3.14 \(\int x^3 \cot ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=80 \[ \frac{x^2}{12 a^2}-\frac{\log \left (a^2 x^2+1\right )}{3 a^4}-\frac{x \cot ^{-1}(a x)}{2 a^3}-\frac{\cot ^{-1}(a x)^2}{4 a^4}+\frac{1}{4} x^4 \cot ^{-1}(a x)^2+\frac{x^3 \cot ^{-1}(a x)}{6 a} \]

[Out]

x^2/(12*a^2) - (x*ArcCot[a*x])/(2*a^3) + (x^3*ArcCot[a*x])/(6*a) - ArcCot[a*x]^2/(4*a^4) + (x^4*ArcCot[a*x]^2)
/4 - Log[1 + a^2*x^2]/(3*a^4)

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Rubi [A]  time = 0.145659, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {4853, 4917, 266, 43, 4847, 260, 4885} \[ \frac{x^2}{12 a^2}-\frac{\log \left (a^2 x^2+1\right )}{3 a^4}-\frac{x \cot ^{-1}(a x)}{2 a^3}-\frac{\cot ^{-1}(a x)^2}{4 a^4}+\frac{1}{4} x^4 \cot ^{-1}(a x)^2+\frac{x^3 \cot ^{-1}(a x)}{6 a} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCot[a*x]^2,x]

[Out]

x^2/(12*a^2) - (x*ArcCot[a*x])/(2*a^3) + (x^3*ArcCot[a*x])/(6*a) - ArcCot[a*x]^2/(4*a^4) + (x^4*ArcCot[a*x]^2)
/4 - Log[1 + a^2*x^2]/(3*a^4)

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

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 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^3 \cot ^{-1}(a x)^2 \, dx &=\frac{1}{4} x^4 \cot ^{-1}(a x)^2+\frac{1}{2} a \int \frac{x^4 \cot ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=\frac{1}{4} x^4 \cot ^{-1}(a x)^2+\frac{\int x^2 \cot ^{-1}(a x) \, dx}{2 a}-\frac{\int \frac{x^2 \cot ^{-1}(a x)}{1+a^2 x^2} \, dx}{2 a}\\ &=\frac{x^3 \cot ^{-1}(a x)}{6 a}+\frac{1}{4} x^4 \cot ^{-1}(a x)^2+\frac{1}{6} \int \frac{x^3}{1+a^2 x^2} \, dx-\frac{\int \cot ^{-1}(a x) \, dx}{2 a^3}+\frac{\int \frac{\cot ^{-1}(a x)}{1+a^2 x^2} \, dx}{2 a^3}\\ &=-\frac{x \cot ^{-1}(a x)}{2 a^3}+\frac{x^3 \cot ^{-1}(a x)}{6 a}-\frac{\cot ^{-1}(a x)^2}{4 a^4}+\frac{1}{4} x^4 \cot ^{-1}(a x)^2+\frac{1}{12} \operatorname{Subst}\left (\int \frac{x}{1+a^2 x} \, dx,x,x^2\right )-\frac{\int \frac{x}{1+a^2 x^2} \, dx}{2 a^2}\\ &=-\frac{x \cot ^{-1}(a x)}{2 a^3}+\frac{x^3 \cot ^{-1}(a x)}{6 a}-\frac{\cot ^{-1}(a x)^2}{4 a^4}+\frac{1}{4} x^4 \cot ^{-1}(a x)^2-\frac{\log \left (1+a^2 x^2\right )}{4 a^4}+\frac{1}{12} \operatorname{Subst}\left (\int \left (\frac{1}{a^2}-\frac{1}{a^2 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac{x^2}{12 a^2}-\frac{x \cot ^{-1}(a x)}{2 a^3}+\frac{x^3 \cot ^{-1}(a x)}{6 a}-\frac{\cot ^{-1}(a x)^2}{4 a^4}+\frac{1}{4} x^4 \cot ^{-1}(a x)^2-\frac{\log \left (1+a^2 x^2\right )}{3 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0203256, size = 61, normalized size = 0.76 \[ \frac{a^2 x^2-4 \log \left (a^2 x^2+1\right )+2 a x \left (a^2 x^2-3\right ) \cot ^{-1}(a x)+3 \left (a^4 x^4-1\right ) \cot ^{-1}(a x)^2}{12 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCot[a*x]^2,x]

[Out]

(a^2*x^2 + 2*a*x*(-3 + a^2*x^2)*ArcCot[a*x] + 3*(-1 + a^4*x^4)*ArcCot[a*x]^2 - 4*Log[1 + a^2*x^2])/(12*a^4)

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Maple [A]  time = 0.051, size = 82, normalized size = 1. \begin{align*}{\frac{{x}^{4} \left ({\rm arccot} \left (ax\right ) \right ) ^{2}}{4}}+{\frac{{x}^{3}{\rm arccot} \left (ax\right )}{6\,a}}-{\frac{x{\rm arccot} \left (ax\right )}{2\,{a}^{3}}}+{\frac{{\rm arccot} \left (ax\right )\arctan \left ( ax \right ) }{2\,{a}^{4}}}+{\frac{{x}^{2}}{12\,{a}^{2}}}-{\frac{\ln \left ({a}^{2}{x}^{2}+1 \right ) }{3\,{a}^{4}}}+{\frac{ \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccot(a*x)^2,x)

[Out]

1/4*x^4*arccot(a*x)^2+1/6*x^3*arccot(a*x)/a-1/2*x*arccot(a*x)/a^3+1/2/a^4*arccot(a*x)*arctan(a*x)+1/12*x^2/a^2
-1/3*ln(a^2*x^2+1)/a^4+1/4/a^4*arctan(a*x)^2

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Maxima [A]  time = 1.54347, size = 104, normalized size = 1.3 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{arccot}\left (a x\right )^{2} + \frac{1}{6} \, a{\left (\frac{a^{2} x^{3} - 3 \, x}{a^{4}} + \frac{3 \, \arctan \left (a x\right )}{a^{5}}\right )} \operatorname{arccot}\left (a x\right ) + \frac{a^{2} x^{2} + 3 \, \arctan \left (a x\right )^{2} - 4 \, \log \left (a^{2} x^{2} + 1\right )}{12 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccot(a*x)^2 + 1/6*a*((a^2*x^3 - 3*x)/a^4 + 3*arctan(a*x)/a^5)*arccot(a*x) + 1/12*(a^2*x^2 + 3*arctan
(a*x)^2 - 4*log(a^2*x^2 + 1))/a^4

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Fricas [A]  time = 1.91668, size = 144, normalized size = 1.8 \begin{align*} \frac{a^{2} x^{2} + 3 \,{\left (a^{4} x^{4} - 1\right )} \operatorname{arccot}\left (a x\right )^{2} + 2 \,{\left (a^{3} x^{3} - 3 \, a x\right )} \operatorname{arccot}\left (a x\right ) - 4 \, \log \left (a^{2} x^{2} + 1\right )}{12 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(a^2*x^2 + 3*(a^4*x^4 - 1)*arccot(a*x)^2 + 2*(a^3*x^3 - 3*a*x)*arccot(a*x) - 4*log(a^2*x^2 + 1))/a^4

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Sympy [A]  time = 1.3269, size = 78, normalized size = 0.98 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{acot}^{2}{\left (a x \right )}}{4} + \frac{x^{3} \operatorname{acot}{\left (a x \right )}}{6 a} + \frac{x^{2}}{12 a^{2}} - \frac{x \operatorname{acot}{\left (a x \right )}}{2 a^{3}} - \frac{\log{\left (a^{2} x^{2} + 1 \right )}}{3 a^{4}} - \frac{\operatorname{acot}^{2}{\left (a x \right )}}{4 a^{4}} & \text{for}\: a \neq 0 \\\frac{\pi ^{2} x^{4}}{16} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acot(a*x)**2,x)

[Out]

Piecewise((x**4*acot(a*x)**2/4 + x**3*acot(a*x)/(6*a) + x**2/(12*a**2) - x*acot(a*x)/(2*a**3) - log(a**2*x**2
+ 1)/(3*a**4) - acot(a*x)**2/(4*a**4), Ne(a, 0)), (pi**2*x**4/16, True))

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Giac [A]  time = 1.13697, size = 146, normalized size = 1.82 \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (\frac{1}{a x}\right )^{2} + \frac{4 \, a^{3} i x^{3} \log \left (\frac{a x - i}{a x + i}\right ) + 4 \, a^{2} x^{2} - 12 \, a i x \log \left (\frac{a x - i}{a x + i}\right ) + 3 \, \log \left (\frac{a x - i}{a x + i}\right )^{2} - 16 \, \log \left (a^{2} x^{2} + 1\right )}{48 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arctan(1/(a*x))^2 + 1/48*(4*a^3*i*x^3*log((a*x - i)/(a*x + i)) + 4*a^2*x^2 - 12*a*i*x*log((a*x - i)/(a
*x + i)) + 3*log((a*x - i)/(a*x + i))^2 - 16*log(a^2*x^2 + 1))/a^4