3.475 \(\int \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=302 \[ \frac {1}{2} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3+\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a}+\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}-\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}-\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{2 a}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{2 a}+\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {3 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {3 i \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {3 i \text {Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {\tanh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {5944, 5952, 4180, 2531, 6609, 2282, 6589, 5950} \[ \frac {3 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}-\frac {3 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}-\frac {3 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{2 a}+\frac {3 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{2 a}+\frac {3 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {3 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {3 i \text {PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {3 i \text {PolyLog}\left (4,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {1}{2} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3+\frac {3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{2 a}+\frac {\tanh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2282

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 2531

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 4180

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 5944

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

Rule 5950

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 5952

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 6589

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 6609

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

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Mathematica [A]  time = 4.30, size = 569, normalized size = 1.88 \[ -\frac {i \left (64 i a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3+192 i \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+192 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )+192 i \pi \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )+384 \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-384 \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )-48 \left (-4 \tanh ^{-1}(a x)^2-4 i \pi \tanh ^{-1}(a x)+\pi ^2+8\right ) \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )+384 \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )-48 \pi ^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )+192 i \pi \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-192 i \pi \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )+384 \text {Li}_4\left (-i e^{-\tanh ^{-1}(a x)}\right )+384 \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )-16 \tanh ^{-1}(a x)^4-32 i \pi \tanh ^{-1}(a x)^3+24 \pi ^2 \tanh ^{-1}(a x)^2+8 i \pi ^3 \tanh ^{-1}(a x)-64 \tanh ^{-1}(a x)^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+64 \tanh ^{-1}(a x)^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )-96 i \pi \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+96 i \pi \tanh ^{-1}(a x)^2 \log \left (1-i e^{\tanh ^{-1}(a x)}\right )-384 \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+48 \pi ^2 \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+384 \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-48 \pi ^2 \tanh ^{-1}(a x) \log \left (1-i e^{\tanh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-8 i \pi ^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (\tan \left (\frac {1}{4} \left (\pi +2 i \tanh ^{-1}(a x)\right )\right )\right )+7 \pi ^4\right )}{128 a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^3,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*(8 + Pi^2 - (4*I)*Pi*ArcTanh[a*x] - 4*ArcTa
nh[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*Arc
Tanh[a*x]*PolyLog[3, (-I)*E^ArcTanh[a*x]] - (192*I)*Pi*PolyLog[3, I*E^ArcTanh[a*x]] + 384*PolyLog[4, (-I)/E^Ar
cTanh[a*x]] + 384*PolyLog[4, (-I)*E^ArcTanh[a*x]]))/a

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.82, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} x^{2}+1}\, \arctanh \left (a x \right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {atanh}\left (a\,x\right )}^3\,\sqrt {1-a^2\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}^{3}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x)**3, x)

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