3.3.41 \(\int \frac {x^3 \tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx\) [241]

Optimal. Leaf size=205 \[ -\frac {3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac {3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac {\tanh ^{-1}(a x)^3}{2 a^4}-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {3 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {3 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \text {PolyLog}\left (4,1-\frac {2}{1-a x}\right )}{4 a^4} \]

[Out]

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

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Rubi [A]
time = 0.33, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6127, 6037, 6021, 6131, 6055, 2449, 2352, 6095, 6205, 6209, 6745} \begin {gather*} \frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{4 a^4}+\frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{2 a^4}-\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{2 a^4}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {\tanh ^{-1}(a x)^3}{2 a^4}-\frac {3 \tanh ^{-1}(a x)^2}{2 a^4}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^4}+\frac {3 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4}-\frac {3 x \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 6021

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

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6055

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

Rule 6095

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 6127

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 6131

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

Rule 6205

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 6209

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

Rule 6745

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 {x^3 \tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx &=-\frac {\int x \tanh ^{-1}(a x)^3 \, dx}{a^2}+\frac {\int \frac {x \tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx}{a^2}\\ &=-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {\int \frac {\tanh ^{-1}(a x)^3}{1-a x} \, dx}{a^3}+\frac {3 \int \frac {x^2 \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{2 a}\\ &=-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}-\frac {3 \int \tanh ^{-1}(a x)^2 \, dx}{2 a^3}+\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{2 a^3}-\frac {3 \int \frac {\tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}\\ &=-\frac {3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac {\tanh ^{-1}(a x)^3}{2 a^4}-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \int \frac {\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}+\frac {3 \int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{a^2}\\ &=-\frac {3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac {3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac {\tanh ^{-1}(a x)^3}{2 a^4}-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \int \frac {\text {Li}_3\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{2 a^3}+\frac {3 \int \frac {\tanh ^{-1}(a x)}{1-a x} \, dx}{a^3}\\ &=-\frac {3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac {3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac {\tanh ^{-1}(a x)^3}{2 a^4}-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {3 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{4 a^4}-\frac {3 \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}\\ &=-\frac {3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac {3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac {\tanh ^{-1}(a x)^3}{2 a^4}-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {3 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{4 a^4}+\frac {3 \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{a^4}\\ &=-\frac {3 \tanh ^{-1}(a x)^2}{2 a^4}-\frac {3 x \tanh ^{-1}(a x)^2}{2 a^3}+\frac {\tanh ^{-1}(a x)^3}{2 a^4}-\frac {x^2 \tanh ^{-1}(a x)^3}{2 a^2}-\frac {\tanh ^{-1}(a x)^4}{4 a^4}+\frac {3 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{4 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 142, normalized size = 0.69 \begin {gather*} -\frac {-6 \tanh ^{-1}(a x)^2+6 a x \tanh ^{-1}(a x)^2-2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-\tanh ^{-1}(a x)^4-12 \tanh ^{-1}(a x) \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )-4 \tanh ^{-1}(a x)^3 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )+6 \left (1+\tanh ^{-1}(a x)^2\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text {PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )}{4 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 52.64, size = 217, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {-\frac {\arctanh \left (a x \right )^{4}}{4}-\frac {\arctanh \left (a x \right )^{2} \left (a x \arctanh \left (a x \right )+\arctanh \left (a x \right )+3\right ) \left (a x -1\right )}{2}+\arctanh \left (a x \right )^{3} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )+\frac {3 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}-\frac {3 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}+\frac {3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{4}-3 \arctanh \left (a x \right )^{2}+3 \arctanh \left (a x \right ) \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )+\frac {3 \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}}{a^{4}}\)	\(217\)
default	\(\frac {-\frac {\arctanh \left (a x \right )^{4}}{4}-\frac {\arctanh \left (a x \right )^{2} \left (a x \arctanh \left (a x \right )+\arctanh \left (a x \right )+3\right ) \left (a x -1\right )}{2}+\arctanh \left (a x \right )^{3} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )+\frac {3 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}-\frac {3 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}+\frac {3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{4}-3 \arctanh \left (a x \right )^{2}+3 \arctanh \left (a x \right ) \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )+\frac {3 \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}}{a^{4}}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

1/64*(4*(a^2*x^2 + log(a*x + 1))*log(-a*x + 1)^3 + log(-a*x + 1)^4)/a^4 - 1/8*integrate(1/2*(2*a^3*x^3*log(a*x
+ 1)^3 - 6*a^3*x^3*log(a*x + 1)^2*log(-a*x + 1) + 3*(a^3*x^3 + a^2*x^2 + (2*a^3*x^3 + a*x + 1)*log(a*x + 1))*
log(-a*x + 1)^2)/(a^5*x^2 - a^3), x)

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x^{3} \operatorname {atanh}^{3}{\left (a x \right )}}{a^{2} x^{2} - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

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