3.5.8 \(\int \frac {\tanh ^{-1}(a x)^3}{x (1-a^2 x^2)^{3/2}} \, dx\) [408]

Optimal. Leaf size=185 \[ -\frac {6 a x}{\sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 a x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3-3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-e^{\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,e^{\tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {PolyLog}\left (3,-e^{\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {PolyLog}\left (3,e^{\tanh ^{-1}(a x)}\right )-6 \text {PolyLog}\left (4,-e^{\tanh ^{-1}(a x)}\right )+6 \text {PolyLog}\left (4,e^{\tanh ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.29, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6177, 6167, 4267, 2611, 6744, 2320, 6724, 6141, 6109, 197} \begin {gather*} -\frac {6 a x}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-\frac {3 a x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )-6 \text {Li}_4\left (-e^{\tanh ^{-1}(a x)}\right )+6 \text {Li}_4\left (e^{\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*x)/Sqrt[1 - a^2*x^2] + (6*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] - (3*a*x*ArcTanh[a*x]^2)/Sqrt[1 - a^2*x^2] + A
rcTanh[a*x]^3/Sqrt[1 - a^2*x^2] - 2*ArcTanh[E^ArcTanh[a*x]]*ArcTanh[a*x]^3 - 3*ArcTanh[a*x]^2*PolyLog[2, -E^Ar
cTanh[a*x]] + 3*ArcTanh[a*x]^2*PolyLog[2, E^ArcTanh[a*x]] + 6*ArcTanh[a*x]*PolyLog[3, -E^ArcTanh[a*x]] - 6*Arc
Tanh[a*x]*PolyLog[3, E^ArcTanh[a*x]] - 6*PolyLog[4, -E^ArcTanh[a*x]] + 6*PolyLog[4, E^ArcTanh[a*x]]

Rule 197

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

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4267

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(Ar
cTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*
fz*x)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 6109

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-b)*p*((a + b*Arc
Tanh[c*x])^(p - 1)/(c*d*Sqrt[d + e*x^2])), x] + (Dist[b^2*p*(p - 1), Int[(a + b*ArcTanh[c*x])^(p - 2)/(d + e*x
^2)^(3/2), x], x] + Simp[x*((a + b*ArcTanh[c*x])^p/(d*Sqrt[d + e*x^2])), x]) /; FreeQ[{a, b, c, d, e}, x] && E
qQ[c^2*d + e, 0] && GtQ[p, 1]

Rule 6141

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

Rule 6167

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Dist[1/Sqrt[d], Su
bst[Int[(a + b*x)^p*Csch[x], x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGt
Q[p, 0] && GtQ[d, 0]

Rule 6177

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 6724

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

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)^3}{x \left (1-a^2 x^2\right )^{3/2}} \, dx &=a^2 \int \frac {x \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx+\int \frac {\tanh ^{-1}(a x)^3}{x \sqrt {1-a^2 x^2}} \, dx\\ &=\frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-(3 a) \int \frac {\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx+\text {Subst}\left (\int x^3 \text {csch}(x) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=\frac {6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 a x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3-3 \text {Subst}\left (\int x^2 \log \left (1-e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )+3 \text {Subst}\left (\int x^2 \log \left (1+e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )-(6 a) \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {6 a x}{\sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 a x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3-3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+6 \text {Subst}\left (\int x \text {Li}_2\left (-e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )-6 \text {Subst}\left (\int x \text {Li}_2\left (e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=-\frac {6 a x}{\sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 a x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3-3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )-6 \text {Subst}\left (\int \text {Li}_3\left (-e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )+6 \text {Subst}\left (\int \text {Li}_3\left (e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=-\frac {6 a x}{\sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 a x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3-3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )-6 \text {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )+6 \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )\\ &=-\frac {6 a x}{\sqrt {1-a^2 x^2}}+\frac {6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 a x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+\frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3-3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )-6 \text {Li}_4\left (-e^{\tanh ^{-1}(a x)}\right )+6 \text {Li}_4\left (e^{\tanh ^{-1}(a x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 230, normalized size = 1.24 \begin {gather*} \frac {1}{8} \left (\pi ^4-\frac {48 a x}{\sqrt {1-a^2 x^2}}+\frac {48 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {24 a x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+\frac {8 \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-2 \tanh ^{-1}(a x)^4-8 \tanh ^{-1}(a x)^3 \log \left (1+e^{-\tanh ^{-1}(a x)}\right )+8 \tanh ^{-1}(a x)^3 \log \left (1-e^{\tanh ^{-1}(a x)}\right )+24 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )+24 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,e^{\tanh ^{-1}(a x)}\right )+48 \tanh ^{-1}(a x) \text {PolyLog}\left (3,-e^{-\tanh ^{-1}(a x)}\right )-48 \tanh ^{-1}(a x) \text {PolyLog}\left (3,e^{\tanh ^{-1}(a x)}\right )+48 \text {PolyLog}\left (4,-e^{-\tanh ^{-1}(a x)}\right )+48 \text {PolyLog}\left (4,e^{\tanh ^{-1}(a x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^4 - (48*a*x)/Sqrt[1 - a^2*x^2] + (48*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] - (24*a*x*ArcTanh[a*x]^2)/Sqrt[1 - a^
2*x^2] + (8*ArcTanh[a*x]^3)/Sqrt[1 - a^2*x^2] - 2*ArcTanh[a*x]^4 - 8*ArcTanh[a*x]^3*Log[1 + E^(-ArcTanh[a*x])]
+ 8*ArcTanh[a*x]^3*Log[1 - E^ArcTanh[a*x]] + 24*ArcTanh[a*x]^2*PolyLog[2, -E^(-ArcTanh[a*x])] + 24*ArcTanh[a*
x]^2*PolyLog[2, E^ArcTanh[a*x]] + 48*ArcTanh[a*x]*PolyLog[3, -E^(-ArcTanh[a*x])] - 48*ArcTanh[a*x]*PolyLog[3,
E^ArcTanh[a*x]] + 48*PolyLog[4, -E^(-ArcTanh[a*x])] + 48*PolyLog[4, E^ArcTanh[a*x]])/8

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Maple [A]
time = 0.73, size = 305, normalized size = 1.65

method	result	size
default	\(-\frac {\left (\arctanh \left (a x \right )^{3}-3 \arctanh \left (a x \right )^{2}+6 \arctanh \left (a x \right )-6\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 \left (a x -1\right )}+\frac {\left (\arctanh \left (a x \right )^{3}+3 \arctanh \left (a x \right )^{2}+6 \arctanh \left (a x \right )+6\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a x +2}+\arctanh \left (a x \right )^{3} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 \arctanh \left (a x \right )^{2} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-6 \arctanh \left (a x \right ) \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+6 \polylog \left (4, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\arctanh \left (a x \right )^{3} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-3 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+6 \arctanh \left (a x \right ) \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-6 \polylog \left (4, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )\)	\(305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*arctanh(a*x)^3/(a^4*x^5 - 2*a^2*x^3 + x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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