3.233 \(\int \frac{\tanh ^{-1}(a x)}{x^3 (1-a^2 x^2)} \, dx\)

Optimal. Leaf size=84 \[ -\frac{1}{2} a^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{2} a^2 \tanh ^{-1}(a x)^2+\frac{1}{2} a^2 \tanh ^{-1}(a x)+a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)}{2 x^2}-\frac{a}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.154584, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {5982, 5916, 325, 206, 5988, 5932, 2447} \[ -\frac{1}{2} a^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{2} a^2 \tanh ^{-1}(a x)^2+\frac{1}{2} a^2 \tanh ^{-1}(a x)+a^2 \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)}{2 x^2}-\frac{a}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(x^3*(1 - a^2*x^2)),x]

[Out]

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

Rule 5982

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[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])

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^3 \left (1-a^2 x^2\right )} \, dx &=a^2 \int \frac{\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx+\int \frac{\tanh ^{-1}(a x)}{x^3} \, dx\\ &=-\frac{\tanh ^{-1}(a x)}{2 x^2}+\frac{1}{2} a^2 \tanh ^{-1}(a x)^2+\frac{1}{2} a \int \frac{1}{x^2 \left (1-a^2 x^2\right )} \, dx+a^2 \int \frac{\tanh ^{-1}(a x)}{x (1+a x)} \, dx\\ &=-\frac{a}{2 x}-\frac{\tanh ^{-1}(a x)}{2 x^2}+\frac{1}{2} a^2 \tanh ^{-1}(a x)^2+a^2 \tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )+\frac{1}{2} a^3 \int \frac{1}{1-a^2 x^2} \, dx-a^3 \int \frac{\log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac{a}{2 x}+\frac{1}{2} a^2 \tanh ^{-1}(a x)-\frac{\tanh ^{-1}(a x)}{2 x^2}+\frac{1}{2} a^2 \tanh ^{-1}(a x)^2+a^2 \tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )-\frac{1}{2} a^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )\\ \end{align*}

Mathematica [A]  time = 0.258635, size = 60, normalized size = 0.71 \[ -\frac{1}{2} a^2 \left (\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x) \left (-\frac{1}{a^2 x^2}+\tanh ^{-1}(a x)+2 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )+1\right )+\frac{1}{a x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]/(x^3*(1 - a^2*x^2)),x]

[Out]

-(a^2*(1/(a*x) - ArcTanh[a*x]*(1 - 1/(a^2*x^2) + 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 = 0.058, size = 209, normalized size = 2.5 \begin{align*} -{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{2}}-{\frac{{\it Artanh} \left ( ax \right ) }{2\,{x}^{2}}}+{a}^{2}{\it Artanh} \left ( ax \right ) \ln \left ( ax \right ) -{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{2}}-{\frac{a}{2\,x}}-{\frac{{a}^{2}\ln \left ( ax-1 \right ) }{4}}+{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{4}}-{\frac{{a}^{2} \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{8}}+{\frac{{a}^{2}}{2}{\it dilog} \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{{a}^{2}\ln \left ( ax-1 \right ) }{4}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{{a}^{2}\ln \left ( ax+1 \right ) }{4}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{{a}^{2}}{4}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{{a}^{2} \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{8}}-{\frac{{a}^{2}{\it dilog} \left ( ax \right ) }{2}}-{\frac{{a}^{2}{\it dilog} \left ( ax+1 \right ) }{2}}-{\frac{{a}^{2}\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^3/(-a^2*x^2+1),x)

[Out]

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

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Maxima [B]  time = 0.969706, size = 219, normalized size = 2.61 \begin{align*} \frac{1}{8} \,{\left (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 - 4 \,{\left (\log \left (a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-a x\right )\right )} a + 4 \,{\left (\log \left (-a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (a x\right )\right )} a + 2 \, a \log \left (a x + 1\right ) - 2 \, a \log \left (a x - 1\right ) + \frac{a x \log \left (a x + 1\right )^{2} - 2 \, a x \log \left (a x + 1\right ) \log \left (a x - 1\right ) - a x \log \left (a x - 1\right )^{2} - 4}{x}\right )} a - \frac{1}{2} \,{\left (a^{2} \log \left (a^{2} x^{2} - 1\right ) - a^{2} \log \left (x^{2}\right ) + \frac{1}{x^{2}}\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^3/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

1/8*(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 + 2*a*log(a*x + 1) - 2*a*log(a*x - 1) + (a*x*log(a*x + 1)^2 - 2*a*x*l
og(a*x + 1)*log(a*x - 1) - a*x*log(a*x - 1)^2 - 4)/x)*a - 1/2*(a^2*log(a^2*x^2 - 1) - a^2*log(x^2) + 1/x^2)*ar
ctanh(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^{5} - x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-arctanh(a*x)/(a^2*x^5 - x^3), 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^{5} - x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/x**3/(-a**2*x**2+1),x)

[Out]

-Integral(atanh(a*x)/(a**2*x**5 - x**3), 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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-arctanh(a*x)/((a^2*x^2 - 1)*x^3), x)