3.1.10 \(\int \frac {\cot ^{-1}(a x)}{x^4} \, dx\) [10]

Optimal. Leaf size=46 \[ \frac {a}{6 x^2}-\frac {\cot ^{-1}(a x)}{3 x^3}+\frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (1+a^2 x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4947, 272, 46} \begin {gather*} \frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (a^2 x^2+1\right )-\frac {\cot ^{-1}(a x)}{3 x^3}+\frac {a}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCot[a*x]/x^4,x]

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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*} \int \frac {\cot ^{-1}(a x)}{x^4} \, dx &=-\frac {\cot ^{-1}(a x)}{3 x^3}-\frac {1}{3} a \int \frac {1}{x^3 \left (1+a^2 x^2\right )} \, dx\\ &=-\frac {\cot ^{-1}(a x)}{3 x^3}-\frac {1}{6} a \text {Subst}\left (\int \frac {1}{x^2 \left (1+a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {\cot ^{-1}(a x)}{3 x^3}-\frac {1}{6} a \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {a^2}{x}+\frac {a^4}{1+a^2 x}\right ) \, dx,x,x^2\right )\\ &=\frac {a}{6 x^2}-\frac {\cot ^{-1}(a x)}{3 x^3}+\frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (1+a^2 x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 44, normalized size = 0.96 \begin {gather*} -\frac {\cot ^{-1}(a x)}{3 x^3}-\frac {1}{6} a \left (-\frac {1}{x^2}-2 a^2 \log (x)+a^2 \log \left (1+a^2 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[a*x]/x^4,x]

[Out]

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

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Maple [A]
time = 0.05, size = 44, normalized size = 0.96

method	result	size
derivativedivides	\(a^{3} \left (-\frac {\mathrm {arccot}\left (a x \right )}{3 a^{3} x^{3}}+\frac {1}{6 a^{2} x^{2}}+\frac {\ln \left (a x \right )}{3}-\frac {\ln \left (a^{2} x^{2}+1\right )}{6}\right )\)	\(44\)
default	\(a^{3} \left (-\frac {\mathrm {arccot}\left (a x \right )}{3 a^{3} x^{3}}+\frac {1}{6 a^{2} x^{2}}+\frac {\ln \left (a x \right )}{3}-\frac {\ln \left (a^{2} x^{2}+1\right )}{6}\right )\)	\(44\)
risch	\(-\frac {i \ln \left (i a x +1\right )}{6 x^{3}}-\frac {-2 a^{3} \ln \left (x \right ) x^{3}+a^{3} \ln \left (-a^{2} x^{2}-1\right ) x^{3}-i \ln \left (-i a x +1\right )-a x +\pi }{6 x^{3}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(a*x)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 42, normalized size = 0.91 \begin {gather*} -\frac {1}{6} \, {\left (a^{2} \log \left (a^{2} x^{2} + 1\right ) - a^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} a - \frac {\operatorname {arccot}\left (a x\right )}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(a*x)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.59, size = 43, normalized size = 0.93 \begin {gather*} -\frac {a^{3} x^{3} \log \left (a^{2} x^{2} + 1\right ) - 2 \, a^{3} x^{3} \log \left (x\right ) - a x + 2 \, \operatorname {arccot}\left (a x\right )}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(a*x)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.25, size = 39, normalized size = 0.85 \begin {gather*} \frac {a^{3} \log {\left (x \right )}}{3} - \frac {a^{3} \log {\left (a^{2} x^{2} + 1 \right )}}{6} + \frac {a}{6 x^{2}} - \frac {\operatorname {acot}{\left (a x \right )}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(a*x)/x**4,x)

[Out]

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

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Giac [A]
time = 0.41, size = 44, normalized size = 0.96 \begin {gather*} \frac {1}{6} \, {\left (a^{2} {\left (\frac {1}{a^{2} x^{2}} - \log \left (\frac {1}{a^{2} x^{2}} + 1\right )\right )} - \frac {2 \, \arctan \left (\frac {1}{a x}\right )}{a x^{3}}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(a*x)/x^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.85, size = 58, normalized size = 1.26 \begin {gather*} \left \{\begin {array}{cl} -\frac {\pi }{6\,x^3} & \text {\ if\ \ }a=0\\ \frac {a^4\,\ln \left (x\right )-\frac {a^4\,\ln \left (a^2\,x^2+1\right )}{2}+\frac {a^2}{2\,x^2}}{3\,a}-\frac {\mathrm {acot}\left (a\,x\right )}{3\,x^3} & \text {\ if\ \ }a\neq 0 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a*x)/x^4,x)

[Out]

piecewise(a == 0, -pi/(6*x^3), a ~= 0, (a^4*log(x) - (a^4*log(a^2*x^2 + 1))/2 + a^2/(2*x^2))/(3*a) - acot(a*x)
/(3*x^3))

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