3.76 \(\int x \cot ^{-1}(a x^2) \, dx\)

Optimal. Leaf size=31 \[ \frac {\log \left (a^2 x^4+1\right )}{4 a}+\frac {1}{2} x^2 \cot ^{-1}\left (a x^2\right ) \]

[Out]

1/2*x^2*arccot(a*x^2)+1/4*ln(a^2*x^4+1)/a

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5034, 260} \[ \frac {\log \left (a^2 x^4+1\right )}{4 a}+\frac {1}{2} x^2 \cot ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*ArcCot[a*x^2])/2 + Log[1 + a^2*x^4]/(4*a)

Rule 260

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 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 (a x^2\right ) \, dx &=\frac {1}{2} x^2 \cot ^{-1}\left (a x^2\right )+a \int \frac {x^3}{1+a^2 x^4} \, dx\\ &=\frac {1}{2} x^2 \cot ^{-1}\left (a x^2\right )+\frac {\log \left (1+a^2 x^4\right )}{4 a}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.00 \[ \frac {\log \left (a^2 x^4+1\right )}{4 a}+\frac {1}{2} x^2 \cot ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*ArcCot[a*x^2])/2 + Log[1 + a^2*x^4]/(4*a)

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fricas [A]  time = 0.98, size = 28, normalized size = 0.90 \[ \frac {2 \, a x^{2} \operatorname {arccot}\left (a x^{2}\right ) + \log \left (a^{2} x^{4} + 1\right )}{4 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*x^2*arccot(a*x^2) + log(a^2*x^4 + 1))/a

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giac [A]  time = 0.13, size = 47, normalized size = 1.52 \[ \frac {1}{4} \, {\left (\frac {2 \, x^{2} \arctan \left (\frac {1}{a x^{2}}\right )}{a} + \frac {\log \left (\frac {1}{a^{2} x^{4}} + 1\right )}{a^{2}} - \frac {\log \left (\frac {1}{a^{2} x^{4}}\right )}{a^{2}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*x^2*arctan(1/(a*x^2))/a + log(1/(a^2*x^4) + 1)/a^2 - log(1/(a^2*x^4))/a^2)*a

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maple [A]  time = 0.04, size = 28, normalized size = 0.90 \[ \frac {x^{2} \mathrm {arccot}\left (a \,x^{2}\right )}{2}+\frac {\ln \left (a^{2} x^{4}+1\right )}{4 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccot(a*x^2),x)

[Out]

1/2*x^2*arccot(a*x^2)+1/4*ln(a^2*x^4+1)/a

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maxima [A]  time = 0.32, size = 28, normalized size = 0.90 \[ \frac {2 \, a x^{2} \operatorname {arccot}\left (a x^{2}\right ) + \log \left (a^{2} x^{4} + 1\right )}{4 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*x^2*arccot(a*x^2) + log(a^2*x^4 + 1))/a

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mupad [B]  time = 0.60, size = 27, normalized size = 0.87 \[ \frac {x^2\,\mathrm {acot}\left (a\,x^2\right )}{2}+\frac {\ln \left (a^2\,x^4+1\right )}{4\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acot(a*x^2),x)

[Out]

(x^2*acot(a*x^2))/2 + log(a^2*x^4 + 1)/(4*a)

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sympy [A]  time = 0.58, size = 31, normalized size = 1.00 \[ \begin {cases} \frac {x^{2} \operatorname {acot}{\left (a x^{2} \right )}}{2} + \frac {\log {\left (a^{2} x^{4} + 1 \right )}}{4 a} & \text {for}\: a \neq 0 \\\frac {\pi x^{2}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acot(a*x**2),x)

[Out]

Piecewise((x**2*acot(a*x**2)/2 + log(a**2*x**4 + 1)/(4*a), Ne(a, 0)), (pi*x**2/4, True))

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