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

Optimal. Leaf size=186 \[ \frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)^2+\frac {1}{6} a^3 x^3 \tanh ^{-1}(a x)+\frac {a^2 x^2}{12}-\frac {2}{3} \log \left (1-a^2 x^2\right )-a^2 x^2 \tanh ^{-1}(a x)^2+\frac {1}{2} \text {Li}_3\left (1-\frac {2}{1-a x}\right )-\frac {1}{2} \text {Li}_3\left (\frac {2}{1-a x}-1\right )-\text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)+\text {Li}_2\left (\frac {2}{1-a x}-1\right ) \tanh ^{-1}(a x)-\frac {3}{2} a x \tanh ^{-1}(a x)+\frac {3}{4} \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right ) \]

[Out]

1/12*a^2*x^2-3/2*a*x*arctanh(a*x)+1/6*a^3*x^3*arctanh(a*x)+3/4*arctanh(a*x)^2-a^2*x^2*arctanh(a*x)^2+1/4*a^4*x
^4*arctanh(a*x)^2-2*arctanh(a*x)^2*arctanh(-1+2/(-a*x+1))-2/3*ln(-a^2*x^2+1)-arctanh(a*x)*polylog(2,1-2/(-a*x+
1))+arctanh(a*x)*polylog(2,-1+2/(-a*x+1))+1/2*polylog(3,1-2/(-a*x+1))-1/2*polylog(3,-1+2/(-a*x+1))

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Rubi [A]  time = 0.53, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {6012, 5914, 6052, 5948, 6058, 6610, 5916, 5980, 5910, 260, 266, 43} \[ \frac {1}{2} \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )-\frac {1}{2} \text {PolyLog}\left (3,\frac {2}{1-a x}-1\right )-\tanh ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )+\tanh ^{-1}(a x) \text {PolyLog}\left (2,\frac {2}{1-a x}-1\right )+\frac {a^2 x^2}{12}-\frac {2}{3} \log \left (1-a^2 x^2\right )+\frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)^2+\frac {1}{6} a^3 x^3 \tanh ^{-1}(a x)-a^2 x^2 \tanh ^{-1}(a x)^2-\frac {3}{2} a x \tanh ^{-1}(a x)+\frac {3}{4} \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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]

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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{x} \, dx &=\int \left (\frac {\tanh ^{-1}(a x)^2}{x}-2 a^2 x \tanh ^{-1}(a x)^2+a^4 x^3 \tanh ^{-1}(a x)^2\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int x \tanh ^{-1}(a x)^2 \, dx\right )+a^4 \int x^3 \tanh ^{-1}(a x)^2 \, dx+\int \frac {\tanh ^{-1}(a x)^2}{x} \, dx\\ &=-a^2 x^2 \tanh ^{-1}(a x)^2+\frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right )-(4 a) \int \frac {\tanh ^{-1}(a x) \tanh ^{-1}\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx+\left (2 a^3\right ) \int \frac {x^2 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx-\frac {1}{2} a^5 \int \frac {x^4 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-a^2 x^2 \tanh ^{-1}(a x)^2+\frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right )-(2 a) \int \tanh ^{-1}(a x) \, dx+(2 a) \int \frac {\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx+(2 a) \int \frac {\tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx-(2 a) \int \frac {\tanh ^{-1}(a x) \log \left (2-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx+\frac {1}{2} a^3 \int x^2 \tanh ^{-1}(a x) \, dx-\frac {1}{2} a^3 \int \frac {x^2 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx\\ &=-2 a x \tanh ^{-1}(a x)+\frac {1}{6} a^3 x^3 \tanh ^{-1}(a x)+\tanh ^{-1}(a x)^2-a^2 x^2 \tanh ^{-1}(a x)^2+\frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right )-\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )+\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-a x}\right )+\frac {1}{2} a \int \tanh ^{-1}(a x) \, dx-\frac {1}{2} a \int \frac {\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx+a \int \frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx-a \int \frac {\text {Li}_2\left (-1+\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx+\left (2 a^2\right ) \int \frac {x}{1-a^2 x^2} \, dx-\frac {1}{6} a^4 \int \frac {x^3}{1-a^2 x^2} \, dx\\ &=-\frac {3}{2} a x \tanh ^{-1}(a x)+\frac {1}{6} a^3 x^3 \tanh ^{-1}(a x)+\frac {3}{4} \tanh ^{-1}(a x)^2-a^2 x^2 \tanh ^{-1}(a x)^2+\frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right )-\log \left (1-a^2 x^2\right )-\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )+\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-a x}\right )+\frac {1}{2} \text {Li}_3\left (1-\frac {2}{1-a x}\right )-\frac {1}{2} \text {Li}_3\left (-1+\frac {2}{1-a x}\right )-\frac {1}{2} a^2 \int \frac {x}{1-a^2 x^2} \, dx-\frac {1}{12} a^4 \operatorname {Subst}\left (\int \frac {x}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {3}{2} a x \tanh ^{-1}(a x)+\frac {1}{6} a^3 x^3 \tanh ^{-1}(a x)+\frac {3}{4} \tanh ^{-1}(a x)^2-a^2 x^2 \tanh ^{-1}(a x)^2+\frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right )-\frac {3}{4} \log \left (1-a^2 x^2\right )-\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )+\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-a x}\right )+\frac {1}{2} \text {Li}_3\left (1-\frac {2}{1-a x}\right )-\frac {1}{2} \text {Li}_3\left (-1+\frac {2}{1-a x}\right )-\frac {1}{12} a^4 \operatorname {Subst}\left (\int \left (-\frac {1}{a^2}-\frac {1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {a^2 x^2}{12}-\frac {3}{2} a x \tanh ^{-1}(a x)+\frac {1}{6} a^3 x^3 \tanh ^{-1}(a x)+\frac {3}{4} \tanh ^{-1}(a x)^2-a^2 x^2 \tanh ^{-1}(a x)^2+\frac {1}{4} a^4 x^4 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right )-\frac {2}{3} \log \left (1-a^2 x^2\right )-\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )+\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-a x}\right )+\frac {1}{2} \text {Li}_3\left (1-\frac {2}{1-a x}\right )-\frac {1}{2} \text {Li}_3\left (-1+\frac {2}{1-a x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 191, normalized size = 1.03 \[ \frac {1}{4} \left (a^4 x^4-1\right ) \tanh ^{-1}(a x)^2+\frac {a^2 x^2}{12}-\frac {2}{3} \log \left (1-a^2 x^2\right )+\frac {1}{6} a x \left (a^2 x^2+3\right ) \tanh ^{-1}(a x)-\left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2-\frac {1}{2} \text {Li}_3\left (\frac {-a x-1}{a x-1}\right )+\frac {1}{2} \text {Li}_3\left (\frac {a x+1}{a x-1}\right )+\text {Li}_2\left (\frac {-a x-1}{a x-1}\right ) \tanh ^{-1}(a x)-\text {Li}_2\left (\frac {a x+1}{a x-1}\right ) \tanh ^{-1}(a x)-2 a x \tanh ^{-1}(a x)+2 \tanh ^{-1}(a x)^2 \tanh ^{-1}\left (1-\frac {2}{1-a x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {artanh}\left (a x\right )^{2}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.80, size = 728, normalized size = 3.91 \[ -a^{2} x^{2} \arctanh \left (a x \right )^{2}-\frac {i \arctanh \left (a x \right )^{2} \pi \,\mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{2}}{2}-\frac {1}{12}+\frac {3 \arctanh \left (a x \right )^{2}}{4}+\frac {a^{4} x^{4} \arctanh \left (a x \right )^{2}}{4}+\frac {a^{2} x^{2}}{12}-\frac {i \arctanh \left (a x \right )^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{2}}{2}+\frac {i \arctanh \left (a x \right )^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )}{2}-2 \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {i \arctanh \left (a x \right )^{2} \pi \mathrm {csgn}\left (\frac {i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{3}}{2}+\arctanh \left (a x \right )^{2} \ln \left (a x \right )-\arctanh \left (a x \right )^{2} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )+\arctanh \left (a x \right )^{2} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+2 \arctanh \left (a x \right ) \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+\arctanh \left (a x \right )^{2} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+2 \arctanh \left (a x \right ) \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\arctanh \left (a x \right ) \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )-\left (a x +1\right ) \arctanh \left (a x \right )-2 \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {4 \ln \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{3}+\frac {\polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}+\frac {\left (a^{2} x^{2}-4 a x +7\right ) \left (a x +1\right ) \arctanh \left (a x \right )}{6}+\frac {\left (a x -3\right ) \left (a x +1\right ) \arctanh \left (a x \right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*x^2*arctanh(a*x)^2-1/2*I*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*((a*x+1)^2/(-a^2*x^2
+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2-1/12+3/4*arctanh(a*x)^2+1/4*a^4*x^4*arctanh(a*x)^2+1/12*a^2*x^2+1/2*I*arc
tanh(a*x)^2*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1))*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*((a*x+1)^2/(-a^2*x
^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))+1/2*I*arctanh(a*x)^2*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-
a^2*x^2+1)))^3-2*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))-1/2*I*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)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2+arctanh(a*x)^2*ln(a*x)-arctanh(a*x)^2*ln((
a*x+1)^2/(-a^2*x^2+1)-1)+arctanh(a*x)^2*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+2*arctanh(a*x)*polylog(2,(a*x+1)/(-a^
2*x^2+1)^(1/2))+arctanh(a*x)^2*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+2*arctanh(a*x)*polylog(2,-(a*x+1)/(-a^2*x^2+1)
^(1/2))-arctanh(a*x)*polylog(2,-(a*x+1)^2/(-a^2*x^2+1))-(a*x+1)*arctanh(a*x)-2*polylog(3,-(a*x+1)/(-a^2*x^2+1)
^(1/2))+4/3*ln(1+(a*x+1)^2/(-a^2*x^2+1))+1/2*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))+1/6*(a^2*x^2-4*a*x+7)*(a*x+1)*
arctanh(a*x)+1/2*(a*x-3)*(a*x+1)*arctanh(a*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, {\left (a^{4} x^{4} - 4 \, a^{2} x^{2}\right )} \log \left (-a x + 1\right )^{2} - \frac {1}{4} \, \int -\frac {2 \, {\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \log \left (a x + 1\right )^{2} - {\left (a^{5} x^{5} - 4 \, a^{3} x^{3} + 4 \, {\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )}{2 \, {\left (a x^{2} - x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(a^4*x^4 - 4*a^2*x^2)*log(-a*x + 1)^2 - 1/4*integrate(-1/2*(2*(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 - (a^5*x^5 - 4*a^3*x^3 + 4*(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))*log(-a*x + 1))/(a*x^2 - x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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