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

Optimal. Leaf size=205 \[ \frac{3 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{3 \text{PolyLog}\left (4,1-\frac{2}{1-a x}\right )}{4 a^4}+\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{\tanh ^{-1}(a x)^3}{2 a^4}-\frac{3 x \tanh ^{-1}(a x)^2}{2 a^3}-\frac{3 \tanh ^{-1}(a x)^2}{2 a^4}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^4}+\frac{3 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4} \]

[Out]

(-3*ArcTanh[a*x]^2)/(2*a^4) - (3*x*ArcTanh[a*x]^2)/(2*a^3) + ArcTanh[a*x]^3/(2*a^4) - (x^2*ArcTanh[a*x]^3)/(2*
a^2) - ArcTanh[a*x]^4/(4*a^4) + (3*ArcTanh[a*x]*Log[2/(1 - a*x)])/a^4 + (ArcTanh[a*x]^3*Log[2/(1 - a*x)])/a^4
+ (3*PolyLog[2, 1 - 2/(1 - a*x)])/(2*a^4) + (3*ArcTanh[a*x]^2*PolyLog[2, 1 - 2/(1 - a*x)])/(2*a^4) - (3*ArcTan
h[a*x]*PolyLog[3, 1 - 2/(1 - a*x)])/(2*a^4) + (3*PolyLog[4, 1 - 2/(1 - a*x)])/(4*a^4)

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Rubi [A]  time = 0.469477, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5980, 5916, 5910, 5984, 5918, 2402, 2315, 5948, 6058, 6062, 6610} \[ \frac{3 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{3 \text{PolyLog}\left (4,1-\frac{2}{1-a x}\right )}{4 a^4}+\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{\tanh ^{-1}(a x)^3}{2 a^4}-\frac{3 x \tanh ^{-1}(a x)^2}{2 a^3}-\frac{3 \tanh ^{-1}(a x)^2}{2 a^4}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^4}+\frac{3 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[a*x]^2)/(2*a^4) - (3*x*ArcTanh[a*x]^2)/(2*a^3) + ArcTanh[a*x]^3/(2*a^4) - (x^2*ArcTanh[a*x]^3)/(2*
a^2) - ArcTanh[a*x]^4/(4*a^4) + (3*ArcTanh[a*x]*Log[2/(1 - a*x)])/a^4 + (ArcTanh[a*x]^3*Log[2/(1 - a*x)])/a^4
+ (3*PolyLog[2, 1 - 2/(1 - a*x)])/(2*a^4) + (3*ArcTanh[a*x]^2*PolyLog[2, 1 - 2/(1 - a*x)])/(2*a^4) - (3*ArcTan
h[a*x]*PolyLog[3, 1 - 2/(1 - a*x)])/(2*a^4) + (3*PolyLog[4, 1 - 2/(1 - a*x)])/(4*a^4)

Rule 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, 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] && GtQ[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 5910

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

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]

Rule 5948

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

Rule 6058

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

Rule 6062

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{x^3 \tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx &=-\frac{\int x \tanh ^{-1}(a x)^3 \, dx}{a^2}+\frac{\int \frac{x \tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx}{a^2}\\ &=-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{\int \frac{\tanh ^{-1}(a x)^3}{1-a x} \, dx}{a^3}+\frac{3 \int \frac{x^2 \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{2 a}\\ &=-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^4}-\frac{3 \int \tanh ^{-1}(a x)^2 \, dx}{2 a^3}+\frac{3 \int \frac{\tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{2 a^3}-\frac{3 \int \frac{\tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}\\ &=-\frac{3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac{\tanh ^{-1}(a x)^3}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{3 \int \frac{\tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}+\frac{3 \int \frac{x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{a^2}\\ &=-\frac{3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac{3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac{\tanh ^{-1}(a x)^3}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{3 \int \frac{\text{Li}_3\left (1-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{2 a^3}+\frac{3 \int \frac{\tanh ^{-1}(a x)}{1-a x} \, dx}{a^3}\\ &=-\frac{3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac{3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac{\tanh ^{-1}(a x)^3}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{3 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{3 \text{Li}_4\left (1-\frac{2}{1-a x}\right )}{4 a^4}-\frac{3 \int \frac{\log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}\\ &=-\frac{3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac{3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac{\tanh ^{-1}(a x)^3}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{3 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{3 \text{Li}_4\left (1-\frac{2}{1-a x}\right )}{4 a^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-a x}\right )}{a^4}\\ &=-\frac{3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac{3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac{\tanh ^{-1}(a x)^3}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^4}+\frac{3 \tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{3 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{3 \text{Li}_4\left (1-\frac{2}{1-a x}\right )}{4 a^4}\\ \end{align*}

Mathematica [A]  time = 0.2687, size = 142, normalized size = 0.69 \[ -\frac{6 \tanh ^{-1}(a x) \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+6 \left (\tanh ^{-1}(a x)^2+1\right ) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text{PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )-2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-\tanh ^{-1}(a x)^4+6 a x \tanh ^{-1}(a x)^2-6 \tanh ^{-1}(a x)^2-4 \tanh ^{-1}(a x)^3 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-12 \tanh ^{-1}(a x) \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{4 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(-6*ArcTanh[a*x]^2 + 6*a*x*ArcTanh[a*x]^2 - 2*(1 - a^2*x^2)*ArcTanh[a*x]^3 - ArcTanh[a*x]^4 - 12*ArcTanh[a*x]
*Log[1 + E^(-2*ArcTanh[a*x])] - 4*ArcTanh[a*x]^3*Log[1 + E^(-2*ArcTanh[a*x])] + 6*(1 + ArcTanh[a*x]^2)*PolyLog
[2, -E^(-2*ArcTanh[a*x])] + 6*ArcTanh[a*x]*PolyLog[3, -E^(-2*ArcTanh[a*x])] + 3*PolyLog[4, -E^(-2*ArcTanh[a*x]
)])/(4*a^4)

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Maple [A]  time = 0.895, size = 248, normalized size = 1.2 \begin{align*} -{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}}{4\,{a}^{4}}}-{\frac{{x}^{2} \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}}{2\,{a}^{2}}}-{\frac{3\,x \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{2\,{a}^{3}}}+{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}}{2\,{a}^{4}}}-{\frac{3\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{2\,{a}^{4}}}+{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}}{{a}^{4}}\ln \left ({\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}}+1 \right ) }+{\frac{3\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{2\,{a}^{4}}{\it polylog} \left ( 2,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }-{\frac{3\,{\it Artanh} \left ( ax \right ) }{2\,{a}^{4}}{\it polylog} \left ( 3,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }+{\frac{3}{4\,{a}^{4}}{\it polylog} \left ( 4,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }+3\,{\frac{{\it Artanh} \left ( ax \right ) }{{a}^{4}}\ln \left ({\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}}+1 \right ) }+{\frac{3}{2\,{a}^{4}}{\it polylog} \left ( 2,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*arctanh(a*x)^4/a^4-1/2*x^2*arctanh(a*x)^3/a^2-3/2*x*arctanh(a*x)^2/a^3+1/2*arctanh(a*x)^3/a^4-3/2*arctanh
(a*x)^2/a^4+1/a^4*arctanh(a*x)^3*ln((a*x+1)^2/(-a^2*x^2+1)+1)+3/2/a^4*arctanh(a*x)^2*polylog(2,-(a*x+1)^2/(-a^
2*x^2+1))-3/2/a^4*arctanh(a*x)*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))+3/4/a^4*polylog(4,-(a*x+1)^2/(-a^2*x^2+1))+3
/a^4*arctanh(a*x)*ln((a*x+1)^2/(-a^2*x^2+1)+1)+3/2/a^4*polylog(2,-(a*x+1)^2/(-a^2*x^2+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{4 \,{\left (a^{2} x^{2} + \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{3} + \log \left (-a x + 1\right )^{4}}{64 \, a^{4}} - \frac{1}{8} \, \int \frac{2 \, a^{3} x^{3} \log \left (a x + 1\right )^{3} - 6 \, a^{3} x^{3} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right ) + 3 \,{\left (a^{3} x^{3} + a^{2} x^{2} +{\left (2 \, a^{3} x^{3} + a x + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2}}{2 \,{\left (a^{5} x^{2} - a^{3}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(4*(a^2*x^2 + log(a*x + 1))*log(-a*x + 1)^3 + log(-a*x + 1)^4)/a^4 - 1/8*integrate(1/2*(2*a^3*x^3*log(a*x
+ 1)^3 - 6*a^3*x^3*log(a*x + 1)^2*log(-a*x + 1) + 3*(a^3*x^3 + a^2*x^2 + (2*a^3*x^3 + a*x + 1)*log(a*x + 1))*
log(-a*x + 1)^2)/(a^5*x^2 - a^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{x^{3} \operatorname{artanh}\left (a x\right )^{3}}{a^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{3} \operatorname{atanh}^{3}{\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**3*atanh(a*x)**3/(-a**2*x**2+1),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{3} \operatorname{artanh}\left (a x\right )^{3}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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