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

Optimal. Leaf size=54 \[ \frac{\text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{\tanh ^{-1}(a x)^2}{2 a^2}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.0706783, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {5984, 5918, 2402, 2315} \[ \frac{\text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{\tanh ^{-1}(a x)^2}{2 a^2}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTanh[a*x])/(1 - a^2*x^2),x]

[Out]

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

Rule 5984

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

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.0485874, size = 44, normalized size = 0.81 \[ -\frac{\text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x) \left (\tanh ^{-1}(a x)+2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )\right )}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*ArcTanh[a*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*a^2)

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Maple [B]  time = 0.046, size = 125, normalized size = 2.3 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{2\,{a}^{2}}}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{2\,{a}^{2}}}-{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{8\,{a}^{2}}}+{\frac{1}{2\,{a}^{2}}{\it dilog} \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{\ln \left ( ax-1 \right ) }{4\,{a}^{2}}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{\ln \left ( ax+1 \right ) }{4\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{1}{4\,{a}^{2}}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{8\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.967308, size = 169, normalized size = 3.13 \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^{3}} - \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^{3}}\right )} + \frac{{\left (\frac{\log \left (a x + 1\right )}{a} - \frac{\log \left (a x - 1\right )}{a}\right )} \log \left (a^{2} x^{2} - 1\right )}{4 \, a} - \frac{\operatorname{artanh}\left (a x\right ) \log \left (a^{2} x^{2} - 1\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctanh(a*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^3 - 4*(log(a*x - 1)*log(1/2*a*x + 1/
2) + dilog(-1/2*a*x + 1/2))/a^3) + 1/4*(log(a*x + 1)/a - log(a*x - 1)/a)*log(a^2*x^2 - 1)/a - 1/2*arctanh(a*x)
*log(a^2*x^2 - 1)/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x*atanh(a*x)/(a**2*x**2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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