3.4.81 \(\int \frac {x^2 \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx\) [381]

Optimal. Leaf size=305 \[ -\frac {6 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}+\frac {\text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}-\frac {3 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{2 a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{2 a^3}-\frac {3 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \text {PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {3 i \text {PolyLog}\left (4,i e^{\tanh ^{-1}(a x)}\right )}{a^3} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 305, 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 = {6163, 6141, 6097, 6099, 4265, 2611, 6744, 2320, 6724} \begin {gather*} \frac {\tanh ^{-1}(a x)^3 \text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {6 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^3}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{2 a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{2 a^3}+\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {3 i \text {Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}-\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/a^3 - (3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(2*a^3) - (x*
Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^3)/(2*a^2) + (ArcTan[E^ArcTanh[a*x]]*ArcTanh[a*x]^3)/a^3 - (((3*I)/2)*ArcTanh[a
*x]^2*PolyLog[2, (-I)*E^ArcTanh[a*x]])/a^3 + (((3*I)/2)*ArcTanh[a*x]^2*PolyLog[2, I*E^ArcTanh[a*x]])/a^3 - ((3
*I)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3 + ((3*I)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/
a^3 + ((3*I)*ArcTanh[a*x]*PolyLog[3, (-I)*E^ArcTanh[a*x]])/a^3 - ((3*I)*ArcTanh[a*x]*PolyLog[3, I*E^ArcTanh[a*
x]])/a^3 - ((3*I)*PolyLog[4, (-I)*E^ArcTanh[a*x]])/a^3 + ((3*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 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[-2*(a + b*ArcTanh[c*x])*(
ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]]/(c*Sqrt[d])), x] + (-Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 - c*x]/Sqrt[1 + c*x
])]/(c*Sqrt[d])), x] + Simp[I*b*(PolyLog[2, I*(Sqrt[1 - c*x]/Sqrt[1 + c*x])]/(c*Sqrt[d])), x]) /; FreeQ[{a, b,
c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 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 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 6163

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(-f)*(f*x)^(m - 1)*Sqrt[d + e*x^2]*((a + b*ArcTanh[c*x])^p/(c^2*d*m)), x] + (Dist[b*f*(p/(c*m)), Int[(f*x)^(m
- 1)*((a + b*ArcTanh[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] + Dist[f^2*((m - 1)/(c^2*m)), Int[(f*x)^(m - 2)*(
(a + b*ArcTanh[c*x])^p/Sqrt[d + e*x^2]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[p
, 0] && GtQ[m, 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}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}+\frac {\int \frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx}{2 a^2}+\frac {3 \int \frac {x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{2 a}\\ &=-\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}+\frac {\text {Subst}\left (\int x^3 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^3}+\frac {3 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac {6 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a^3}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {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 )}{2 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 )}{2 a^3}\\ &=-\frac {6 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}+\frac {\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 )}{2 a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{2 a^3}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {(3 i) \text {Subst}\left (\int x \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}-\frac {(3 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 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}+\frac {\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 )}{2 a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{2 a^3}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {(3 i) \text {Subst}\left (\int \text {Li}_3\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}+\frac {(3 i) \text {Subst}\left (\int \text {Li}_3\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^3}\\ &=-\frac {6 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}+\frac {\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 )}{2 a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{2 a^3}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {(3 i) \text {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {(3 i) \text {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ &=-\frac {6 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{2 a^2}+\frac {\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 )}{2 a^3}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{2 a^3}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^3}+\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}-\frac {3 i \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a^3}+\frac {3 i \text {Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a^3}\\ \end {align*}

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Mathematica [A]
time = 2.77, size = 570, normalized size = 1.87 \begin {gather*} -\frac {i \left (7 \pi ^4+8 i \pi ^3 \tanh ^{-1}(a x)+24 \pi ^2 \tanh ^{-1}(a x)^2-192 i \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-32 i \pi \tanh ^{-1}(a x)^3-64 i a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3-16 \tanh ^{-1}(a x)^4+384 \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-384 \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+48 \pi ^2 \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-96 i \pi \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-64 \tanh ^{-1}(a x)^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-48 \pi ^2 \tanh ^{-1}(a x) \log \left (1-i e^{\tanh ^{-1}(a x)}\right )+96 i \pi \tanh ^{-1}(a x)^2 \log \left (1-i e^{\tanh ^{-1}(a x)}\right )-8 i \pi ^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )+64 \tanh ^{-1}(a x)^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (\tan \left (\frac {1}{4} \left (\pi +2 i \tanh ^{-1}(a x)\right )\right )\right )-48 \left (\pi ^2-4 i \pi \tanh ^{-1}(a x)-4 \left (2+\tanh ^{-1}(a x)^2\right )\right ) \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-384 \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+192 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )-48 \pi ^2 \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )+192 i \pi \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )+192 i \pi \text {PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )+384 \tanh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-384 \tanh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )-192 i \pi \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )+384 \text {PolyLog}\left (4,-i e^{-\tanh ^{-1}(a x)}\right )+384 \text {PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )\right )}{128 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/128*I)*(7*Pi^4 + (8*I)*Pi^3*ArcTanh[a*x] + 24*Pi^2*ArcTanh[a*x]^2 - (192*I)*Sqrt[1 - a^2*x^2]*ArcTanh[a*x
]^2 - (32*I)*Pi*ArcTanh[a*x]^3 - (64*I)*a*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^3 - 16*ArcTanh[a*x]^4 + 384*ArcTanh
[a*x]*Log[1 - I/E^ArcTanh[a*x]] + (8*I)*Pi^3*Log[1 + I/E^ArcTanh[a*x]] - 384*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[
a*x]] + 48*Pi^2*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] - (96*I)*Pi*ArcTanh[a*x]^2*Log[1 + I/E^ArcTanh[a*x]] -
64*ArcTanh[a*x]^3*Log[1 + I/E^ArcTanh[a*x]] - 48*Pi^2*ArcTanh[a*x]*Log[1 - I*E^ArcTanh[a*x]] + (96*I)*Pi*ArcTa
nh[a*x]^2*Log[1 - I*E^ArcTanh[a*x]] - (8*I)*Pi^3*Log[1 + I*E^ArcTanh[a*x]] + 64*ArcTanh[a*x]^3*Log[1 + I*E^Arc
Tanh[a*x]] + (8*I)*Pi^3*Log[Tan[(Pi + (2*I)*ArcTanh[a*x])/4]] - 48*(Pi^2 - (4*I)*Pi*ArcTanh[a*x] - 4*(2 + ArcT
anh[a*x]^2))*PolyLog[2, (-I)/E^ArcTanh[a*x]] - 384*PolyLog[2, I/E^ArcTanh[a*x]] + 192*ArcTanh[a*x]^2*PolyLog[2
, (-I)*E^ArcTanh[a*x]] - 48*Pi^2*PolyLog[2, I*E^ArcTanh[a*x]] + (192*I)*Pi*ArcTanh[a*x]*PolyLog[2, I*E^ArcTanh
[a*x]] + (192*I)*Pi*PolyLog[3, (-I)/E^ArcTanh[a*x]] + 384*ArcTanh[a*x]*PolyLog[3, (-I)/E^ArcTanh[a*x]] - 384*A
rcTanh[a*x]*PolyLog[3, (-I)*E^ArcTanh[a*x]] - (192*I)*Pi*PolyLog[3, I*E^ArcTanh[a*x]] + 384*PolyLog[4, (-I)/E^
ArcTanh[a*x]] + 384*PolyLog[4, (-I)*E^ArcTanh[a*x]]))/a^3

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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \arctanh \left (a x \right )^{3}}{\sqrt {-a^{2} x^{2}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

integrate(x^2*arctanh(a*x)^3/sqrt(-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^2\,{\mathrm {atanh}\left (a\,x\right )}^3}{\sqrt {1-a^2\,x^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)^(1/2),x)

[Out]

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

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