3.10 \(\int \frac {\coth ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=47 \[ \frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (1-a^2 x^2\right )-\frac {\coth ^{-1}(a x)}{3 x^3}-\frac {a}{6 x^2} \]

[Out]

-1/6*a/x^2-1/3*arccoth(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 = 0.03, antiderivative size = 47, 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 = {5917, 266, 44} \[ -\frac {1}{6} a^3 \log \left (1-a^2 x^2\right )+\frac {1}{3} a^3 \log (x)-\frac {a}{6 x^2}-\frac {\coth ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 266

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 5917

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
oth[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCoth[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(a x)}{x^4} \, dx &=-\frac {\coth ^{-1}(a x)}{3 x^3}+\frac {1}{3} a \int \frac {1}{x^3 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {\coth ^{-1}(a x)}{3 x^3}+\frac {1}{6} a \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {\coth ^{-1}(a x)}{3 x^3}+\frac {1}{6} a \operatorname {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 {\coth ^{-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 = 47, normalized size = 1.00 \[ \frac {1}{3} a^3 \log (x)-\frac {1}{6} a^3 \log \left (1-a^2 x^2\right )-\frac {\coth ^{-1}(a x)}{3 x^3}-\frac {a}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.04, size = 50, normalized size = 1.06 \[ -\frac {a^{3} x^{3} \log \left (a^{2} x^{2} - 1\right ) - 2 \, a^{3} x^{3} \log \relax (x) + a x + \log \left (\frac {a x + 1}{a x - 1}\right )}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(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 + log((a*x + 1)/(a*x - 1)))/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (a x\right )}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(a*x)/x^4, x)

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maple [A]  time = 0.04, size = 48, normalized size = 1.02 \[ -\frac {\mathrm {arccoth}\left (a x \right )}{3 x^{3}}-\frac {a}{6 x^{2}}+\frac {a^{3} \ln \left (a x \right )}{3}-\frac {a^{3} \ln \left (a x -1\right )}{6}-\frac {a^{3} \ln \left (a x +1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 40, normalized size = 0.85 \[ -\frac {1}{6} \, {\left (a^{2} \log \left (a^{2} x^{2} - 1\right ) - a^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} a - \frac {\operatorname {arcoth}\left (a x\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(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*arccoth(a*x)/x^3

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mupad [B]  time = 1.18, size = 39, normalized size = 0.83 \[ \frac {a^3\,\ln \relax (x)}{3}-\frac {\frac {\mathrm {acoth}\left (a\,x\right )}{3}+\frac {a\,x}{6}}{x^3}-\frac {a^3\,\ln \left (a^2\,x^2-1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.74, size = 46, normalized size = 0.98 \[ \frac {a^{3} \log {\relax (x )}}{3} - \frac {a^{3} \log {\left (a x + 1 \right )}}{3} + \frac {a^{3} \operatorname {acoth}{\left (a x \right )}}{3} - \frac {a}{6 x^{2}} - \frac {\operatorname {acoth}{\left (a x \right )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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