\(\int x \cot ^{-1}(a x^2) \, dx\) [76]

Optimal result

Integrand size = 8, antiderivative size = 31 \[ \int x \cot ^{-1}\left (a x^2\right ) \, dx=\frac {1}{2} x^2 \cot ^{-1}\left (a x^2\right )+\frac {\log \left (1+a^2 x^4\right )}{4 a} \]

[Out]

1/2*x^2*arccot(a*x^2)+1/4*ln(a^2*x^4+1)/a

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4947, 266} \[ \int x \cot ^{-1}\left (a x^2\right ) \, dx=\frac {\log \left (a^2 x^4+1\right )}{4 a}+\frac {1}{2} x^2 \cot ^{-1}\left (a x^2\right ) \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int x \cot ^{-1}\left (a x^2\right ) \, dx=\frac {1}{2} x^2 \cot ^{-1}\left (a x^2\right )+\frac {\log \left (1+a^2 x^4\right )}{4 a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90

method	result	size
parts	\(\frac {x^{2} \operatorname {arccot}\left (a \,x^{2}\right )}{2}+\frac {\ln \left (a^{2} x^{4}+1\right )}{4 a}\)	\(28\)
parallelrisch	\(\frac {2 \,\operatorname {arccot}\left (a \,x^{2}\right ) a \,x^{2}+\ln \left (a^{2} x^{4}+1\right )}{4 a}\)	\(29\)
derivativedivides	\(\frac {\operatorname {arccot}\left (a \,x^{2}\right ) a \,x^{2}+\frac {\ln \left (a^{2} x^{4}+1\right )}{2}}{2 a}\)	\(30\)
default	\(\frac {\operatorname {arccot}\left (a \,x^{2}\right ) a \,x^{2}+\frac {\ln \left (a^{2} x^{4}+1\right )}{2}}{2 a}\)	\(30\)
risch	\(\frac {i x^{2} \ln \left (i a \,x^{2}+1\right )}{4}-\frac {i x^{2} \ln \left (-i a \,x^{2}+1\right )}{4}+\frac {\pi \,x^{2}}{4}+\frac {\ln \left (-a^{2} x^{4}-1\right )}{4 a}\)	\(56\)

[In]

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

[Out]

1/2*x^2*arccot(a*x^2)+1/4*ln(a^2*x^4+1)/a

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int x \cot ^{-1}\left (a x^2\right ) \, dx=\frac {2 \, a x^{2} \operatorname {arccot}\left (a x^{2}\right ) + \log \left (a^{2} x^{4} + 1\right )}{4 \, a} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int x \cot ^{-1}\left (a x^2\right ) \, dx=\begin {cases} \frac {x^{2} \operatorname {acot}{\left (a x^{2} \right )}}{2} + \frac {\log {\left (a^{2} x^{4} + 1 \right )}}{4 a} & \text {for}\: a \neq 0 \\\frac {\pi x^{2}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int x \cot ^{-1}\left (a x^2\right ) \, dx=\frac {2 \, a x^{2} \operatorname {arccot}\left (a x^{2}\right ) + \log \left (a^{2} x^{4} + 1\right )}{4 \, a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int x \cot ^{-1}\left (a x^2\right ) \, dx=\frac {1}{4} \, {\left (\frac {2 \, x^{2} \arctan \left (\frac {1}{a x^{2}}\right )}{a} + \frac {\log \left (\frac {1}{a^{2} x^{4}} + 1\right )}{a^{2}} - \frac {\log \left (\frac {1}{a^{2} x^{4}}\right )}{a^{2}}\right )} a \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int x \cot ^{-1}\left (a x^2\right ) \, dx=\frac {x^2\,\mathrm {acot}\left (a\,x^2\right )}{2}+\frac {\ln \left (a^2\,x^4+1\right )}{4\,a} \]

[In]

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

[Out]

(x^2*acot(a*x^2))/2 + log(a^2*x^4 + 1)/(4*a)