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

Optimal. Leaf size=248 \[ \frac{4 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{5 a}-\frac{8 \tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{5 a}-\frac{1-a^2 x^2}{20 a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}-\frac{1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{8}{15} x \tanh ^{-1}(a x)^3+\frac{8 \tanh ^{-1}(a x)^3}{15 a}-x \tanh ^{-1}(a x)-\frac{8 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{5 a} \]

[Out]

-(1 - a^2*x^2)/(20*a) - x*ArcTanh[a*x] - (x*(1 - a^2*x^2)*ArcTanh[a*x])/10 + (2*(1 - a^2*x^2)*ArcTanh[a*x]^2)/
(5*a) + (3*(1 - a^2*x^2)^2*ArcTanh[a*x]^2)/(20*a) + (8*ArcTanh[a*x]^3)/(15*a) + (8*x*ArcTanh[a*x]^3)/15 + (4*x
*(1 - a^2*x^2)*ArcTanh[a*x]^3)/15 + (x*(1 - a^2*x^2)^2*ArcTanh[a*x]^3)/5 - (8*ArcTanh[a*x]^2*Log[2/(1 - a*x)])
/(5*a) - Log[1 - a^2*x^2]/(2*a) - (8*ArcTanh[a*x]*PolyLog[2, 1 - 2/(1 - a*x)])/(5*a) + (4*PolyLog[3, 1 - 2/(1
- a*x)])/(5*a)

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Rubi [A]  time = 0.251565, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {5944, 5910, 5984, 5918, 5948, 6058, 6610, 260, 5942} \[ \frac{4 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{5 a}-\frac{8 \tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{5 a}-\frac{1-a^2 x^2}{20 a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}-\frac{1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{8}{15} x \tanh ^{-1}(a x)^3+\frac{8 \tanh ^{-1}(a x)^3}{15 a}-x \tanh ^{-1}(a x)-\frac{8 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^2*ArcTanh[a*x]^3,x]

[Out]

-(1 - a^2*x^2)/(20*a) - x*ArcTanh[a*x] - (x*(1 - a^2*x^2)*ArcTanh[a*x])/10 + (2*(1 - a^2*x^2)*ArcTanh[a*x]^2)/
(5*a) + (3*(1 - a^2*x^2)^2*ArcTanh[a*x]^2)/(20*a) + (8*ArcTanh[a*x]^3)/(15*a) + (8*x*ArcTanh[a*x]^3)/15 + (4*x
*(1 - a^2*x^2)*ArcTanh[a*x]^3)/15 + (x*(1 - a^2*x^2)^2*ArcTanh[a*x]^3)/5 - (8*ArcTanh[a*x]^2*Log[2/(1 - a*x)])
/(5*a) - Log[1 - a^2*x^2]/(2*a) - (8*ArcTanh[a*x]*PolyLog[2, 1 - 2/(1 - a*x)])/(5*a) + (4*PolyLog[3, 1 - 2/(1
- a*x)])/(5*a)

Rule 5944

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5942

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

Rubi steps

\begin{align*} \int \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3 \, dx &=\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac{3}{10} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx+\frac{4}{5} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3 \, dx\\ &=-\frac{1-a^2 x^2}{20 a}-\frac{1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac{1}{5} \int \tanh ^{-1}(a x) \, dx+\frac{8}{15} \int \tanh ^{-1}(a x)^3 \, dx-\frac{4}{5} \int \tanh ^{-1}(a x) \, dx\\ &=-\frac{1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac{1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{8}{15} x \tanh ^{-1}(a x)^3+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3+\frac{1}{5} a \int \frac{x}{1-a^2 x^2} \, dx+\frac{1}{5} (4 a) \int \frac{x}{1-a^2 x^2} \, dx-\frac{1}{5} (8 a) \int \frac{x \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx\\ &=-\frac{1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac{1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{8 \tanh ^{-1}(a x)^3}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^3+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac{\log \left (1-a^2 x^2\right )}{2 a}-\frac{8}{5} \int \frac{\tanh ^{-1}(a x)^2}{1-a x} \, dx\\ &=-\frac{1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac{1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{8 \tanh ^{-1}(a x)^3}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^3+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac{8 \tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{5 a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}+\frac{16}{5} \int \frac{\tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac{1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac{1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{8 \tanh ^{-1}(a x)^3}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^3+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac{8 \tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{5 a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}-\frac{8 \tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{5 a}+\frac{8}{5} \int \frac{\text{Li}_2\left (1-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac{1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac{1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac{2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac{8 \tanh ^{-1}(a x)^3}{15 a}+\frac{8}{15} x \tanh ^{-1}(a x)^3+\frac{4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac{8 \tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{5 a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}-\frac{8 \tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{5 a}+\frac{4 \text{Li}_3\left (1-\frac{2}{1-a x}\right )}{5 a}\\ \end{align*}

Mathematica [A]  time = 0.602464, size = 183, normalized size = 0.74 \[ \frac{96 \tanh ^{-1}(a x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+48 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+3 a^2 x^2-30 \log \left (1-a^2 x^2\right )+12 a^5 x^5 \tanh ^{-1}(a x)^3+9 a^4 x^4 \tanh ^{-1}(a x)^2-40 a^3 x^3 \tanh ^{-1}(a x)^3+6 a^3 x^3 \tanh ^{-1}(a x)-42 a^2 x^2 \tanh ^{-1}(a x)^2+60 a x \tanh ^{-1}(a x)^3-66 a x \tanh ^{-1}(a x)-32 \tanh ^{-1}(a x)^3+33 \tanh ^{-1}(a x)^2-96 \tanh ^{-1}(a x)^2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-3}{60 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3 + 3*a^2*x^2 - 66*a*x*ArcTanh[a*x] + 6*a^3*x^3*ArcTanh[a*x] + 33*ArcTanh[a*x]^2 - 42*a^2*x^2*ArcTanh[a*x]^2
+ 9*a^4*x^4*ArcTanh[a*x]^2 - 32*ArcTanh[a*x]^3 + 60*a*x*ArcTanh[a*x]^3 - 40*a^3*x^3*ArcTanh[a*x]^3 + 12*a^5*x
^5*ArcTanh[a*x]^3 - 96*ArcTanh[a*x]^2*Log[1 + E^(-2*ArcTanh[a*x])] - 30*Log[1 - a^2*x^2] + 96*ArcTanh[a*x]*Pol
yLog[2, -E^(-2*ArcTanh[a*x])] + 48*PolyLog[3, -E^(-2*ArcTanh[a*x])])/(60*a)

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Maple [C]  time = 1.285, size = 883, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20/a+11/20*arctanh(a*x)^2/a+8/15*arctanh(a*x)^3/a+x*arctanh(a*x)^3-11/10*x*arctanh(a*x)+1/10*a^2*arctanh(a*
x)*x^3+1/20*a*x^2-2/3*a^2*arctanh(a*x)^3*x^3-7/10*a*arctanh(a*x)^2*x^2+4/5/a*arctanh(a*x)^2*ln(a*x-1)+4/5/a*ar
ctanh(a*x)^2*ln(a*x+1)-8/5/a*arctanh(a*x)^2*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+3/20*a^3*arctanh(a*x)^2*x^4+1/5*a^4
*arctanh(a*x)^3*x^5-8/5/a*arctanh(a*x)^2*ln(2)-8/5/a*arctanh(a*x)*polylog(2,-(a*x+1)^2/(-a^2*x^2+1))-arctanh(a
*x)/a+2/5*I/a*arctanh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2
/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))*Pi+1/a*ln((a*x+1)^2/(-a^2*x^2+1)+1)+4/5/a*polylog(3,-(a*x+1)^2/(-a^2*
x^2+1))-4/5*I/a*arctanh(a*x)^2*Pi-2/5*I/a*arctanh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(
a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*Pi-2/5*I/a*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1
)/(-a^2*x^2+1)^(1/2))^2*Pi-4/5*I/a*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^
(1/2))*Pi+2/5*I/a*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x
^2+1)+1))^2*Pi-4/5*I/a*arctanh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*Pi-2/5*I/a*arctanh(a*x)^2*csgn(I*(a
*x+1)^2/(a^2*x^2-1))^3*Pi+4/5*I/a*arctanh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2*Pi-2/5*I/a*arctanh(a*x)^
2*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*Pi

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2400*(36*a^5*x^5 - 45*a^4*x^4 - 140*a^3*x^3 + 210*a^2*x^2 + 480*a*x - 60*(3*a^5*x^5 - 10*a^3*x^3 + 15*a*x +
8)*log(a*x + 1))*log(-a*x + 1)^2/a - 1/8*(log(-a*x + 1)^3 - 3*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 6)*(a*x - 1
)/a - 1/1440000*(288*(125*log(-a*x + 1)^3 - 75*log(-a*x + 1)^2 + 30*log(-a*x + 1) - 6)*(a*x - 1)^5 + 5625*(32*
log(-a*x + 1)^3 - 24*log(-a*x + 1)^2 + 12*log(-a*x + 1) - 3)*(a*x - 1)^4 + 40000*(9*log(-a*x + 1)^3 - 9*log(-a
*x + 1)^2 + 6*log(-a*x + 1) - 2)*(a*x - 1)^3 + 90000*(4*log(-a*x + 1)^3 - 6*log(-a*x + 1)^2 + 6*log(-a*x + 1)
- 3)*(a*x - 1)^2 + 180000*(log(-a*x + 1)^3 - 3*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 6)*(a*x - 1))/a + 1/432*(4*
(9*log(-a*x + 1)^3 - 9*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 2)*(a*x - 1)^3 + 27*(4*log(-a*x + 1)^3 - 6*log(-a*x
+ 1)^2 + 6*log(-a*x + 1) - 3)*(a*x - 1)^2 + 108*(log(-a*x + 1)^3 - 3*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 6)*(
a*x - 1))/a - 1/8*integrate(-1/150*(150*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*log(a*x + 1)^3 +
(36*a^5*x^5 - 45*a^4*x^4 - 140*a^3*x^3 + 210*a^2*x^2 - 450*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x -
1)*log(a*x + 1)^2 + 480*a*x - 60*(3*a^5*x^5 - 10*a^3*x^3 + 15*a*x + 8)*log(a*x + 1))*log(-a*x + 1))/(a*x - 1)
, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname{artanh}\left (a x\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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