3.1.14 \(\int x^3 \cot ^{-1}(a x)^2 \, dx\) [14]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4947, 5037, 272, 45, 4931, 266, 5005} \begin {gather*} -\frac {\cot ^{-1}(a x)^2}{4 a^4}-\frac {x \cot ^{-1}(a x)}{2 a^3}+\frac {x^2}{12 a^2}-\frac {\log \left (a^2 x^2+1\right )}{3 a^4}+\frac {1}{4} x^4 \cot ^{-1}(a x)^2+\frac {x^3 \cot ^{-1}(a x)}{6 a} \end {gather*}

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 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 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 272

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 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 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]

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 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} \text {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} \text {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*}

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Mathematica [A]
time = 0.01, size = 61, normalized size = 0.76 \begin {gather*} \frac {a^2 x^2+2 a x \left (-3+a^2 x^2\right ) \cot ^{-1}(a x)+3 \left (-1+a^4 x^4\right ) \cot ^{-1}(a x)^2-4 \log \left (1+a^2 x^2\right )}{12 a^4} \end {gather*}

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 = 0.10, size = 78, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {\frac {a^{4} x^{4} \mathrm {arccot}\left (a x \right )^{2}}{4}+\frac {a^{3} x^{3} \mathrm {arccot}\left (a x \right )}{6}-\frac {a x \,\mathrm {arccot}\left (a x \right )}{2}+\frac {\mathrm {arccot}\left (a x \right ) \arctan \left (a x \right )}{2}+\frac {a^{2} x^{2}}{12}-\frac {\ln \left (a^{2} x^{2}+1\right )}{3}+\frac {\arctan \left (a x \right )^{2}}{4}}{a^{4}}\)	\(78\)
default	\(\frac {\frac {a^{4} x^{4} \mathrm {arccot}\left (a x \right )^{2}}{4}+\frac {a^{3} x^{3} \mathrm {arccot}\left (a x \right )}{6}-\frac {a x \,\mathrm {arccot}\left (a x \right )}{2}+\frac {\mathrm {arccot}\left (a x \right ) \arctan \left (a x \right )}{2}+\frac {a^{2} x^{2}}{12}-\frac {\ln \left (a^{2} x^{2}+1\right )}{3}+\frac {\arctan \left (a x \right )^{2}}{4}}{a^{4}}\)	\(78\)
risch	\(-\frac {\left (a^{4} x^{4}-1\right ) \ln \left (i a x +1\right )^{2}}{16 a^{4}}+\frac {\left (3 i \pi \,a^{4} x^{4}+3 x^{4} \ln \left (-i a x +1\right ) a^{4}+2 i a^{3} x^{3}-6 i a x -3 \ln \left (-i a x +1\right )\right ) \ln \left (i a x +1\right )}{24 a^{4}}-\frac {i \pi \,x^{4} \ln \left (-i a x +1\right )}{8}+\frac {\pi ^{2} x^{4}}{16}-\frac {x^{4} \ln \left (-i a x +1\right )^{2}}{16}-\frac {i x^{3} \ln \left (-i a x +1\right )}{12 a}+\frac {\pi \,x^{3}}{12 a}+\frac {i x \ln \left (-i a x +1\right )}{4 a^{3}}+\frac {x^{2}}{12 a^{2}}-\frac {\pi x}{4 a^{3}}+\frac {\pi \arctan \left (a x \right )}{4 a^{4}}+\frac {\ln \left (-i a x +1\right )^{2}}{16 a^{4}}-\frac {\ln \left (a^{2} x^{2}+1\right )}{3 a^{4}}\)	\(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 77, normalized size = 0.96 \begin {gather*} \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 {gather*}

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 = 2.92, size = 60, normalized size = 0.75 \begin {gather*} \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 {gather*}

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 = 0.28, size = 78, normalized size = 0.98 \begin {gather*} \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 {gather*}

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

[Out]

integrate(x^3*arccot(a*x)^2, x)

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Mupad [B]
time = 0.20, size = 66, normalized size = 0.82 \begin {gather*} \frac {x^4\,{\mathrm {acot}\left (a\,x\right )}^2}{4}-\frac {\frac {\ln \left (a^2\,x^2+1\right )}{3}-\frac {a^2\,x^2}{12}+\frac {{\mathrm {acot}\left (a\,x\right )}^2}{4}-\frac {a^3\,x^3\,\mathrm {acot}\left (a\,x\right )}{6}+\frac {a\,x\,\mathrm {acot}\left (a\,x\right )}{2}}{a^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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