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

Optimal. Leaf size=98 \[ -\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{x}-6 a \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+6 a \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+6 a \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-6 a \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )-6 a \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2 \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6008, 6020, 4182, 2531, 2282, 6589} \[ -6 a \tanh ^{-1}(a x) \text {PolyLog}\left (2,-e^{\tanh ^{-1}(a x)}\right )+6 a \tanh ^{-1}(a x) \text {PolyLog}\left (2,e^{\tanh ^{-1}(a x)}\right )+6 a \text {PolyLog}\left (3,-e^{\tanh ^{-1}(a x)}\right )-6 a \text {PolyLog}\left (3,e^{\tanh ^{-1}(a x)}\right )-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{x}-6 a \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-6*a*ArcTanh[E^ArcTanh[a*x]]*ArcTanh[a*x]^2 - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^3)/x - 6*a*ArcTanh[a*x]*PolyLog[
2, -E^ArcTanh[a*x]] + 6*a*ArcTanh[a*x]*PolyLog[2, E^ArcTanh[a*x]] + 6*a*PolyLog[3, -E^ArcTanh[a*x]] - 6*a*Poly
Log[3, E^ArcTanh[a*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 4182

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 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 6020

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

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Mathematica [A]  time = 0.39, size = 131, normalized size = 1.34 \[ a \left (-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^3}{a x}+6 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{-\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {Li}_2\left (e^{-\tanh ^{-1}(a x)}\right )+6 \text {Li}_3\left (-e^{-\tanh ^{-1}(a x)}\right )-6 \text {Li}_3\left (e^{-\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-3 \tanh ^{-1}(a x)^2 \log \left (e^{-\tanh ^{-1}(a x)}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{4} - x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.43, size = 190, normalized size = 1.94 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )^{3}}{x}-3 a \arctanh \left (a x \right )^{2} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-6 a \arctanh \left (a x \right ) \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+6 a \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 a \arctanh \left (a x \right )^{2} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+6 a \arctanh \left (a x \right ) \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-6 a \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(a*x-1)*(a*x+1))^(1/2)*arctanh(a*x)^3/x-3*a*arctanh(a*x)^2*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-6*a*arctanh(a*x
)*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+6*a*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a*arctanh(a*x)^2*ln(1-(a
*x+1)/(-a^2*x^2+1)^(1/2))+6*a*arctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-6*a*polylog(3,(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 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^3}{x^2\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(atanh(a*x)^3/(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 \[ \int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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