\(\int x^5 \cot ^{-1}(a x^2) \, dx\) [74]

Optimal result

Integrand size = 10, antiderivative size = 41 \[ \int x^5 \cot ^{-1}\left (a x^2\right ) \, dx=\frac {x^4}{12 a}+\frac {1}{6} x^6 \cot ^{-1}\left (a x^2\right )-\frac {\log \left (1+a^2 x^4\right )}{12 a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4947, 272, 45} \[ \int x^5 \cot ^{-1}\left (a x^2\right ) \, dx=-\frac {\log \left (a^2 x^4+1\right )}{12 a^3}+\frac {x^4}{12 a}+\frac {1}{6} x^6 \cot ^{-1}\left (a x^2\right ) \]

[In]

Int[x^5*ArcCot[a*x^2],x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} x^6 \cot ^{-1}\left (a x^2\right )+\frac {1}{3} a \int \frac {x^7}{1+a^2 x^4} \, dx \\ & = \frac {1}{6} x^6 \cot ^{-1}\left (a x^2\right )+\frac {1}{12} a \text {Subst}\left (\int \frac {x}{1+a^2 x} \, dx,x,x^4\right ) \\ & = \frac {1}{6} x^6 \cot ^{-1}\left (a x^2\right )+\frac {1}{12} a \text {Subst}\left (\int \left (\frac {1}{a^2}-\frac {1}{a^2 \left (1+a^2 x\right )}\right ) \, dx,x,x^4\right ) \\ & = \frac {x^4}{12 a}+\frac {1}{6} x^6 \cot ^{-1}\left (a x^2\right )-\frac {\log \left (1+a^2 x^4\right )}{12 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int x^5 \cot ^{-1}\left (a x^2\right ) \, dx=\frac {x^4}{12 a}+\frac {1}{6} x^6 \cot ^{-1}\left (a x^2\right )-\frac {\log \left (1+a^2 x^4\right )}{12 a^3} \]

[In]

Integrate[x^5*ArcCot[a*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(-\frac {-2 x^{6} \operatorname {arccot}\left (a \,x^{2}\right ) a^{3}-a^{2} x^{4}+\ln \left (a^{2} x^{4}+1\right )}{12 a^{3}}\)	\(39\)
default	\(\frac {x^{6} \operatorname {arccot}\left (a \,x^{2}\right )}{6}+\frac {a \left (\frac {x^{4}}{4 a^{2}}-\frac {\ln \left (a^{2} x^{4}+1\right )}{4 a^{4}}\right )}{3}\)	\(40\)
parts	\(\frac {x^{6} \operatorname {arccot}\left (a \,x^{2}\right )}{6}+\frac {a \left (\frac {x^{4}}{4 a^{2}}-\frac {\ln \left (a^{2} x^{4}+1\right )}{4 a^{4}}\right )}{3}\)	\(40\)
risch	\(\frac {i x^{6} \ln \left (i a \,x^{2}+1\right )}{12}-\frac {i x^{6} \ln \left (-i a \,x^{2}+1\right )}{12}+\frac {\pi \,x^{6}}{12}+\frac {x^{4}}{12 a}-\frac {\ln \left (-a^{2} x^{4}-1\right )}{12 a^{3}}\)	\(64\)

[In]

int(x^5*arccot(a*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int x^5 \cot ^{-1}\left (a x^2\right ) \, dx=\frac {2 \, a^{3} x^{6} \operatorname {arccot}\left (a x^{2}\right ) + a^{2} x^{4} - \log \left (a^{2} x^{4} + 1\right )}{12 \, a^{3}} \]

[In]

integrate(x^5*arccot(a*x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int x^5 \cot ^{-1}\left (a x^2\right ) \, dx=\begin {cases} \frac {x^{6} \operatorname {acot}{\left (a x^{2} \right )}}{6} + \frac {x^{4}}{12 a} - \frac {\log {\left (a^{2} x^{4} + 1 \right )}}{12 a^{3}} & \text {for}\: a \neq 0 \\\frac {\pi x^{6}}{12} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5*acot(a*x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int x^5 \cot ^{-1}\left (a x^2\right ) \, dx=\frac {1}{6} \, x^{6} \operatorname {arccot}\left (a x^{2}\right ) + \frac {1}{12} \, {\left (\frac {x^{4}}{a^{2}} - \frac {\log \left (a^{2} x^{4} + 1\right )}{a^{4}}\right )} a \]

[In]

integrate(x^5*arccot(a*x^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int x^5 \cot ^{-1}\left (a x^2\right ) \, dx=\frac {1}{6} \, x^{6} \arctan \left (\frac {1}{a x^{2}}\right ) + \frac {1}{12} \, {\left (\frac {x^{4}}{a^{2}} - \frac {\log \left (a^{2} x^{4} + 1\right )}{a^{4}}\right )} a \]

[In]

integrate(x^5*arccot(a*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int x^5 \cot ^{-1}\left (a x^2\right ) \, dx=\frac {x^6\,\mathrm {acot}\left (a\,x^2\right )}{6}-\frac {\ln \left (a^2\,x^4+1\right )}{12\,a^3}+\frac {x^4}{12\,a} \]

[In]

int(x^5*acot(a*x^2),x)

[Out]

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