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

Optimal. Leaf size=205 \[ \frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{4 a^4}+\frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{2 a^4}-\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{2 a^4}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {\tanh ^{-1}(a x)^3}{2 a^4}-\frac {3 \tanh ^{-1}(a x)^2}{2 a^4}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^4}+\frac {3 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4}-\frac {3 x \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.47, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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 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 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 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 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 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 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 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 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 = 0.28, size = 142, normalized size = 0.69 \[ -\frac {-2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+6 \tanh ^{-1}(a x) \text {Li}_3\left (-e^{-2 \tanh ^{-1}(a x)}\right )+6 \left (\tanh ^{-1}(a x)^2+1\right ) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text {Li}_4\left (-e^{-2 \tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x)^4+6 a x \tanh ^{-1}(a x)^2-6 \tanh ^{-1}(a x)^2-4 \tanh ^{-1}(a x)^3 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-12 \tanh ^{-1}(a x) \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{4 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*(-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)*Pol
yLog[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])])/a^4

________________________________________________________________________________________

fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x^{3} \operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{2} - 1}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{3} \operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{2} - 1}\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 3.15, size = 248, normalized size = 1.21 \[ -\frac {\arctanh \left (a x \right )^{4}}{4 a^{4}}-\frac {x^{2} \arctanh \left (a x \right )^{3}}{2 a^{2}}-\frac {3 x \arctanh \left (a x \right )^{2}}{2 a^{3}}+\frac {\arctanh \left (a x \right )^{3}}{2 a^{4}}-\frac {3 \arctanh \left (a x \right )^{2}}{2 a^{4}}+\frac {\arctanh \left (a x \right )^{3} \ln \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{4}}+\frac {3 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 a^{4}}-\frac {3 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 a^{4}}+\frac {3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{4 a^{4}}+\frac {3 \arctanh \left (a x \right ) \ln \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{4}}+\frac {3 \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 a^{4}} \]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {4 \, {\left (a^{2} x^{2} + \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{3} + \log \left (-a x + 1\right )^{4}}{64 \, a^{4}} - \frac {1}{8} \, \int \frac {2 \, a^{3} x^{3} \log \left (a x + 1\right )^{3} - 6 \, a^{3} x^{3} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right ) + 3 \, {\left (a^{3} x^{3} + a^{2} x^{2} + {\left (2 \, a^{3} x^{3} + a x + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2}}{2 \, {\left (a^{5} x^{2} - a^{3}\right )}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ -\int \frac {x^3\,{\mathrm {atanh}\left (a\,x\right )}^3}{a^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{3} \operatorname {atanh}^{3}{\left (a x \right )}}{a^{2} x^{2} - 1}\, dx \]

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)

________________________________________________________________________________________