3.1.37 \(\int \frac {x^4 \cot ^{-1}(x)}{1+x^2} \, dx\) [37]

Optimal. Leaf size=40 \[ \frac {x^2}{6}-x \cot ^{-1}(x)+\frac {1}{3} x^3 \cot ^{-1}(x)-\frac {1}{2} \cot ^{-1}(x)^2-\frac {2}{3} \log \left (1+x^2\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {5037, 4947, 272, 45, 4931, 266, 5005} \begin {gather*} \frac {1}{3} x^3 \cot ^{-1}(x)+\frac {x^2}{6}-\frac {2}{3} \log \left (x^2+1\right )-x \cot ^{-1}(x)-\frac {1}{2} \cot ^{-1}(x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*ArcCot[x])/(1 + x^2),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 32, normalized size = 0.80 \begin {gather*} \frac {1}{6} \left (x^2+2 x \left (-3+x^2\right ) \cot ^{-1}(x)-3 \cot ^{-1}(x)^2-4 \log \left (1+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*ArcCot[x])/(1 + x^2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.16, size = 38, normalized size = 0.95

method	result	size
default	\(\frac {x^{3} \mathrm {arccot}\left (x \right )}{3}-x \,\mathrm {arccot}\left (x \right )+\mathrm {arccot}\left (x \right ) \arctan \left (x \right )+\frac {x^{2}}{6}-\frac {2 \ln \left (x^{2}+1\right )}{3}+\frac {\arctan \left (x \right )^{2}}{2}\)	\(38\)
risch	\(\frac {\ln \left (i x +1\right )^{2}}{8}+\left (\frac {i x^{3}}{6}-\frac {i x}{2}-\frac {\ln \left (-i x +1\right )}{4}\right ) \ln \left (i x +1\right )+\frac {\ln \left (-i x +1\right )^{2}}{8}-\frac {i x^{3} \ln \left (-i x +1\right )}{6}+\frac {i \ln \left (-i x +1\right ) x}{2}+\frac {\pi \,x^{3}}{6}-\frac {\pi x}{2}+\frac {x^{2}}{6}+\frac {\pi \arctan \left (x \right )}{2}-\frac {2 \ln \left (x^{2}+1\right )}{3}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arccot(x)/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 35, normalized size = 0.88 \begin {gather*} \frac {1}{6} \, x^{2} + \frac {1}{3} \, {\left (x^{3} - 3 \, x + 3 \, \arctan \left (x\right )\right )} \operatorname {arccot}\left (x\right ) + \frac {1}{2} \, \arctan \left (x\right )^{2} - \frac {2}{3} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccot(x)/(x^2+1),x, algorithm="maxima")

[Out]

1/6*x^2 + 1/3*(x^3 - 3*x + 3*arctan(x))*arccot(x) + 1/2*arctan(x)^2 - 2/3*log(x^2 + 1)

________________________________________________________________________________________

Fricas [A]
time = 1.68, size = 31, normalized size = 0.78 \begin {gather*} \frac {1}{6} \, x^{2} + \frac {1}{3} \, {\left (x^{3} - 3 \, x\right )} \operatorname {arccot}\left (x\right ) - \frac {1}{2} \, \operatorname {arccot}\left (x\right )^{2} - \frac {2}{3} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arccot(x)/(x^2+1),x, algorithm="fricas")

[Out]

1/6*x^2 + 1/3*(x^3 - 3*x)*arccot(x) - 1/2*arccot(x)^2 - 2/3*log(x^2 + 1)

________________________________________________________________________________________

Sympy [A]
time = 0.16, size = 34, normalized size = 0.85 \begin {gather*} \frac {x^{3} \operatorname {acot}{\left (x \right )}}{3} + \frac {x^{2}}{6} - x \operatorname {acot}{\left (x \right )} - \frac {2 \log {\left (x^{2} + 1 \right )}}{3} - \frac {\operatorname {acot}^{2}{\left (x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*acot(x)/(x**2+1),x)

[Out]

x**3*acot(x)/3 + x**2/6 - x*acot(x) - 2*log(x**2 + 1)/3 - acot(x)**2/2

________________________________________________________________________________________

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^4*arccot(x)/(x^2+1),x, algorithm="giac")

[Out]

integrate(x^4*arccot(x)/(x^2 + 1), x)

________________________________________________________________________________________

Mupad [B]
time = 0.73, size = 32, normalized size = 0.80 \begin {gather*} \frac {x^3\,\mathrm {acot}\left (x\right )}{3}-\frac {2\,\ln \left (x^2+1\right )}{3}-\frac {{\mathrm {acot}\left (x\right )}^2}{2}-x\,\mathrm {acot}\left (x\right )+\frac {x^2}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*acot(x))/(x^2 + 1),x)

[Out]

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

________________________________________________________________________________________