3.438 \(\int x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=336 \[ -\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}-\frac {i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}-\frac {19 \sin ^{-1}(a x)}{360 a^5}+\frac {\tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}+\frac {x^3 \sqrt {1-a^2 x^2}}{60 a^2}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{360 a^5}+\frac {x \sqrt {1-a^2 x^2}}{18 a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{16 a^4}+\frac {11 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{180 a^3} \]

[Out]

-19/360*arcsin(a*x)/a^5+1/8*arctan((a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2/a^5-1/8*I*arctanh(a*x)*polylog(2
,-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^5+1/8*I*arctanh(a*x)*polylog(2,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^5+1/8*I*polyl
og(3,-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^5-1/8*I*polylog(3,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^5+1/18*x*(-a^2*x^2+1)^
(1/2)/a^4+1/60*x^3*(-a^2*x^2+1)^(1/2)/a^2-1/360*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^5+11/180*x^2*arctanh(a*x)*(-
a^2*x^2+1)^(1/2)/a^3+1/15*x^4*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a-1/16*x*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/a^4-1
/24*x^3*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/a^2+1/6*x^5*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.40, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 45, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6014, 6016, 321, 216, 5994, 5952, 4180, 2531, 2282, 6589} \[ -\frac {i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}-\frac {i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {x^3 \sqrt {1-a^2 x^2}}{60 a^2}+\frac {x \sqrt {1-a^2 x^2}}{18 a^4}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}+\frac {11 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{180 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{16 a^4}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{360 a^5}-\frac {19 \sin ^{-1}(a x)}{360 a^5}+\frac {\tanh ^{-1}(a x)^2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{8 a^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2,x]

[Out]

(x*Sqrt[1 - a^2*x^2])/(18*a^4) + (x^3*Sqrt[1 - a^2*x^2])/(60*a^2) - (19*ArcSin[a*x])/(360*a^5) - (Sqrt[1 - a^2
*x^2]*ArcTanh[a*x])/(360*a^5) + (11*x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(180*a^3) + (x^4*Sqrt[1 - a^2*x^2]*Arc
Tanh[a*x])/(15*a) - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(16*a^4) - (x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(24
*a^2) + (x^5*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/6 + (ArcTan[E^ArcTanh[a*x]]*ArcTanh[a*x]^2)/(8*a^5) - ((I/8)*Ar
cTanh[a*x]*PolyLog[2, (-I)*E^ArcTanh[a*x]])/a^5 + ((I/8)*ArcTanh[a*x]*PolyLog[2, I*E^ArcTanh[a*x]])/a^5 + ((I/
8)*PolyLog[3, (-I)*E^ArcTanh[a*x]])/a^5 - ((I/8)*PolyLog[3, I*E^ArcTanh[a*x]])/a^5

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 321

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

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 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 5994

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 6014

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

Rule 6016

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Sim
p[(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 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]

Rubi steps

\begin {align*} \int x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx &=-\left (a^2 \int \frac {x^6 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\right )+\int \frac {x^4 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{4 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {5}{6} \int \frac {x^4 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx+\frac {3 \int \frac {x^2 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{4 a^2}+\frac {\int \frac {x^3 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{2 a}-\frac {1}{3} a \int \frac {x^5 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{6 a^3}+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{8 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {1}{15} \int \frac {x^4}{\sqrt {1-a^2 x^2}} \, dx+\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{8 a^4}+\frac {\int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{3 a^3}+\frac {3 \int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{4 a^3}+\frac {\int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{6 a^2}-\frac {5 \int \frac {x^2 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}-\frac {4 \int \frac {x^3 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{15 a}-\frac {5 \int \frac {x^3 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{12 a}\\ &=-\frac {x \sqrt {1-a^2 x^2}}{12 a^4}+\frac {x^3 \sqrt {1-a^2 x^2}}{60 a^2}-\frac {13 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^5}+\frac {11 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{180 a^3}+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {3 \operatorname {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^5}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{12 a^4}-\frac {5 \int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{16 a^4}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{3 a^4}+\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{4 a^4}-\frac {8 \int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{45 a^3}-\frac {5 \int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{18 a^3}-\frac {5 \int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{8 a^3}-\frac {\int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{20 a^2}-\frac {4 \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{45 a^2}-\frac {5 \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{36 a^2}\\ &=\frac {x \sqrt {1-a^2 x^2}}{18 a^4}+\frac {x^3 \sqrt {1-a^2 x^2}}{60 a^2}+\frac {7 \sin ^{-1}(a x)}{6 a^5}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{360 a^5}+\frac {11 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{180 a^3}+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{4 a^5}-\frac {(3 i) \operatorname {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^5}+\frac {(3 i) \operatorname {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^5}-\frac {5 \operatorname {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{16 a^5}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{40 a^4}-\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{45 a^4}-\frac {5 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{72 a^4}-\frac {8 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{45 a^4}-\frac {5 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{18 a^4}-\frac {5 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a^4}\\ &=\frac {x \sqrt {1-a^2 x^2}}{18 a^4}+\frac {x^3 \sqrt {1-a^2 x^2}}{60 a^2}-\frac {19 \sin ^{-1}(a x)}{360 a^5}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{360 a^5}+\frac {11 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{180 a^3}+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{8 a^5}-\frac {3 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^5}+\frac {3 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^5}+\frac {(5 i) \operatorname {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^5}-\frac {(5 i) \operatorname {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^5}+\frac {(3 i) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^5}-\frac {(3 i) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^5}\\ &=\frac {x \sqrt {1-a^2 x^2}}{18 a^4}+\frac {x^3 \sqrt {1-a^2 x^2}}{60 a^2}-\frac {19 \sin ^{-1}(a x)}{360 a^5}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{360 a^5}+\frac {11 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{180 a^3}+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{8 a^5}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}-\frac {(5 i) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^5}+\frac {(5 i) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^5}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{4 a^5}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{4 a^5}\\ &=\frac {x \sqrt {1-a^2 x^2}}{18 a^4}+\frac {x^3 \sqrt {1-a^2 x^2}}{60 a^2}-\frac {19 \sin ^{-1}(a x)}{360 a^5}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{360 a^5}+\frac {11 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{180 a^3}+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{8 a^5}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {3 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{4 a^5}-\frac {3 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{4 a^5}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{8 a^5}\\ &=\frac {x \sqrt {1-a^2 x^2}}{18 a^4}+\frac {x^3 \sqrt {1-a^2 x^2}}{60 a^2}-\frac {19 \sin ^{-1}(a x)}{360 a^5}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{360 a^5}+\frac {11 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{180 a^3}+\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {\tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{8 a^5}-\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}+\frac {i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}-\frac {i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{8 a^5}\\ \end {align*}

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Mathematica [A]  time = 1.69, size = 268, normalized size = 0.80 \[ \frac {\sqrt {1-a^2 x^2} \left (-\frac {i \left (90 \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-90 \tanh ^{-1}(a x) \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+90 \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-90 \text {Li}_3\left (i e^{-\tanh ^{-1}(a x)}\right )+45 \tanh ^{-1}(a x)^2 \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-45 \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-76 i \tan ^{-1}\left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )\right )}{\sqrt {1-a^2 x^2}}+120 a x \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2+48 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)+140 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)+6 a x \left (a^2 x^2-1\right ) \left (35 \tanh ^{-1}(a x)^2+2\right )+90 \tanh ^{-1}(a x)+a x \left (45 \tanh ^{-1}(a x)^2+52\right )\right )}{720 a^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2,x]

[Out]

(Sqrt[1 - a^2*x^2]*(90*ArcTanh[a*x] + 140*(-1 + a^2*x^2)*ArcTanh[a*x] + 48*(-1 + a^2*x^2)^2*ArcTanh[a*x] + 120
*a*x*(-1 + a^2*x^2)^2*ArcTanh[a*x]^2 + 6*a*x*(-1 + a^2*x^2)*(2 + 35*ArcTanh[a*x]^2) + a*x*(52 + 45*ArcTanh[a*x
]^2) - (I*((-76*I)*ArcTan[Tanh[ArcTanh[a*x]/2]] + 45*ArcTanh[a*x]^2*Log[1 - I/E^ArcTanh[a*x]] - 45*ArcTanh[a*x
]^2*Log[1 + I/E^ArcTanh[a*x]] + 90*ArcTanh[a*x]*PolyLog[2, (-I)/E^ArcTanh[a*x]] - 90*ArcTanh[a*x]*PolyLog[2, I
/E^ArcTanh[a*x]] + 90*PolyLog[3, (-I)/E^ArcTanh[a*x]] - 90*PolyLog[3, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/
(720*a^5)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a^{2} x^{2} + 1} x^{4} \operatorname {artanh}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} x^{2} + 1} x^{4} \operatorname {artanh}\left (a x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-a^2*x^2 + 1)*x^4*arctanh(a*x)^2, x)

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \[ \int x^{4} \arctanh \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2),x)

[Out]

int(x^4*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} x^{2} + 1} x^{4} \operatorname {artanh}\left (a x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*x^2 + 1)*x^4*arctanh(a*x)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,{\mathrm {atanh}\left (a\,x\right )}^2\,\sqrt {1-a^2\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*atanh(a*x)^2*(1 - a^2*x^2)^(1/2),x)

[Out]

int(x^4*atanh(a*x)^2*(1 - a^2*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}^{2}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*atanh(a*x)**2*(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**4*sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x)**2, x)

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