3.85 \(\int x^2 \cot ^{-1}(\sqrt{x}) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0124204, 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 = {5034, 50, 63, 203} \[ \frac{x^{5/2}}{15}-\frac{x^{3/2}}{9}+\frac{1}{3} x^3 \cot ^{-1}\left (\sqrt{x}\right )+\frac{\sqrt{x}}{3}-\frac{1}{3} \tan ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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])

Rubi steps

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

Mathematica [A]  time = 0.0146515, size = 40, normalized size = 0.78 \[ \frac{1}{45} \left (\left (3 x^2-5 x+15\right ) \sqrt{x}+15 x^3 \cot ^{-1}\left (\sqrt{x}\right )-15 \tan ^{-1}\left (\sqrt{x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(15 - 5*x + 3*x^2) + 15*x^3*ArcCot[Sqrt[x]] - 15*ArcTan[Sqrt[x]])/45

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Maple [A]  time = 0.023, size = 32, normalized size = 0.6 \begin{align*} -{\frac{1}{9}{x}^{{\frac{3}{2}}}}+{\frac{1}{15}{x}^{{\frac{5}{2}}}}+{\frac{{x}^{3}}{3}{\rm arccot} \left (\sqrt{x}\right )}-{\frac{1}{3}\arctan \left ( \sqrt{x} \right ) }+{\frac{1}{3}\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45556, size = 42, normalized size = 0.82 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{arccot}\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}{3} \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccot(sqrt(x)) + 1/15*x^(5/2) - 1/9*x^(3/2) + 1/3*sqrt(x) - 1/3*arctan(sqrt(x))

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Fricas [A]  time = 2.22925, size = 88, normalized size = 1.73 \begin{align*} \frac{1}{3} \,{\left (x^{3} + 1\right )} \operatorname{arccot}\left (\sqrt{x}\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*arccot(x^(1/2)),x, algorithm="fricas")

[Out]

1/3*(x^3 + 1)*arccot(sqrt(x)) + 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{acot}{\left (\sqrt{x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*acot(sqrt(x)), x)

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Giac [A]  time = 1.10455, size = 42, normalized size = 0.82 \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (\frac{1}{\sqrt{x}}\right ) + \frac{1}{15} \, x^{\frac{5}{2}} - \frac{1}{9} \, x^{\frac{3}{2}} + \frac{1}{3} \, \sqrt{x} - \frac{1}{3} \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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