\mathrm {atan}\left (\sqrt {x}\right )}{2}+\frac {x^2\,\mathrm {acot}\left (\sqrt {x}\right )}{2}-\frac {\sqrt {x}}{2}+\frac {x^{3/2}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.29, size = 32, normalized size = 0.76 \[ \frac {x^{\frac {3}{2}}}{6} - \frac {\sqrt {x}}{2} + \frac {x^{2} \operatorname {acot}{\left (\sqrt {x} \right )}}{2} + \frac {\operatorname {atan}{\left (\sqrt {x} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**(3/2)/6 - sqrt(x)/2 + x**2*acot(sqrt(x))/2 + atan(sqrt(x))/2

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