3.19 \(\int \frac{\coth ^{-1}(a x)^2}{x^2} \, dx\)

Optimal. Leaf size=55 \[ -a \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+a \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{x}+2 a \log \left (2-\frac{2}{a x+1}\right ) \coth ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.108329, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5917, 5989, 5933, 2447} \[ -a \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+a \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{x}+2 a \log \left (2-\frac{2}{a x+1}\right ) \coth ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]^2/x^2,x]

[Out]

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

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 5989

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcCoth[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcCoth[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]

Rule 5933

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

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{\coth ^{-1}(a x)^2}{x^2} \, dx &=-\frac{\coth ^{-1}(a x)^2}{x}+(2 a) \int \frac{\coth ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx\\ &=a \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{x}+(2 a) \int \frac{\coth ^{-1}(a x)}{x (1+a x)} \, dx\\ &=a \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{x}+2 a \coth ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )-\left (2 a^2\right ) \int \frac{\log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=a \coth ^{-1}(a x)^2-\frac{\coth ^{-1}(a x)^2}{x}+2 a \coth ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )-a \text{Li}_2\left (-1+\frac{2}{1+a x}\right )\\ \end{align*}

Mathematica [A]  time = 0.100025, size = 49, normalized size = 0.89 \[ -a \text{PolyLog}\left (2,-e^{-2 \coth ^{-1}(a x)}\right )+\frac{(a x-1) \coth ^{-1}(a x)^2}{x}+2 a \coth ^{-1}(a x) \log \left (e^{-2 \coth ^{-1}(a x)}+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCoth[a*x]^2/x^2,x]

[Out]

((-1 + a*x)*ArcCoth[a*x]^2)/x + 2*a*ArcCoth[a*x]*Log[1 + E^(-2*ArcCoth[a*x])] - a*PolyLog[2, -E^(-2*ArcCoth[a*
x])]

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Maple [B]  time = 0.056, size = 159, normalized size = 2.9 \begin{align*} -{\frac{ \left ({\rm arccoth} \left (ax\right ) \right ) ^{2}}{x}}-a{\rm arccoth} \left (ax\right )\ln \left ( ax-1 \right ) +2\,a{\rm arccoth} \left (ax\right )\ln \left ( ax \right ) -a{\rm arccoth} \left (ax\right )\ln \left ( ax+1 \right ) -{\frac{a \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{4}}+a{\it dilog} \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) +{\frac{a\ln \left ( ax-1 \right ) }{2}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{a}{2}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{a\ln \left ( ax+1 \right ) }{2}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{a \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{4}}-a{\it dilog} \left ( ax \right ) -a{\it dilog} \left ( ax+1 \right ) -a\ln \left ( ax \right ) \ln \left ( ax+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)^2/x^2,x)

[Out]

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

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Maxima [B]  time = 0.981925, size = 197, normalized size = 3.58 \begin{align*} \frac{1}{4} \, a^{2}{\left (\frac{\log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) - \log \left (a x - 1\right )^{2}}{a} + \frac{4 \,{\left (\log \left (a x - 1\right ) \log \left (\frac{1}{2} \, a x + \frac{1}{2}\right ) +{\rm Li}_2\left (-\frac{1}{2} \, a x + \frac{1}{2}\right )\right )}}{a} - \frac{4 \,{\left (\log \left (a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-a x\right )\right )}}{a} + \frac{4 \,{\left (\log \left (-a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (a x\right )\right )}}{a}\right )} - a{\left (\log \left (a^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} \operatorname{arcoth}\left (a x\right ) - \frac{\operatorname{arcoth}\left (a x\right )^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcoth}\left (a x\right )^{2}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccoth(a*x)^2/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acoth}^{2}{\left (a x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)**2/x**2,x)

[Out]

Integral(acoth(a*x)**2/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcoth}\left (a x\right )^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(a*x)^2/x^2, x)