3.1.39 \(\int \frac {x^2 \cot ^{-1}(x)}{1+x^2} \, dx\) [39]

Optimal. Leaf size=23 \[ x \cot ^{-1}(x)+\frac {1}{2} \cot ^{-1}(x)^2+\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

x*arccot(x)+1/2*arccot(x)^2+1/2*ln(x^2+1)

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Rubi [A]
time = 0.04, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5037, 4931, 266, 5005} \begin {gather*} \frac {1}{2} \log \left (x^2+1\right )+\frac {1}{2} \cot ^{-1}(x)^2+x \cot ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcCot[x])/(1 + x^2),x]

[Out]

x*ArcCot[x] + ArcCot[x]^2/2 + Log[1 + x^2]/2

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4931

Int[((a_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCot[c*x^n])^p, x] + Dist[b*c
*n*p, Int[x^n*((a + b*ArcCot[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 5005

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[-(a + b*ArcCot[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 5037

Int[(((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcCot[c*x])^p, x], x] - Dist[d*(f^2/e), Int[(f*x)^(m - 2)*((a + b*ArcCot[c*x])^p/
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {x^2 \cot ^{-1}(x)}{1+x^2} \, dx &=\int \cot ^{-1}(x) \, dx-\int \frac {\cot ^{-1}(x)}{1+x^2} \, dx\\ &=x \cot ^{-1}(x)+\frac {1}{2} \cot ^{-1}(x)^2+\int \frac {x}{1+x^2} \, dx\\ &=x \cot ^{-1}(x)+\frac {1}{2} \cot ^{-1}(x)^2+\frac {1}{2} \log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 23, normalized size = 1.00 \begin {gather*} x \cot ^{-1}(x)+\frac {1}{2} \cot ^{-1}(x)^2+\frac {1}{2} \log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*ArcCot[x])/(1 + x^2),x]

[Out]

x*ArcCot[x] + ArcCot[x]^2/2 + Log[1 + x^2]/2

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Maple [A]
time = 0.15, size = 26, normalized size = 1.13

method	result	size
default	\(-\mathrm {arccot}\left (x \right ) \arctan \left (x \right )+x \,\mathrm {arccot}\left (x \right )+\frac {\ln \left (x^{2}+1\right )}{2}-\frac {\arctan \left (x \right )^{2}}{2}\)	\(26\)
risch	\(-\frac {\ln \left (i x +1\right )^{2}}{8}+\left (\frac {i x}{2}+\frac {\ln \left (-i x +1\right )}{4}\right ) \ln \left (i x +1\right )-\frac {\ln \left (-i x +1\right )^{2}}{8}-\frac {i \ln \left (-i x +1\right ) x}{2}+\frac {\pi x}{2}-\frac {\pi \arctan \left (x \right )}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 24, normalized size = 1.04 \begin {gather*} {\left (x - \arctan \left (x\right )\right )} \operatorname {arccot}\left (x\right ) - \frac {1}{2} \, \arctan \left (x\right )^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.89, size = 19, normalized size = 0.83 \begin {gather*} x \operatorname {arccot}\left (x\right ) + \frac {1}{2} \, \operatorname {arccot}\left (x\right )^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccot(x) + 1/2*arccot(x)^2 + 1/2*log(x^2 + 1)

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Sympy [A]
time = 0.14, size = 19, normalized size = 0.83 \begin {gather*} x \operatorname {acot}{\left (x \right )} + \frac {\log {\left (x^{2} + 1 \right )}}{2} + \frac {\operatorname {acot}^{2}{\left (x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*acot(x) + log(x**2 + 1)/2 + acot(x)**2/2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccot(x)/(x^2 + 1), x)

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Mupad [B]
time = 0.64, size = 19, normalized size = 0.83 \begin {gather*} \frac {{\mathrm {acot}\left (x\right )}^2}{2}+x\,\mathrm {acot}\left (x\right )+\frac {\ln \left (x^2+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*acot(x))/(x^2 + 1),x)

[Out]

log(x^2 + 1)/2 + acot(x)^2/2 + x*acot(x)

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