3.245 \(\int \frac{\tanh ^{-1}(a x)^3}{x (1-a^2 x^2)} \, dx\)
Optimal. Leaf size=91 \[ -\frac{3}{4} \text{PolyLog}\left (4,\frac{2}{a x+1}-1\right )-\frac{3}{2} \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-\frac{3}{2} \tanh ^{-1}(a x) \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )+\frac{1}{4} \tanh ^{-1}(a x)^4+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3 \]
[Out]
ArcTanh[a*x]^4/4 + ArcTanh[a*x]^3*Log[2 - 2/(1 + a*x)] - (3*ArcTanh[a*x]^2*PolyLog[2, -1 + 2/(1 + a*x)])/2 - (
3*ArcTanh[a*x]*PolyLog[3, -1 + 2/(1 + a*x)])/2 - (3*PolyLog[4, -1 + 2/(1 + a*x)])/4
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Rubi [A] time = 0.221187, antiderivative size = 91, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used =
{5988, 5932, 5948, 6056, 6060, 6610} \[ -\frac{3}{4} \text{PolyLog}\left (4,\frac{2}{a x+1}-1\right )-\frac{3}{2} \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-\frac{3}{2} \tanh ^{-1}(a x) \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )+\frac{1}{4} \tanh ^{-1}(a x)^4+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3 \]
Antiderivative was successfully verified.
[In]
Int[ArcTanh[a*x]^3/(x*(1 - a^2*x^2)),x]
[Out]
ArcTanh[a*x]^4/4 + ArcTanh[a*x]^3*Log[2 - 2/(1 + a*x)] - (3*ArcTanh[a*x]^2*PolyLog[2, -1 + 2/(1 + a*x)])/2 - (
3*ArcTanh[a*x]*PolyLog[3, -1 + 2/(1 + a*x)])/2 - (3*PolyLog[4, -1 + 2/(1 + a*x)])/4
Rule 5988
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]
Rule 5932
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 2/(1 + (e*x)
/d)])/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]
Rule 5948
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
Rule 6056
Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] - Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[2, 1 - u])
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2
/(1 + c*x))^2, 0]
Rule 6060
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a
+ b*ArcTanh[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[k
+ 1, u])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[u^2 -
(1 - 2/(1 + c*x))^2, 0]
Rule 6610
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
; !FalseQ[w]] /; FreeQ[n, x]
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(a x)^3}{x \left (1-a^2 x^2\right )} \, dx &=\frac{1}{4} \tanh ^{-1}(a x)^4+\int \frac{\tanh ^{-1}(a x)^3}{x (1+a x)} \, dx\\ &=\frac{1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )-(3 a) \int \frac{\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac{1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )-\frac{3}{2} \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )+(3 a) \int \frac{\tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac{1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )-\frac{3}{2} \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )-\frac{3}{2} \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )+\frac{1}{2} (3 a) \int \frac{\text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac{1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )-\frac{3}{2} \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )-\frac{3}{2} \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )-\frac{3}{4} \text{Li}_4\left (-1+\frac{2}{1+a x}\right )\\ \end{align*}
Mathematica [A] time = 0.0734301, size = 83, normalized size = 0.91 \[ \frac{3}{2} \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )-\frac{3}{2} \tanh ^{-1}(a x) \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )+\frac{3}{4} \text{PolyLog}\left (4,e^{2 \tanh ^{-1}(a x)}\right )-\frac{1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right ) \]
Warning: Unable to verify antiderivative.
[In]
Integrate[ArcTanh[a*x]^3/(x*(1 - a^2*x^2)),x]
[Out]
-ArcTanh[a*x]^4/4 + ArcTanh[a*x]^3*Log[1 - E^(2*ArcTanh[a*x])] + (3*ArcTanh[a*x]^2*PolyLog[2, E^(2*ArcTanh[a*x
])])/2 - (3*ArcTanh[a*x]*PolyLog[3, E^(2*ArcTanh[a*x])])/2 + (3*PolyLog[4, E^(2*ArcTanh[a*x])])/4
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Maple [C] time = 0.349, size = 1245, normalized size = 13.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(arctanh(a*x)^3/x/(-a^2*x^2+1),x)
[Out]
-1/4*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x
)^3+1/4*I*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*arctanh(a*x)^3+1/2*I*Pi*csgn(I
*(a*x+1)/(-a^2*x^2+1)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*arctanh(a*x)^3-1/2*I*arctanh(a*x)^3*Pi*csgn(I/((a
*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2+1/4*I*Pi*csgn(I/((a*x
+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^3+1/2*I*Pi*csgn
(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^3+1/2*I*arctanh(a*x)^3*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*
x+1)^2/(-a^2*x^2+1)+1))^3+1/4*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^3+1
/4*I*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^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))+1/2*I*Pi*arctanh(a*x)^3+arctanh(a*x)^3*ln(1-(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))+ln(2)*arctanh(a*x)^3-1/4*arctanh(a*x)^4-1/2*I*P
i*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^3+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))-1/4*I*arctanh(a*x)^3*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(a^2*x^2-
1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))+1/2*I*arctanh(a*x)^3*Pi*csgn(I*((a*x+1)^2/(-a^2*x
^2+1)-1))*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))-1/2
*arctanh(a*x)^3*ln(a*x-1)-1/2*arctanh(a*x)^3*ln(a*x+1)+arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))-1/2*I*arc
tanh(a*x)^3*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1)
)^2-arctanh(a*x)^3*ln((a*x+1)^2/(-a^2*x^2+1)-1)+arctanh(a*x)^3*ln(a*x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{16} \, \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{3} + \frac{1}{64} \, \log \left (-a x + 1\right )^{4} - \frac{1}{8} \, \int \frac{3 \,{\left (a^{2} x^{2} + a x + 2\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2} + 2 \, \log \left (a x + 1\right )^{3} - 6 \, \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )}{2 \,{\left (a^{2} x^{3} - x\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^3/x/(-a^2*x^2+1),x, algorithm="maxima")
[Out]
1/16*log(a*x + 1)*log(-a*x + 1)^3 + 1/64*log(-a*x + 1)^4 - 1/8*integrate(1/2*(3*(a^2*x^2 + a*x + 2)*log(a*x +
1)*log(-a*x + 1)^2 + 2*log(a*x + 1)^3 - 6*log(a*x + 1)^2*log(-a*x + 1))/(a^2*x^3 - x), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\operatorname{artanh}\left (a x\right )^{3}}{a^{2} x^{3} - x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^3/x/(-a^2*x^2+1),x, algorithm="fricas")
[Out]
integral(-arctanh(a*x)^3/(a^2*x^3 - x), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\operatorname{atanh}^{3}{\left (a x \right )}}{a^{2} x^{3} - x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(atanh(a*x)**3/x/(-a**2*x**2+1),x)
[Out]
-Integral(atanh(a*x)**3/(a**2*x**3 - x), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\operatorname{artanh}\left (a x\right )^{3}}{{\left (a^{2} x^{2} - 1\right )} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^3/x/(-a^2*x^2+1),x, algorithm="giac")
[Out]
integrate(-arctanh(a*x)^3/((a^2*x^2 - 1)*x), x)