3.231 \(\int \frac{\tanh ^{-1}(a x)}{x (1-a^2 x^2)} \, dx\)

Optimal. Leaf size=45 \[ -\frac{1}{2} \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{2} \tanh ^{-1}(a x)^2+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.0870035, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {5988, 5932, 2447} \[ -\frac{1}{2} \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{2} \tanh ^{-1}(a x)^2+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(x*(1 - a^2*x^2)),x]

[Out]

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

Rule 5988

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcTanh[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 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[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{\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx &=\frac{1}{2} \tanh ^{-1}(a x)^2+\int \frac{\tanh ^{-1}(a x)}{x (1+a x)} \, dx\\ &=\frac{1}{2} \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )-a \int \frac{\log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac{1}{2} \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )-\frac{1}{2} \text{Li}_2\left (-1+\frac{2}{1+a x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0669345, size = 42, normalized size = 0.93 \[ \frac{1}{2} \left (\tanh ^{-1}(a x) \left (\tanh ^{-1}(a x)+2 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )\right )-\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]/(x*(1 - a^2*x^2)),x]

[Out]

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

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Maple [B]  time = 0.052, size = 130, normalized size = 2.9 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{2}}+{\it Artanh} \left ( ax \right ) \ln \left ( ax \right ) -{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{2}}-{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{8}}+{\frac{1}{2}{\it dilog} \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{\ln \left ( ax-1 \right ) }{4}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{1}{4} \left ( \ln \left ( ax+1 \right ) -\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) \right ) \ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{8}}-{\frac{{\it dilog} \left ( ax \right ) }{2}}-{\frac{{\it dilog} \left ( ax+1 \right ) }{2}}-{\frac{\ln \left ( ax \right ) \ln \left ( ax+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)/x/(-a^2*x^2+1),x)

[Out]

-1/2*arctanh(a*x)*ln(a*x-1)+arctanh(a*x)*ln(a*x)-1/2*arctanh(a*x)*ln(a*x+1)-1/8*ln(a*x-1)^2+1/2*dilog(1/2+1/2*
a*x)+1/4*ln(a*x-1)*ln(1/2+1/2*a*x)-1/4*(ln(a*x+1)-ln(1/2+1/2*a*x))*ln(-1/2*a*x+1/2)+1/8*ln(a*x+1)^2-1/2*dilog(
a*x)-1/2*dilog(a*x+1)-1/2*ln(a*x)*ln(a*x+1)

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Maxima [B]  time = 0.970119, size = 178, normalized size = 3.96 \begin{align*} \frac{1}{8} \, a{\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 )} - \frac{1}{2} \,{\left (\log \left (a^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

1/8*a*((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
) - 1/2*(log(a^2*x^2 - 1) - log(x^2))*arctanh(a*x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\operatorname{artanh}\left (a x\right )}{a^{2} x^{3} - x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-arctanh(a*x)/(a^2*x^3 - x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\operatorname{atanh}{\left (a x \right )}}{a^{2} x^{3} - x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/x/(-a**2*x**2+1),x)

[Out]

-Integral(atanh(a*x)/(a**2*x**3 - x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\operatorname{artanh}\left (a x\right )}{{\left (a^{2} x^{2} - 1\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-arctanh(a*x)/((a^2*x^2 - 1)*x), x)