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

Optimal. Leaf size=123 \[ -\frac{7 a}{16 \left (1-a^2 x^2\right )}-\frac{a}{16 \left (1-a^2 x^2\right )^2}-\frac{1}{2} a \log \left (1-a^2 x^2\right )+\frac{7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+a \log (x)+\frac{15}{16} a \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)}{x} \]

[Out]

-a/(16*(1 - a^2*x^2)^2) - (7*a)/(16*(1 - a^2*x^2)) - ArcTanh[a*x]/x + (a^2*x*ArcTanh[a*x])/(4*(1 - a^2*x^2)^2)
+ (7*a^2*x*ArcTanh[a*x])/(8*(1 - a^2*x^2)) + (15*a*ArcTanh[a*x]^2)/16 + a*Log[x] - (a*Log[1 - a^2*x^2])/2

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Rubi [A]  time = 0.235148, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.55, Rules used = {6030, 5982, 5916, 266, 36, 29, 31, 5948, 5956, 261, 5960} \[ -\frac{7 a}{16 \left (1-a^2 x^2\right )}-\frac{a}{16 \left (1-a^2 x^2\right )^2}-\frac{1}{2} a \log \left (1-a^2 x^2\right )+\frac{7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+a \log (x)+\frac{15}{16} a \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(16*(1 - a^2*x^2)^2) - (7*a)/(16*(1 - a^2*x^2)) - ArcTanh[a*x]/x + (a^2*x*ArcTanh[a*x])/(4*(1 - a^2*x^2)^2)
+ (7*a^2*x*ArcTanh[a*x])/(8*(1 - a^2*x^2)) + (15*a*ArcTanh[a*x]^2)/16 + a*Log[x] - (a*Log[1 - a^2*x^2])/2

Rule 6030

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/d, Int
[x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/d, Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTanh
[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegersQ[p, 2*q] && LtQ[q, -1] && ILtQ[
m, 0] && NeQ[p, -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 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 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 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 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 5960

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q + 1))
/(4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x] -
Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*
d + e, 0] && LtQ[q, -1] && NeQ[q, -3/2]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )^3} \, dx &=a^2 \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx+\int \frac{\tanh ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac{a}{16 \left (1-a^2 x^2\right )^2}+\frac{a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{1}{4} \left (3 a^2\right ) \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx+a^2 \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac{\tanh ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{a}{16 \left (1-a^2 x^2\right )^2}+\frac{a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{7}{16} a \tanh ^{-1}(a x)^2+a^2 \int \frac{\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx-\frac{1}{8} \left (3 a^3\right ) \int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx-\frac{1}{2} a^3 \int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac{\tanh ^{-1}(a x)}{x^2} \, dx\\ &=-\frac{a}{16 \left (1-a^2 x^2\right )^2}-\frac{7 a}{16 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{x}+\frac{a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{15}{16} a \tanh ^{-1}(a x)^2+a \int \frac{1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac{a}{16 \left (1-a^2 x^2\right )^2}-\frac{7 a}{16 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{x}+\frac{a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{15}{16} a \tanh ^{-1}(a x)^2+\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{a}{16 \left (1-a^2 x^2\right )^2}-\frac{7 a}{16 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{x}+\frac{a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{15}{16} a \tanh ^{-1}(a x)^2+\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} a^3 \operatorname{Subst}\left (\int \frac{1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac{a}{16 \left (1-a^2 x^2\right )^2}-\frac{7 a}{16 \left (1-a^2 x^2\right )}-\frac{\tanh ^{-1}(a x)}{x}+\frac{a^2 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac{7 a^2 x \tanh ^{-1}(a x)}{8 \left (1-a^2 x^2\right )}+\frac{15}{16} a \tanh ^{-1}(a x)^2+a \log (x)-\frac{1}{2} a \log \left (1-a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.161525, size = 94, normalized size = 0.76 \[ \frac{1}{16} \left (a \left (\frac{7 a^2 x^2-8}{\left (a^2 x^2-1\right )^2}-8 \log \left (1-a^2 x^2\right )+16 \log (x)\right )-\frac{2 \left (15 a^4 x^4-25 a^2 x^2+8\right ) \tanh ^{-1}(a x)}{x \left (a^2 x^2-1\right )^2}+15 a \tanh ^{-1}(a x)^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(8 - 25*a^2*x^2 + 15*a^4*x^4)*ArcTanh[a*x])/(x*(-1 + a^2*x^2)^2) + 15*a*ArcTanh[a*x]^2 + a*((-8 + 7*a^2*x
^2)/(-1 + a^2*x^2)^2 + 16*Log[x] - 8*Log[1 - a^2*x^2]))/16

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Maple [B]  time = 0.06, size = 228, normalized size = 1.9 \begin{align*}{\frac{a{\it Artanh} \left ( ax \right ) }{16\, \left ( ax-1 \right ) ^{2}}}-{\frac{7\,a{\it Artanh} \left ( ax \right ) }{16\,ax-16}}-{\frac{15\,a{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{16}}-{\frac{{\it Artanh} \left ( ax \right ) }{x}}-{\frac{a{\it Artanh} \left ( ax \right ) }{16\, \left ( ax+1 \right ) ^{2}}}-{\frac{7\,a{\it Artanh} \left ( ax \right ) }{16\,ax+16}}+{\frac{15\,a{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{16}}-{\frac{15\,a \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{64}}+{\frac{15\,a\ln \left ( ax-1 \right ) }{32}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{15\,a \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{64}}+{\frac{15\,a\ln \left ( ax+1 \right ) }{32}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }-{\frac{15\,a}{32}\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }-{\frac{a}{64\, \left ( ax-1 \right ) ^{2}}}+{\frac{15\,a}{64\,ax-64}}-{\frac{a\ln \left ( ax-1 \right ) }{2}}+a\ln \left ( ax \right ) -{\frac{a}{64\, \left ( ax+1 \right ) ^{2}}}-{\frac{15\,a}{64\,ax+64}}-{\frac{a\ln \left ( ax+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a*arctanh(a*x)/(a*x-1)^2-7/16*a*arctanh(a*x)/(a*x-1)-15/16*a*arctanh(a*x)*ln(a*x-1)-arctanh(a*x)/x-1/16*a
*arctanh(a*x)/(a*x+1)^2-7/16*a*arctanh(a*x)/(a*x+1)+15/16*a*arctanh(a*x)*ln(a*x+1)-15/64*a*ln(a*x-1)^2+15/32*a
*ln(a*x-1)*ln(1/2+1/2*a*x)-15/64*a*ln(a*x+1)^2+15/32*a*ln(-1/2*a*x+1/2)*ln(a*x+1)-15/32*a*ln(-1/2*a*x+1/2)*ln(
1/2+1/2*a*x)-1/64*a/(a*x-1)^2+15/64*a/(a*x-1)-1/2*a*ln(a*x-1)+a*ln(a*x)-1/64*a/(a*x+1)^2-15/64*a/(a*x+1)-1/2*a
*ln(a*x+1)

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Maxima [A]  time = 0.99672, size = 275, normalized size = 2.24 \begin{align*} \frac{1}{64} \, a{\left (\frac{28 \, a^{2} x^{2} - 15 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 30 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 15 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 32}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - 32 \, \log \left (a x + 1\right ) - 32 \, \log \left (a x - 1\right ) + 64 \, \log \left (x\right )\right )} + \frac{1}{16} \,{\left (15 \, a \log \left (a x + 1\right ) - 15 \, a \log \left (a x - 1\right ) - \frac{2 \,{\left (15 \, a^{4} x^{4} - 25 \, a^{2} x^{2} + 8\right )}}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*a*((28*a^2*x^2 - 15*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 30*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)*
log(a*x - 1) - 15*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 - 32)/(a^4*x^4 - 2*a^2*x^2 + 1) - 32*log(a*x + 1) -
32*log(a*x - 1) + 64*log(x)) + 1/16*(15*a*log(a*x + 1) - 15*a*log(a*x - 1) - 2*(15*a^4*x^4 - 25*a^2*x^2 + 8)/
(a^4*x^5 - 2*a^2*x^3 + x))*arctanh(a*x)

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Fricas [A]  time = 2.06662, size = 358, normalized size = 2.91 \begin{align*} \frac{28 \, a^{3} x^{3} + 15 \,{\left (a^{5} x^{5} - 2 \, a^{3} x^{3} + a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 32 \, a x - 32 \,{\left (a^{5} x^{5} - 2 \, a^{3} x^{3} + a x\right )} \log \left (a^{2} x^{2} - 1\right ) + 64 \,{\left (a^{5} x^{5} - 2 \, a^{3} x^{3} + a x\right )} \log \left (x\right ) - 4 \,{\left (15 \, a^{4} x^{4} - 25 \, a^{2} x^{2} + 8\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{64 \,{\left (a^{4} x^{5} - 2 \, a^{2} x^{3} + x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(28*a^3*x^3 + 15*(a^5*x^5 - 2*a^3*x^3 + a*x)*log(-(a*x + 1)/(a*x - 1))^2 - 32*a*x - 32*(a^5*x^5 - 2*a^3*x
^3 + a*x)*log(a^2*x^2 - 1) + 64*(a^5*x^5 - 2*a^3*x^3 + a*x)*log(x) - 4*(15*a^4*x^4 - 25*a^2*x^2 + 8)*log(-(a*x
+ 1)/(a*x - 1)))/(a^4*x^5 - 2*a^2*x^3 + x)

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Sympy [A]  time = 8.62253, size = 578, normalized size = 4.7 \begin{align*} \begin{cases} \frac{64 a^{5} x^{5} \log{\left (x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{64 a^{5} x^{5} \log{\left (x - \frac{1}{a} \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} + \frac{60 a^{5} x^{5} \operatorname{atanh}^{2}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{64 a^{5} x^{5} \operatorname{atanh}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} + \frac{7 a^{5} x^{5}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{120 a^{4} x^{4} \operatorname{atanh}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{128 a^{3} x^{3} \log{\left (x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} + \frac{128 a^{3} x^{3} \log{\left (x - \frac{1}{a} \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{120 a^{3} x^{3} \operatorname{atanh}^{2}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} + \frac{128 a^{3} x^{3} \operatorname{atanh}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} + \frac{14 a^{3} x^{3}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} + \frac{200 a^{2} x^{2} \operatorname{atanh}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} + \frac{64 a x \log{\left (x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{64 a x \log{\left (x - \frac{1}{a} \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} + \frac{60 a x \operatorname{atanh}^{2}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{64 a x \operatorname{atanh}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{25 a x}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} - \frac{64 \operatorname{atanh}{\left (a x \right )}}{64 a^{4} x^{5} - 128 a^{2} x^{3} + 64 x} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((64*a**5*x**5*log(x)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) - 64*a**5*x**5*log(x - 1/a)/(64*a**4*x**5
- 128*a**2*x**3 + 64*x) + 60*a**5*x**5*atanh(a*x)**2/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) - 64*a**5*x**5*ata
nh(a*x)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) + 7*a**5*x**5/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) - 120*a**4*x
**4*atanh(a*x)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) - 128*a**3*x**3*log(x)/(64*a**4*x**5 - 128*a**2*x**3 + 64
*x) + 128*a**3*x**3*log(x - 1/a)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) - 120*a**3*x**3*atanh(a*x)**2/(64*a**4*
x**5 - 128*a**2*x**3 + 64*x) + 128*a**3*x**3*atanh(a*x)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) + 14*a**3*x**3/(
64*a**4*x**5 - 128*a**2*x**3 + 64*x) + 200*a**2*x**2*atanh(a*x)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) + 64*a*x
*log(x)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) - 64*a*x*log(x - 1/a)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) + 60
*a*x*atanh(a*x)**2/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) - 64*a*x*atanh(a*x)/(64*a**4*x**5 - 128*a**2*x**3 + 6
4*x) - 25*a*x/(64*a**4*x**5 - 128*a**2*x**3 + 64*x) - 64*atanh(a*x)/(64*a**4*x**5 - 128*a**2*x**3 + 64*x), Ne(
a, 0)), (0, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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