\(\int \frac {\cot ^{-1}(a x^2)}{x^3} \, dx\) [78]

Optimal result

Integrand size = 10, antiderivative size = 34 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=-\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \log (x)+\frac {1}{4} a \log \left (1+a^2 x^4\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4947, 272, 36, 29, 31} \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=\frac {1}{4} a \log \left (a^2 x^4+1\right )-\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \log (x) \]

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \int \frac {1}{x \left (1+a^2 x^4\right )} \, dx \\ & = -\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-\frac {1}{4} a \text {Subst}\left (\int \frac {1}{x \left (1+a^2 x\right )} \, dx,x,x^4\right ) \\ & = -\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-\frac {1}{4} a \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^4\right )+\frac {1}{4} a^3 \text {Subst}\left (\int \frac {1}{1+a^2 x} \, dx,x,x^4\right ) \\ & = -\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \log (x)+\frac {1}{4} a \log \left (1+a^2 x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=-\frac {\cot ^{-1}\left (a x^2\right )}{2 x^2}-a \log (x)+\frac {1}{4} a \log \left (1+a^2 x^4\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {\operatorname {arccot}\left (a \,x^{2}\right )}{2 x^{2}}-a \left (\ln \left (x \right )-\frac {\ln \left (a^{2} x^{4}+1\right )}{4}\right )\)	\(31\)
parts	\(-\frac {\operatorname {arccot}\left (a \,x^{2}\right )}{2 x^{2}}-a \left (\ln \left (x \right )-\frac {\ln \left (a^{2} x^{4}+1\right )}{4}\right )\)	\(31\)
parallelrisch	\(-\frac {4 a \ln \left (x \right ) x^{2}-a \ln \left (a^{2} x^{4}+1\right ) x^{2}+2 \,\operatorname {arccot}\left (a \,x^{2}\right )}{4 x^{2}}\)	\(39\)
risch	\(-\frac {i \ln \left (i a \,x^{2}+1\right )}{4 x^{2}}-\frac {4 a \ln \left (x \right ) x^{2}-a \ln \left (a^{2} x^{4}+1\right ) x^{2}-i \ln \left (-i a \,x^{2}+1\right )+\pi }{4 x^{2}}\)	\(62\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=\frac {a x^{2} \log \left (a^{2} x^{4} + 1\right ) - 4 \, a x^{2} \log \left (x\right ) - 2 \, \operatorname {arccot}\left (a x^{2}\right )}{4 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=- a \log {\left (x \right )} + \frac {a \log {\left (a^{2} x^{4} + 1 \right )}}{4} - \frac {\operatorname {acot}{\left (a x^{2} \right )}}{2 x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=\frac {1}{4} \, a {\left (\log \left (a^{2} x^{4} + 1\right ) - \log \left (x^{4}\right )\right )} - \frac {\operatorname {arccot}\left (a x^{2}\right )}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=-\frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {1}{a x^{2}}\right )}{a x^{2}} - \log \left (\frac {1}{a^{2} x^{4}} + 1\right )\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {\cot ^{-1}\left (a x^2\right )}{x^3} \, dx=\frac {a\,\ln \left (-a^2\,x^4-1\right )}{4}-\frac {\mathrm {acot}\left (a\,x^2\right )}{2\,x^2}-a\,\ln \left (x\right ) \]

[In]

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

[Out]

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