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

Optimal. Leaf size=153 \[ -\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {6 i \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 i \text {Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 \tanh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5952, 4180, 2531, 6609, 2282, 6589} \[ -\frac {3 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {3 i \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {6 i \tanh ^{-1}(a x) \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {6 i \text {PolyLog}\left (4,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 i \text {PolyLog}\left (4,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 \tanh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[E^ArcTanh[a*x]]*ArcTanh[a*x]^3)/a - ((3*I)*ArcTanh[a*x]^2*PolyLog[2, (-I)*E^ArcTanh[a*x]])/a + ((3*I
)*ArcTanh[a*x]^2*PolyLog[2, I*E^ArcTanh[a*x]])/a + ((6*I)*ArcTanh[a*x]*PolyLog[3, (-I)*E^ArcTanh[a*x]])/a - ((
6*I)*ArcTanh[a*x]*PolyLog[3, I*E^ArcTanh[a*x]])/a - ((6*I)*PolyLog[4, (-I)*E^ArcTanh[a*x]])/a + ((6*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 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 \frac {\tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \, dx &=\frac {\operatorname {Subst}\left (\int x^3 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^3}{a}-\frac {(3 i) \operatorname {Subst}\left (\int x^2 \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-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 )}{a}\\ &=\frac {2 \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 )}{a}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {(6 i) \operatorname {Subst}\left (\int x \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}-\frac {(6 i) \operatorname {Subst}\left (\int x \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {2 \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 )}{a}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_3\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {(6 i) \operatorname {Subst}\left (\int \text {Li}_3\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {2 \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 )}{a}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a}\\ &=\frac {2 \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 )}{a}+\frac {3 i \tanh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {6 i \tanh ^{-1}(a x) \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {6 i \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {6 i \text {Li}_4\left (i e^{\tanh ^{-1}(a x)}\right )}{a}\\ \end {align*}

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Mathematica [B]  time = 0.44, size = 451, normalized size = 2.95 \[ -\frac {i \left (192 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )+192 i \pi \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )+384 \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-384 \tanh ^{-1}(a x) \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )-48 \left (\pi -2 i \tanh ^{-1}(a x)\right )^2 \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-48 \pi ^2 \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )+192 i \pi \text {Li}_3\left (-i e^{-\tanh ^{-1}(a x)}\right )-192 i \pi \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )+384 \text {Li}_4\left (-i e^{-\tanh ^{-1}(a x)}\right )+384 \text {Li}_4\left (-i e^{\tanh ^{-1}(a x)}\right )-16 \tanh ^{-1}(a x)^4-32 i \pi \tanh ^{-1}(a x)^3+24 \pi ^2 \tanh ^{-1}(a x)^2+8 i \pi ^3 \tanh ^{-1}(a x)-64 \tanh ^{-1}(a x)^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+64 \tanh ^{-1}(a x)^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )-96 i \pi \tanh ^{-1}(a x)^2 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )+96 i \pi \tanh ^{-1}(a x)^2 \log \left (1-i e^{\tanh ^{-1}(a x)}\right )+48 \pi ^2 \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-48 \pi ^2 \tanh ^{-1}(a x) \log \left (1-i e^{\tanh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-8 i \pi ^3 \log \left (1+i e^{\tanh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (\tan \left (\frac {1}{4} \left (\pi +2 i \tanh ^{-1}(a x)\right )\right )\right )+7 \pi ^4\right )}{64 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/64*I)*(7*Pi^4 + (8*I)*Pi^3*ArcTanh[a*x] + 24*Pi^2*ArcTanh[a*x]^2 - (32*I)*Pi*ArcTanh[a*x]^3 - 16*ArcTanh[
a*x]^4 + (8*I)*Pi^3*Log[1 + I/E^ArcTanh[a*x]] + 48*Pi^2*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] - (96*I)*Pi*Arc
Tanh[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*ArcTanh[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^ArcTanh[a*x]] + (8*I)*Pi^3*Log[Tan[(Pi + (2*I)*ArcTanh[a*x])/4]] - 48*(Pi
- (2*I)*ArcTanh[a*x])^2*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*Pol
yLog[3, (-I)/E^ArcTanh[a*x]] + 384*ArcTanh[a*x]*PolyLog[3, (-I)/E^ArcTanh[a*x]] - 384*ArcTanh[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*Pol
yLog[4, (-I)*E^ArcTanh[a*x]]))/a

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {\arctanh \left (a x \right )^{3}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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