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

Optimal. Leaf size=102 \[ -3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )-6 \text {Li}_4\left (-e^{\tanh ^{-1}(a x)}\right )+6 \text {Li}_4\left (e^{\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3 \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6020, 4182, 2531, 6609, 2282, 6589} \[ -3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-e^{\tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,e^{\tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {PolyLog}\left (3,-e^{\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \text {PolyLog}\left (3,e^{\tanh ^{-1}(a x)}\right )-6 \text {PolyLog}\left (4,-e^{\tanh ^{-1}(a x)}\right )+6 \text {PolyLog}\left (4,e^{\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*ArcTanh[E^ArcTanh[a*x]]*ArcTanh[a*x]^3 - 3*ArcTanh[a*x]^2*PolyLog[2, -E^ArcTanh[a*x]] + 3*ArcTanh[a*x]^2*Po
lyLog[2, E^ArcTanh[a*x]] + 6*ArcTanh[a*x]*PolyLog[3, -E^ArcTanh[a*x]] - 6*ArcTanh[a*x]*PolyLog[3, E^ArcTanh[a*
x]] - 6*PolyLog[4, -E^ArcTanh[a*x]] + 6*PolyLog[4, 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 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]

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

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Mathematica [A]  time = 0.14, size = 146, normalized size = 1.43 \[ \frac {1}{8} \left (24 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-e^{-\tanh ^{-1}(a x)}\right )+24 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+48 \tanh ^{-1}(a x) \text {Li}_3\left (-e^{-\tanh ^{-1}(a x)}\right )-48 \tanh ^{-1}(a x) \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )+48 \text {Li}_4\left (-e^{-\tanh ^{-1}(a x)}\right )+48 \text {Li}_4\left (e^{\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}(a x)^4-8 \tanh ^{-1}(a x)^3 \log \left (e^{-\tanh ^{-1}(a x)}+1\right )+8 \tanh ^{-1}(a x)^3 \log \left (1-e^{\tanh ^{-1}(a x)}\right )+\pi ^4\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Pi^4 - 2*ArcTanh[a*x]^4 - 8*ArcTanh[a*x]^3*Log[1 + E^(-ArcTanh[a*x])] + 8*ArcTanh[a*x]^3*Log[1 - E^ArcTanh[a*
x]] + 24*ArcTanh[a*x]^2*PolyLog[2, -E^(-ArcTanh[a*x])] + 24*ArcTanh[a*x]^2*PolyLog[2, E^ArcTanh[a*x]] + 48*Arc
Tanh[a*x]*PolyLog[3, -E^(-ArcTanh[a*x])] - 48*ArcTanh[a*x]*PolyLog[3, E^ArcTanh[a*x]] + 48*PolyLog[4, -E^(-Arc
Tanh[a*x])] + 48*PolyLog[4, E^ArcTanh[a*x]])/8

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{3} - x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.46, size = 215, normalized size = 2.11 \[ -\arctanh \left (a x \right )^{3} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-3 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+6 \arctanh \left (a x \right ) \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-6 \polylog \left (4, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+\arctanh \left (a x \right )^{3} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 \arctanh \left (a x \right )^{2} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-6 \arctanh \left (a x \right ) \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+6 \polylog \left (4, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctanh(a*x)^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-3*arctanh(a*x)^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+6*arc
tanh(a*x)*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))-6*polylog(4,-(a*x+1)/(-a^2*x^2+1)^(1/2))+arctanh(a*x)^3*ln(1-
(a*x+1)/(-a^2*x^2+1)^(1/2))+3*arctanh(a*x)^2*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-6*arctanh(a*x)*polylog(3,(a
*x+1)/(-a^2*x^2+1)^(1/2))+6*polylog(4,(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 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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