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

Optimal. Leaf size=107 \[ -\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^4}-\frac{x^4}{2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac{x^3}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.65472, antiderivative size = 160, normalized size of antiderivative = 1.5, number of steps used = 25, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {6028, 5996, 6034, 5448, 12, 3298, 6032, 5966} \[ -\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^4}+\frac{x}{2 a^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{x}{2 a^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{a^2 x^2+1}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{3}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{2}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(2*a^3*(1 - a^2*x^2)^2*ArcTanh[a*x]^2) + x/(2*a^3*(1 - a^2*x^2)*ArcTanh[a*x]^2) - 2/(a^4*(1 - a^2*x^2)^2*Ar
cTanh[a*x]) + 3/(2*a^4*(1 - a^2*x^2)*ArcTanh[a*x]) + (1 + a^2*x^2)/(2*a^4*(1 - a^2*x^2)*ArcTanh[a*x]) - SinhIn
tegral[2*ArcTanh[a*x]]/(2*a^4) + SinhIntegral[4*ArcTanh[a*x]]/a^4

Rule 6028

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int
[x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*A
rcTanh[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] &&
IGtQ[m, 1] && NeQ[p, -1]

Rule 5996

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

Rule 6034

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(
m + 1), Subst[Int[((a + b*x)^p*Sinh[x]^m)/Cosh[x]^(m + 2*(q + 1)), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c,
d, e, p}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 6032

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

Rule 5966

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

Rubi steps

\begin{align*} \int \frac{x^3}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx &=\frac{\int \frac{x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx}{a^2}-\frac{\int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3} \, dx}{a^2}\\ &=-\frac{x}{2 a^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{x}{2 a^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}+\frac{1+a^2 x^2}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{\int \frac{1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a^3}-\frac{2 \int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx}{a^2}+\frac{3 \int \frac{x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a}\\ &=-\frac{x}{2 a^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{x}{2 a^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{1}{2 a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{1+a^2 x^2}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{2 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac{3 \int \frac{1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a^3}-\frac{3 \int \frac{1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx}{2 a^3}+\frac{2 \int \frac{x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac{x}{2 a^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{x}{2 a^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{2}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{1+a^2 x^2}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac{2 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac{3 \int \frac{x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx}{a^2}+\frac{6 \int \frac{x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac{x}{2 a^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{x}{2 a^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{2}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{1+a^2 x^2}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac{2 \operatorname{Subst}\left (\int \left (\frac{\sinh (2 x)}{4 x}+\frac{\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac{6 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=-\frac{x}{2 a^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{x}{2 a^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{2}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{1+a^2 x^2}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^4}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^4}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac{6 \operatorname{Subst}\left (\int \left (\frac{\sinh (2 x)}{4 x}+\frac{\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=-\frac{x}{2 a^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{x}{2 a^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{2}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{1+a^2 x^2}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^4}\\ &=-\frac{x}{2 a^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac{x}{2 a^3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac{2}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac{1+a^2 x^2}{2 a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^4}\\ \end{align*}

Mathematica [A]  time = 0.259509, size = 66, normalized size = 0.62 \[ -\frac{\frac{a^2 x^2 \left (\left (a^2 x^2+3\right ) \tanh ^{-1}(a x)+a x\right )}{\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2}+\text{Shi}\left (2 \tanh ^{-1}(a x)\right )-2 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.072, size = 82, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{8\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}+{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{4\,{\it Artanh} \left ( ax \right ) }}-{\frac{{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{2}}-{\frac{\sinh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{16\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{4\,{\it Artanh} \left ( ax \right ) }}+{\it Shi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(1/8/arctanh(a*x)^2*sinh(2*arctanh(a*x))+1/4/arctanh(a*x)*cosh(2*arctanh(a*x))-1/2*Shi(2*arctanh(a*x))-1
/16/arctanh(a*x)^2*sinh(4*arctanh(a*x))-1/4/arctanh(a*x)*cosh(4*arctanh(a*x))+Shi(4*arctanh(a*x)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, a x^{3} +{\left (a^{2} x^{4} + 3 \, x^{2}\right )} \log \left (a x + 1\right ) -{\left (a^{2} x^{4} + 3 \, x^{2}\right )} \log \left (-a x + 1\right )}{{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (a x + 1\right )^{2} - 2 \,{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) +{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (-a x + 1\right )^{2}} + \int -\frac{2 \,{\left (5 \, a^{2} x^{3} + 3 \, x\right )}}{{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (a x + 1\right ) -{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (-a x + 1\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*x^3 + (a^2*x^4 + 3*x^2)*log(a*x + 1) - (a^2*x^4 + 3*x^2)*log(-a*x + 1))/((a^6*x^4 - 2*a^4*x^2 + a^2)*log
(a*x + 1)^2 - 2*(a^6*x^4 - 2*a^4*x^2 + a^2)*log(a*x + 1)*log(-a*x + 1) + (a^6*x^4 - 2*a^4*x^2 + a^2)*log(-a*x
+ 1)^2) + integrate(-2*(5*a^2*x^3 + 3*x)/((a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)*log(a*x + 1) - (a^8*x^6 - 3*
a^6*x^4 + 3*a^4*x^2 - a^2)*log(-a*x + 1)), x)

________________________________________________________________________________________

Fricas [B]  time = 1.99358, size = 621, normalized size = 5.8 \begin{align*} -\frac{8 \, a^{3} x^{3} -{\left (2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (\frac{a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (\frac{a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) -{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) +{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (-\frac{a x - 1}{a x + 1}\right )\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} + 4 \,{\left (a^{4} x^{4} + 3 \, a^{2} x^{2}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{4 \,{\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*a^3*x^3 - (2*(a^4*x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) - 2*(
a^4*x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) - (a^4*x^4 - 2*a^2*x^2 + 1)
*log_integral(-(a*x + 1)/(a*x - 1)) + (a^4*x^4 - 2*a^2*x^2 + 1)*log_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x
+ 1)/(a*x - 1))^2 + 4*(a^4*x^4 + 3*a^2*x^2)*log(-(a*x + 1)/(a*x - 1)))/((a^8*x^4 - 2*a^6*x^2 + a^4)*log(-(a*x
+ 1)/(a*x - 1))^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{3}}{a^{6} x^{6} \operatorname{atanh}^{3}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname{atanh}^{3}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname{atanh}^{3}{\left (a x \right )} - \operatorname{atanh}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**3/(a**6*x**6*atanh(a*x)**3 - 3*a**4*x**4*atanh(a*x)**3 + 3*a**2*x**2*atanh(a*x)**3 - atanh(a*x)**
3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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