3.5.5 \(\int \frac {x^2 \tanh ^{-1}(a x)^3}{(1-a^2 x^2)^{3/2}} \, dx\) [405]

Optimal. Leaf size=246 \[ -\frac {6}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{a^2 \sqrt {1-a^2 x^2}}-\frac {2 \text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {6 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {6 i \text {PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 i \text {PolyLog}\left (4,i e^{\tanh ^{-1}(a x)}\right )}{a^3} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6175, 6099, 4265, 2611, 6744, 2320, 6724, 6109, 6105} \begin {gather*} -\frac {2 \tanh ^{-1}(a x)^3 \text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {6 i \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 i \text {Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {x \tanh ^{-1}(a x)^3}{a^2 \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {6}{a^3 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-6/(a^3*Sqrt[1 - a^2*x^2]) + (6*x*ArcTanh[a*x])/(a^2*Sqrt[1 - a^2*x^2]) - (3*ArcTanh[a*x]^2)/(a^3*Sqrt[1 - a^2
*x^2]) + (x*ArcTanh[a*x]^3)/(a^2*Sqrt[1 - a^2*x^2]) - (2*ArcTan[E^ArcTanh[a*x]]*ArcTanh[a*x]^3)/a^3 + ((3*I)*A
rcTanh[a*x]^2*PolyLog[2, (-I)*E^ArcTanh[a*x]])/a^3 - ((3*I)*ArcTanh[a*x]^2*PolyLog[2, I*E^ArcTanh[a*x]])/a^3 -
((6*I)*ArcTanh[a*x]*PolyLog[3, (-I)*E^ArcTanh[a*x]])/a^3 + ((6*I)*ArcTanh[a*x]*PolyLog[3, I*E^ArcTanh[a*x]])/
a^3 + ((6*I)*PolyLog[4, (-I)*E^ArcTanh[a*x]])/a^3 - ((6*I)*PolyLog[4, I*E^ArcTanh[a*x]])/a^3

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 4265

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

Rule 6099

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

Rule 6105

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-b/(c*d*Sqrt[d + e*x^2]
), x] + Simp[x*((a + b*ArcTanh[c*x])/(d*Sqrt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 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 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 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 {x^2 \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\frac {\int \frac {\tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^2}-\frac {\int \frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac {3 \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{a^2 \sqrt {1-a^2 x^2}}-\frac {\text {Subst}\left (\int x^3 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}+\frac {6 \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^2}\\ &=-\frac {6}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{a^2 \sqrt {1-a^2 x^2}}-\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac {(3 i) \text {Subst}\left (\int x^2 \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac {(3 i) \text {Subst}\left (\int x^2 \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac {6}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{a^2 \sqrt {1-a^2 x^2}}-\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {(6 i) \text {Subst}\left (\int x \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}+\frac {(6 i) \text {Subst}\left (\int x \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac {6}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{a^2 \sqrt {1-a^2 x^2}}-\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {(6 i) \text {Subst}\left (\int \text {Li}_3\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac {(6 i) \text {Subst}\left (\int \text {Li}_3\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac {6}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{a^2 \sqrt {1-a^2 x^2}}-\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {(6 i) \text {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {(6 i) \text {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ &=-\frac {6}{a^3 \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{a^2 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a^3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{a^2 \sqrt {1-a^2 x^2}}-\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {6 i \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 i \text {Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(541\) vs. \(2(246)=492\).
time = 0.63, size = 541, normalized size = 2.20 \begin {gather*} \frac {7 i \pi ^4-\frac {384}{\sqrt {1-a^2 x^2}}-8 \pi ^3 \tanh ^{-1}(a x)+\frac {384 a x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}+24 i \pi ^2 \tanh ^{-1}(a x)^2-\frac {192 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}+32 \pi \tanh ^{-1}(a x)^3+\frac {64 a x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-16 i \tanh ^{-1}(a x)^4-8 \pi ^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+48 i \pi ^2 \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+96 \pi \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-64 i \tanh ^{-1}(a x)^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-48 i \pi ^2 \tanh ^{-1}(a x) \log \left (1-i e^{\tanh ^{-1}(a x)}\right )-96 \pi \tanh ^{-1}(a x)^2 \log \left (1-i e^{\tanh ^{-1}(a x)}\right )+8 \pi ^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )+64 i \tanh ^{-1}(a x)^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )-8 \pi ^3 \log \left (\tan \left (\frac {1}{4} \left (\pi +2 i \tanh ^{-1}(a x)\right )\right )\right )-48 i \left (\pi -2 i \tanh ^{-1}(a x)\right )^2 \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )+192 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )-48 i \pi ^2 \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )-192 \pi \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )-192 \pi \text {PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )+384 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-384 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )+192 \pi \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )+384 i \text {PolyLog}\left (4,-i e^{-\tanh ^{-1}(a x)}\right )+384 i \text {PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{64 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((7*I)*Pi^4 - 384/Sqrt[1 - a^2*x^2] - 8*Pi^3*ArcTanh[a*x] + (384*a*x*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] + (24*I)*
Pi^2*ArcTanh[a*x]^2 - (192*ArcTanh[a*x]^2)/Sqrt[1 - a^2*x^2] + 32*Pi*ArcTanh[a*x]^3 + (64*a*x*ArcTanh[a*x]^3)/
Sqrt[1 - a^2*x^2] - (16*I)*ArcTanh[a*x]^4 - 8*Pi^3*Log[1 + I/E^ArcTanh[a*x]] + (48*I)*Pi^2*ArcTanh[a*x]*Log[1
+ I/E^ArcTanh[a*x]] + 96*Pi*ArcTanh[a*x]^2*Log[1 + I/E^ArcTanh[a*x]] - (64*I)*ArcTanh[a*x]^3*Log[1 + I/E^ArcTa
nh[a*x]] - (48*I)*Pi^2*ArcTanh[a*x]*Log[1 - I*E^ArcTanh[a*x]] - 96*Pi*ArcTanh[a*x]^2*Log[1 - I*E^ArcTanh[a*x]]
+ 8*Pi^3*Log[1 + I*E^ArcTanh[a*x]] + (64*I)*ArcTanh[a*x]^3*Log[1 + I*E^ArcTanh[a*x]] - 8*Pi^3*Log[Tan[(Pi + (
2*I)*ArcTanh[a*x])/4]] - (48*I)*(Pi - (2*I)*ArcTanh[a*x])^2*PolyLog[2, (-I)/E^ArcTanh[a*x]] + (192*I)*ArcTanh[
a*x]^2*PolyLog[2, (-I)*E^ArcTanh[a*x]] - (48*I)*Pi^2*PolyLog[2, I*E^ArcTanh[a*x]] - 192*Pi*ArcTanh[a*x]*PolyLo
g[2, I*E^ArcTanh[a*x]] - 192*Pi*PolyLog[3, (-I)/E^ArcTanh[a*x]] + (384*I)*ArcTanh[a*x]*PolyLog[3, (-I)/E^ArcTa
nh[a*x]] - (384*I)*ArcTanh[a*x]*PolyLog[3, (-I)*E^ArcTanh[a*x]] + 192*Pi*PolyLog[3, I*E^ArcTanh[a*x]] + (384*I
)*PolyLog[4, (-I)/E^ArcTanh[a*x]] + (384*I)*PolyLog[4, (-I)*E^ArcTanh[a*x]])/(64*a^3)

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \arctanh \left (a x \right )^{3}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*x^2*arctanh(a*x)^3/(a^4*x^4 - 2*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^{2} \operatorname {atanh}^{3}{\left (a x \right )}}{\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(x**2*atanh(a*x)**3/(-a**2*x**2+1)**(3/2),x)

[Out]

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

[Out]

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

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

[Out]

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

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