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

Optimal. Leaf size=91 \[ -\frac {3}{4} \text {Li}_4\left (\frac {2}{a x+1}-1\right )-\frac {3}{2} \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2-\frac {3}{2} \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)+\frac {1}{4} \tanh ^{-1}(a x)^4+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3 \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [A]  time = 0.07, size = 83, normalized size = 0.91 \[ \frac {3}{2} \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )-\frac {3}{2} \tanh ^{-1}(a x) \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {3}{4} \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )-\frac {1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*ArcTanh[a*x]^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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fricas [F]  time = 1.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{3} - x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.57, size = 1245, normalized size = 13.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*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))+arcta
nh(a*x)^3*ln(a*x)-1/4*arctanh(a*x)^4+1/2*I*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*arctanh(a*x)^3-1/2*I*Pi*csg
n(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*arctanh(a*x)^3+1/4*I*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3+1/4*I
*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+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)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3-1/4*I*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*
x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))*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/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/
(1+(a*x+1)^2/(-a^2*x^2+1)))+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))-6
*arctanh(a*x)*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))-arctanh(a*x)^3*ln((a*x+1)^2/(-a^2*x^2+1)-1)+ln(2)*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))+1/2*I*
Pi*arctanh(a*x)^3-6*arctanh(a*x)*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+1/4*I*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+
1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+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*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1
)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2-1/4*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(
1+(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)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2-1/2*I*arctanh(a*x)^3*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2
+1)))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2+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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{3} + \frac {1}{64} \, \log \left (-a x + 1\right )^{4} - \frac {1}{8} \, \int \frac {3 \, {\left (a^{2} x^{2} + a x + 2\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2} + 2 \, \log \left (a x + 1\right )^{3} - 6 \, \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )}{2 \, {\left (a^{2} x^{3} - x\right )}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{a^{2} x^{3} - x}\, dx \]

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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