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

Optimal. Leaf size=162 \[ -a^2 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )+a^2 \text{PolyLog}\left (3,\frac{2}{1-a x}-1\right )+2 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )-2 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{1-a x}-1\right )+\frac{1}{2} a^4 x^2 \tanh ^{-1}(a x)^2+a^2 \log (x)+a^3 x \tanh ^{-1}(a x)-4 a^2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac{2}{1-a x}\right )-\frac{\tanh ^{-1}(a x)^2}{2 x^2}-\frac{a \tanh ^{-1}(a x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.461116, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 15, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.682, Rules used = {6012, 5916, 5982, 266, 36, 29, 31, 5948, 5914, 6052, 6058, 6610, 5980, 5910, 260} \[ -a^2 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )+a^2 \text{PolyLog}\left (3,\frac{2}{1-a x}-1\right )+2 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )-2 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{1-a x}-1\right )+\frac{1}{2} a^4 x^2 \tanh ^{-1}(a x)^2+a^2 \log (x)+a^3 x \tanh ^{-1}(a x)-4 a^2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac{2}{1-a x}\right )-\frac{\tanh ^{-1}(a x)^2}{2 x^2}-\frac{a \tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6012

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Int[E
xpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[
c^2*d + e, 0] && IGtQ[p, 0] && IGtQ[q, 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 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 266

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

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 5914

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

Rule 6052

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

Rubi steps

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

Mathematica [A]  time = 0.0699802, size = 183, normalized size = 1.13 \[ a^2 \text{PolyLog}\left (3,\frac{-a x-1}{a x-1}\right )-a^2 \text{PolyLog}\left (3,\frac{a x+1}{a x-1}\right )-2 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{-a x-1}{a x-1}\right )+2 a^2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{a x+1}{a x-1}\right )+\frac{1}{2} a^2 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2+\frac{\left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2}{2 x^2}+a^2 \log (x)+a^3 x \tanh ^{-1}(a x)-4 a^2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac{2}{1-a x}\right )-\frac{a \tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 1.15, size = 774, normalized size = 4.8 \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)^2/x^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{16} \,{\left (2 \, x^{2} \log \left (-a x + 1\right ) - a{\left (\frac{a x^{2} + 2 \, x}{a^{2}} + \frac{2 \, \log \left (a x - 1\right )}{a^{3}}\right )}\right )} a^{4} - \frac{1}{2} \, a^{4} \int x \log \left (a x + 1\right ) \log \left (-a x + 1\right )\,{d x} + \frac{1}{4} \, a^{3} \int a x \log \left (a x + 1\right )^{2}\,{d x} + \frac{1}{4} \, a^{3} \int \frac{\log \left (a x + 1\right )^{2}}{a^{3} x^{3}}\,{d x} + \frac{1}{4} \,{\left (a x -{\left (a x - 1\right )} \log \left (-a x + 1\right ) - 1\right )} a^{2} - \frac{1}{2} \, a^{2} \int \frac{\log \left (a x + 1\right )^{2}}{x}\,{d x} + a^{2} \int \frac{\log \left (a x + 1\right ) \log \left (-a x + 1\right )}{x}\,{d x} - \frac{1}{4} \, a^{2} \int \frac{\log \left (-a x + 1\right )}{x}\,{d x} - \frac{1}{4} \,{\left (a{\left (\log \left (a x - 1\right ) - \log \left (x\right )\right )} - \frac{\log \left (-a x + 1\right )}{x}\right )} a + \frac{{\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right )^{2}}{8 \, x^{2}} - \frac{1}{2} \, \int \frac{\log \left (a x + 1\right ) \log \left (-a x + 1\right )}{x^{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)^2/x^3,x, algorithm="maxima")

[Out]

-1/16*(2*x^2*log(-a*x + 1) - a*((a*x^2 + 2*x)/a^2 + 2*log(a*x - 1)/a^3))*a^4 - 1/2*a^4*integrate(x*log(a*x + 1
)*log(-a*x + 1), x) + 1/4*a^3*integrate(a*x*log(a*x + 1)^2, x) + 1/4*a^3*integrate(log(a*x + 1)^2/(a^3*x^3), x
) + 1/4*(a*x - (a*x - 1)*log(-a*x + 1) - 1)*a^2 - 1/2*a^2*integrate(log(a*x + 1)^2/x, x) + a^2*integrate(log(a
*x + 1)*log(-a*x + 1)/x, x) - 1/4*a^2*integrate(log(-a*x + 1)/x, x) - 1/4*(a*(log(a*x - 1) - log(x)) - log(-a*
x + 1)/x)*a + 1/8*(a^4*x^4 - 1)*log(-a*x + 1)^2/x^2 - 1/2*integrate(log(a*x + 1)*log(-a*x + 1)/x^3, x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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