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

Optimal. Leaf size=121 \[ -\frac {\tanh ^{-1}(a x)^4}{8 a^3}+\frac {3 \tanh ^{-1}(a x)^2}{8 a^3}+\frac {x \tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}+\frac {3 x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {3}{8 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )} \]

[Out]

-3/8/a^3/(-a^2*x^2+1)+3/4*x*arctanh(a*x)/a^2/(-a^2*x^2+1)+3/8*arctanh(a*x)^2/a^3-3/4*arctanh(a*x)^2/a^3/(-a^2*
x^2+1)+1/2*x*arctanh(a*x)^3/a^2/(-a^2*x^2+1)-1/8*arctanh(a*x)^4/a^3

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Rubi [A]  time = 0.13, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6000, 5994, 5956, 261} \[ -\frac {3}{8 a^3 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}+\frac {3 x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^4}{8 a^3}+\frac {3 \tanh ^{-1}(a x)^2}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3/(8*a^3*(1 - a^2*x^2)) + (3*x*ArcTanh[a*x])/(4*a^2*(1 - a^2*x^2)) + (3*ArcTanh[a*x]^2)/(8*a^3) - (3*ArcTanh[
a*x]^2)/(4*a^3*(1 - a^2*x^2)) + (x*ArcTanh[a*x]^3)/(2*a^2*(1 - a^2*x^2)) - ArcTanh[a*x]^4/(8*a^3)

Rule 261

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

Rule 5956

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

Rule 5994

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

Rule 6000

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> -Simp[(a + b*ArcT
anh[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (-Dist[(b*p)/(2*c), Int[(x*(a + b*ArcTanh[c*x])^(p - 1))/(d + e*
x^2)^2, x], x] + Simp[(x*(a + b*ArcTanh[c*x])^p)/(2*c^2*d*(d + e*x^2)), x]) /; FreeQ[{a, b, c, d, e}, x] && Eq
Q[c^2*d + e, 0] && GtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^2} \, dx &=\frac {x \tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^4}{8 a^3}-\frac {3 \int \frac {x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx}{2 a}\\ &=-\frac {3 \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^4}{8 a^3}+\frac {3 \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx}{2 a^2}\\ &=\frac {3 x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)^2}{8 a^3}-\frac {3 \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^4}{8 a^3}-\frac {3 \int \frac {x}{\left (1-a^2 x^2\right )^2} \, dx}{4 a}\\ &=-\frac {3}{8 a^3 \left (1-a^2 x^2\right )}+\frac {3 x \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)^2}{8 a^3}-\frac {3 \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}+\frac {x \tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^4}{8 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 72, normalized size = 0.60 \[ \frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4+3 \left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^2-4 a x \tanh ^{-1}(a x)^3-6 a x \tanh ^{-1}(a x)+3}{8 a^3 \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 - 6*a*x*ArcTanh[a*x] + 3*(1 + a^2*x^2)*ArcTanh[a*x]^2 - 4*a*x*ArcTanh[a*x]^3 + (1 - a^2*x^2)*ArcTanh[a*x]^4
)/(8*a^3*(-1 + a^2*x^2))

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fricas [A]  time = 1.03, size = 114, normalized size = 0.94 \[ -\frac {8 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + {\left (a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} + 48 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 12 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 48}{128 \, {\left (a^{5} x^{2} - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(8*a*x*log(-(a*x + 1)/(a*x - 1))^3 + (a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^4 + 48*a*x*log(-(a*x + 1)/
(a*x - 1)) - 12*(a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 - 48)/(a^5*x^2 - a^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.82, size = 1771, normalized size = 14.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16/a/(a*x-1)/(a*x+1)*x^2+1/8/a^3/(a*x-1)/(a*x+1)*arctanh(a*x)^4+3/8/a^3/(a*x-1)/(a*x+1)*arctanh(a*x)^2-1/4*I
/a^3/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^3*Pi+1/8*
I/a^3/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))
*arctanh(a*x)^3*Pi-1/8*I/a^3/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^
2*arctanh(a*x)^3*Pi-1/8*I/a^3/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*csgn(
I/(1+(a*x+1)^2/(-a^2*x^2+1)))*arctanh(a*x)^3*Pi+1/8*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^
2*x^2-1))^3*x^2+1/8*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+
1)))^3*x^2-1/4*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*x^2+1/4*I/a/(a*x-1)/
(a*x+1)*arctanh(a*x)^3*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*x^2+3/16/a^3/(a*x-1)/(a*x+1)-1/4/a^3*arctanh(a*
x)^3/(a*x-1)+1/4/a^3*arctanh(a*x)^3*ln(a*x-1)-1/4/a^3*arctanh(a*x)^3/(a*x+1)-1/4/a^3*arctanh(a*x)^3*ln(a*x+1)+
1/2/a^3*arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))-1/8*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)
^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*x^2+1/8*I/a/(a*x-1)/(a*x+1)*arctanh
(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*x^2+1/8*I/a^3/(a*x-1)/(a*x+1)*cs
gn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I/(1+(a*x+1)^2/(-a^2
*x^2+1)))*arctanh(a*x)^3*Pi+1/8*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^
2/(-a^2*x^2+1)))^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*x^2+1/4*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*csgn(I*(a*
x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*x^2-1/8*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*csgn(I*
(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+
1)))*x^2+1/4*I/a^3/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi+1/4*I/a^3/(a*x-1)/(a*x+1)*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1
)))^3*arctanh(a*x)^3*Pi-1/4*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*x^2-1/8*I/a^3/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)
^2/(a^2*x^2-1))^3*arctanh(a*x)^3*Pi-1/8*I/a^3/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*
x^2+1)))^3*arctanh(a*x)^3*Pi-1/4*I/a^3/(a*x-1)/(a*x+1)*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*arctanh(a*x)^3*Pi-
1/8/a/(a*x-1)/(a*x+1)*arctanh(a*x)^4*x^2+3/8/a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*x^2-3/4/a^2/(a*x-1)/(a*x+1)*arct
anh(a*x)*x

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maxima [B]  time = 0.35, size = 465, normalized size = 3.84 \[ -\frac {1}{4} \, {\left (\frac {2 \, x}{a^{4} x^{2} - a^{2}} + \frac {\log \left (a x + 1\right )}{a^{3}} - \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname {artanh}\left (a x\right )^{3} + \frac {3 \, {\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} + 4\right )} a \operatorname {artanh}\left (a x\right )^{2}}{16 \, {\left (a^{6} x^{2} - a^{4}\right )}} + \frac {1}{128} \, {\left (\frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{4} - 4 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{4} - 6 \, {\left (2 \, a^{2} x^{2} - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2\right )} \log \left (a x + 1\right )^{2} - 12 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4 \, {\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} - 6 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} \log \left (a x + 1\right ) + 48\right )} a^{2}}{a^{8} x^{2} - a^{6}} - \frac {8 \, {\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} - 3 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} + 12 \, a x - 3 \, {\left (2 \, a^{2} x^{2} - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2\right )} \log \left (a x + 1\right ) + 6 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} a \operatorname {artanh}\left (a x\right )}{a^{7} x^{2} - a^{5}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x/(a^4*x^2 - a^2) + log(a*x + 1)/a^3 - log(a*x - 1)/a^3)*arctanh(a*x)^3 + 3/16*((a^2*x^2 - 1)*log(a*x
+ 1)^2 - 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) + (a^2*x^2 - 1)*log(a*x - 1)^2 + 4)*a*arctanh(a*x)^2/(a^6*x
^2 - a^4) + 1/128*(((a^2*x^2 - 1)*log(a*x + 1)^4 - 4*(a^2*x^2 - 1)*log(a*x + 1)^3*log(a*x - 1) + (a^2*x^2 - 1)
*log(a*x - 1)^4 - 6*(2*a^2*x^2 - (a^2*x^2 - 1)*log(a*x - 1)^2 - 2)*log(a*x + 1)^2 - 12*(a^2*x^2 - 1)*log(a*x -
1)^2 - 4*((a^2*x^2 - 1)*log(a*x - 1)^3 - 6*(a^2*x^2 - 1)*log(a*x - 1))*log(a*x + 1) + 48)*a^2/(a^8*x^2 - a^6)
- 8*((a^2*x^2 - 1)*log(a*x + 1)^3 - 3*(a^2*x^2 - 1)*log(a*x + 1)^2*log(a*x - 1) - (a^2*x^2 - 1)*log(a*x - 1)^
3 + 12*a*x - 3*(2*a^2*x^2 - (a^2*x^2 - 1)*log(a*x - 1)^2 - 2)*log(a*x + 1) + 6*(a^2*x^2 - 1)*log(a*x - 1))*a*a
rctanh(a*x)/(a^7*x^2 - a^5))*a

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mupad [B]  time = 1.71, size = 410, normalized size = 3.39 \[ \frac {3\,{\ln \left (a\,x+1\right )}^2}{32\,a^3}-\frac {3\,{\ln \left (a\,x+1\right )}^2}{16\,\left (a^3-a^5\,x^2\right )}+\frac {3\,{\ln \left (1-a\,x\right )}^2}{32\,a^3}-\frac {{\ln \left (a\,x+1\right )}^4}{128\,a^3}-\frac {{\ln \left (1-a\,x\right )}^4}{128\,a^3}-\frac {3\,{\ln \left (1-a\,x\right )}^2}{16\,a^3-16\,a^5\,x^2}-\frac {3}{2\,\left (4\,a^3-4\,a^5\,x^2\right )}-\frac {x\,{\ln \left (1-a\,x\right )}^3}{2\,\left (8\,a^2-8\,a^4\,x^2\right )}-\frac {3\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )}{16\,a^3}+\frac {3\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )}{8\,a^3-8\,a^5\,x^2}+\frac {3\,x\,\ln \left (a\,x+1\right )}{8\,\left (a^2-a^4\,x^2\right )}+\frac {\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^3}{32\,a^3}+\frac {{\ln \left (a\,x+1\right )}^3\,\ln \left (1-a\,x\right )}{32\,a^3}-\frac {6\,x\,\ln \left (1-a\,x\right )}{16\,a^2-16\,a^4\,x^2}+\frac {x\,{\ln \left (a\,x+1\right )}^3}{16\,\left (a^2-a^4\,x^2\right )}-\frac {3\,{\ln \left (a\,x+1\right )}^2\,{\ln \left (1-a\,x\right )}^2}{64\,a^3}+\frac {6\,x\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2}{32\,a^2-32\,a^4\,x^2}-\frac {6\,x\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )}{32\,a^2-32\,a^4\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*log(a*x + 1)^2)/(32*a^3) - (3*log(a*x + 1)^2)/(16*(a^3 - a^5*x^2)) + (3*log(1 - a*x)^2)/(32*a^3) - log(a*x
+ 1)^4/(128*a^3) - log(1 - a*x)^4/(128*a^3) - (3*log(1 - a*x)^2)/(16*a^3 - 16*a^5*x^2) - 3/(2*(4*a^3 - 4*a^5*x
^2)) - (x*log(1 - a*x)^3)/(2*(8*a^2 - 8*a^4*x^2)) - (3*log(a*x + 1)*log(1 - a*x))/(16*a^3) + (3*log(a*x + 1)*l
og(1 - a*x))/(8*a^3 - 8*a^5*x^2) + (3*x*log(a*x + 1))/(8*(a^2 - a^4*x^2)) + (log(a*x + 1)*log(1 - a*x)^3)/(32*
a^3) + (log(a*x + 1)^3*log(1 - a*x))/(32*a^3) - (6*x*log(1 - a*x))/(16*a^2 - 16*a^4*x^2) + (x*log(a*x + 1)^3)/
(16*(a^2 - a^4*x^2)) - (3*log(a*x + 1)^2*log(1 - a*x)^2)/(64*a^3) + (6*x*log(a*x + 1)*log(1 - a*x)^2)/(32*a^2
- 32*a^4*x^2) - (6*x*log(a*x + 1)^2*log(1 - a*x))/(32*a^2 - 32*a^4*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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