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

Optimal. Leaf size=53 \[ \frac{\log \left (a^2 x^2+1\right )}{2 a^2}+\frac{\cot ^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \cot ^{-1}(a x)^2+\frac{x \cot ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.0719063, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4853, 4917, 4847, 260, 4885} \[ \frac{\log \left (a^2 x^2+1\right )}{2 a^2}+\frac{\cot ^{-1}(a x)^2}{2 a^2}+\frac{1}{2} x^2 \cot ^{-1}(a x)^2+\frac{x \cot ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4853

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
t[c*x])^p)/(d*(m + 1)), x] + Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCot[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4917

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]

Rule 4847

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

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 4885

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]

Rubi steps

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

Mathematica [A]  time = 0.0131157, size = 42, normalized size = 0.79 \[ \frac{\log \left (a^2 x^2+1\right )+\left (a^2 x^2+1\right ) \cot ^{-1}(a x)^2+2 a x \cot ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.048, size = 61, normalized size = 1.2 \begin{align*}{\frac{{x}^{2} \left ({\rm arccot} \left (ax\right ) \right ) ^{2}}{2}}-{\frac{{\rm arccot} \left (ax\right )\arctan \left ( ax \right ) }{{a}^{2}}}+{\frac{x{\rm arccot} \left (ax\right )}{a}}+{\frac{\ln \left ({a}^{2}{x}^{2}+1 \right ) }{2\,{a}^{2}}}-{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,{a}^{2}}} \end{align*}

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

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Maxima [A]  time = 1.53683, size = 77, normalized size = 1.45 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arccot}\left (a x\right )^{2} + a{\left (\frac{x}{a^{2}} - \frac{\arctan \left (a x\right )}{a^{3}}\right )} \operatorname{arccot}\left (a x\right ) - \frac{\arctan \left (a x\right )^{2} - \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.74942, size = 105, normalized size = 1.98 \begin{align*} \frac{2 \, a x \operatorname{arccot}\left (a x\right ) +{\left (a^{2} x^{2} + 1\right )} \operatorname{arccot}\left (a x\right )^{2} + \log \left (a^{2} x^{2} + 1\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.645709, size = 54, normalized size = 1.02 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{acot}^{2}{\left (a x \right )}}{2} + \frac{x \operatorname{acot}{\left (a x \right )}}{a} + \frac{\log{\left (a^{2} x^{2} + 1 \right )}}{2 a^{2}} + \frac{\operatorname{acot}^{2}{\left (a x \right )}}{2 a^{2}} & \text{for}\: a \neq 0 \\\frac{\pi ^{2} x^{2}}{8} & \text{otherwise} \end{cases} \end{align*}

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 + x*acot(a*x)/a + log(a**2*x**2 + 1)/(2*a**2) + acot(a*x)**2/(2*a**2), Ne(a, 0)
), (pi**2*x**2/8, True))

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Giac [A]  time = 1.12203, size = 101, normalized size = 1.91 \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (\frac{1}{a x}\right )^{2} + \frac{4 \, a i x \log \left (\frac{a x - i}{a x + i}\right ) - \log \left (\frac{a x - i}{a x + i}\right )^{2} + 4 \, \log \left (a^{2} x^{2} + 1\right )}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arctan(1/(a*x))^2 + 1/8*(4*a*i*x*log((a*x - i)/(a*x + i)) - log((a*x - i)/(a*x + i))^2 + 4*log(a^2*x^2
 + 1))/a^2