3.9 \(\int \frac{\coth ^{-1}(a x)}{x^3} \, dx\)

Optimal. Leaf size=31 \[ \frac{1}{2} a^2 \tanh ^{-1}(a x)-\frac{\coth ^{-1}(a x)}{2 x^2}-\frac{a}{2 x} \]

[Out]

-a/(2*x) - ArcCoth[a*x]/(2*x^2) + (a^2*ArcTanh[a*x])/2

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Rubi [A]  time = 0.0163453, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5917, 325, 206} \[ \frac{1}{2} a^2 \tanh ^{-1}(a x)-\frac{\coth ^{-1}(a x)}{2 x^2}-\frac{a}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]/x^3,x]

[Out]

-a/(2*x) - ArcCoth[a*x]/(2*x^2) + (a^2*ArcTanh[a*x])/2

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]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0082258, size = 47, normalized size = 1.52 \[ -\frac{1}{4} a^2 \log (1-a x)+\frac{1}{4} a^2 \log (a x+1)-\frac{\coth ^{-1}(a x)}{2 x^2}-\frac{a}{2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x]/x^3,x]

[Out]

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

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Maple [A]  time = 0.039, size = 39, normalized size = 1.3 \begin{align*} -{\frac{{\rm arccoth} \left (ax\right )}{2\,{x}^{2}}}-{\frac{a}{2\,x}}-{\frac{{a}^{2}\ln \left ( ax-1 \right ) }{4}}+{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)/x^3,x)

[Out]

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

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Maxima [A]  time = 0.972809, size = 49, normalized size = 1.58 \begin{align*} \frac{1}{4} \,{\left (a \log \left (a x + 1\right ) - a \log \left (a x - 1\right ) - \frac{2}{x}\right )} a - \frac{\operatorname{arcoth}\left (a x\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.51869, size = 80, normalized size = 2.58 \begin{align*} -\frac{2 \, a x -{\left (a^{2} x^{2} - 1\right )} \log \left (\frac{a x + 1}{a x - 1}\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.57081, size = 24, normalized size = 0.77 \begin{align*} \frac{a^{2} \operatorname{acoth}{\left (a x \right )}}{2} - \frac{a}{2 x} - \frac{\operatorname{acoth}{\left (a x \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)/x**3,x)

[Out]

a**2*acoth(a*x)/2 - a/(2*x) - acoth(a*x)/(2*x**2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (a x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(a*x)/x^3, x)