3.4.37 \(\int \frac {x^3}{(1-a^2 x^2)^3 \tanh ^{-1}(a x)^3} \, dx\) [337]

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

[Out]

-1/2*x^3/a/(-a^2*x^2+1)^2/arctanh(a*x)^2-3/2*x^2/a^2/(-a^2*x^2+1)^2/arctanh(a*x)-1/2*x^4/(-a^2*x^2+1)^2/arctan
h(a*x)-1/2*Shi(2*arctanh(a*x))/a^4+Shi(4*arctanh(a*x))/a^4

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Rubi [A]
time = 0.45, antiderivative size = 160, normalized size of antiderivative = 1.50, 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 = {6175, 6143, 6181, 5556, 12, 3379, 6179, 6113} \begin {gather*} -\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 {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)}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 3379

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 5556

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 6113

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]

Rule 6143

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 6175

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 6179

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 6181

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])

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 \text {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 \text {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}-\frac {2 \text {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 {\text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {2 \text {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 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {6 \text {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 {\text {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^4}+\frac {\text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^4}-\frac {3 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}+\frac {6 \text {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 \text {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*}

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Mathematica [A]
time = 0.17, size = 66, normalized size = 0.62 \begin {gather*} -\frac {\frac {a^2 x^2 \left (a x+\left (3+a^2 x^2\right ) \tanh ^{-1}(a x)\right )}{\left (-1+a^2 x^2\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((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]])/a^4

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Maple [A]
time = 2.85, size = 82, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\hyperbolicSineIntegral \left (4 \arctanh \left (a x \right )\right )+\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )^{2}}+\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}-\frac {\hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{2}}{a^{4}}\)	\(82\)
default	\(\frac {-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\hyperbolicSineIntegral \left (4 \arctanh \left (a x \right )\right )+\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )^{2}}+\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}-\frac {\hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{2}}{a^{4}}\)	\(82\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (96) = 192\).
time = 0.42, size = 267, normalized size = 2.50 \begin {gather*} -\frac {8 \, a^{3} x^{3} - {\left (2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\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 )} \operatorname {log\_integral}\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 )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\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 {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {gather*}

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {x^3}{{\mathrm {atanh}\left (a\,x\right )}^3\,{\left (a^2\,x^2-1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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