3.1.91 \(\int x^{3/2} \cot ^{-1}(\sqrt {x}) \, dx\) [91]

Optimal. Leaf size=36 \[ -\frac {x}{5}+\frac {x^2}{10}+\frac {2}{5} x^{5/2} \cot ^{-1}\left (\sqrt {x}\right )+\frac {1}{5} \log (1+x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, 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 = {4947, 45} \begin {gather*} \frac {2}{5} x^{5/2} \cot ^{-1}\left (\sqrt {x}\right )+\frac {x^2}{10}-\frac {x}{5}+\frac {1}{5} \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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 4947

Int[((a_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCot[c*x^
n])^p/(m + 1)), x] + Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcCot[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^{3/2} \cot ^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {2}{5} x^{5/2} \cot ^{-1}\left (\sqrt {x}\right )+\frac {1}{5} \int \frac {x^2}{1+x} \, dx\\ &=\frac {2}{5} x^{5/2} \cot ^{-1}\left (\sqrt {x}\right )+\frac {1}{5} \int \left (-1+x+\frac {1}{1+x}\right ) \, dx\\ &=-\frac {x}{5}+\frac {x^2}{10}+\frac {2}{5} x^{5/2} \cot ^{-1}\left (\sqrt {x}\right )+\frac {1}{5} \log (1+x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 29, normalized size = 0.81 \begin {gather*} \frac {1}{10} \left ((-2+x) x+4 x^{5/2} \cot ^{-1}\left (\sqrt {x}\right )+2 \log (1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.01, size = 25, normalized size = 0.69

method result size
derivativedivides \(-\frac {x}{5}+\frac {x^{2}}{10}+\frac {2 x^{\frac {5}{2}} \mathrm {arccot}\left (\sqrt {x}\right )}{5}+\frac {\ln \left (1+x \right )}{5}\) \(25\)
default \(-\frac {x}{5}+\frac {x^{2}}{10}+\frac {2 x^{\frac {5}{2}} \mathrm {arccot}\left (\sqrt {x}\right )}{5}+\frac {\ln \left (1+x \right )}{5}\) \(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*arccot(x^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 24, normalized size = 0.67 \begin {gather*} \frac {2}{5} \, x^{\frac {5}{2}} \operatorname {arccot}\left (\sqrt {x}\right ) + \frac {1}{10} \, x^{2} - \frac {1}{5} \, x + \frac {1}{5} \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.37, size = 24, normalized size = 0.67 \begin {gather*} \frac {2}{5} \, x^{\frac {5}{2}} \operatorname {arccot}\left (\sqrt {x}\right ) + \frac {1}{10} \, x^{2} - \frac {1}{5} \, x + \frac {1}{5} \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (29) = 58\).
time = 1.04, size = 85, normalized size = 2.36 \begin {gather*} \frac {4 x^{\frac {7}{2}} \operatorname {acot}{\left (\sqrt {x} \right )}}{10 x + 10} + \frac {4 x^{\frac {5}{2}} \operatorname {acot}{\left (\sqrt {x} \right )}}{10 x + 10} + \frac {x^{3}}{10 x + 10} - \frac {x^{2}}{10 x + 10} + \frac {2 x \log {\left (x + 1 \right )}}{10 x + 10} + \frac {2 \log {\left (x + 1 \right )}}{10 x + 10} + \frac {2}{10 x + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 39, normalized size = 1.08 \begin {gather*} \frac {2}{5} \, x^{\frac {5}{2}} \arctan \left (\frac {1}{\sqrt {x}}\right ) - \frac {1}{10} \, x^{2} {\left (\frac {2}{x} - \frac {3}{x^{2}} - 1\right )} + \frac {1}{5} \, \log \left (x\right ) + \frac {1}{5} \, \log \left (\frac {1}{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^(5/2)*arctan(1/sqrt(x)) - 1/10*x^2*(2/x - 3/x^2 - 1) + 1/5*log(x) + 1/5*log(1/x + 1)

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Mupad [B]
time = 0.65, size = 24, normalized size = 0.67 \begin {gather*} \frac {\ln \left (x+1\right )}{5}-\frac {x}{5}+\frac {2\,x^{5/2}\,\mathrm {acot}\left (\sqrt {x}\right )}{5}+\frac {x^2}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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