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

Optimal. Leaf size=200 \[ -\frac {3}{2} a^2 \text {Li}_2\left (\frac {2}{a x+1}-1\right )-\frac {3}{4} a^2 \text {Li}_4\left (\frac {2}{a x+1}-1\right )-\frac {3}{2} a^2 \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2-\frac {3}{2} a^2 \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)+\frac {1}{4} a^2 \tanh ^{-1}(a x)^4+\frac {1}{2} a^2 \tanh ^{-1}(a x)^3+\frac {3}{2} a^2 \tanh ^{-1}(a x)^2+a^2 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3+3 a^2 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)^3}{2 x^2}-\frac {3 a \tanh ^{-1}(a x)^2}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.50, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {5982, 5916, 5988, 5932, 2447, 5948, 6056, 6060, 6610} \[ -\frac {3}{2} a^2 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )-\frac {3}{4} a^2 \text {PolyLog}\left (4,\frac {2}{a x+1}-1\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x) \text {PolyLog}\left (3,\frac {2}{a x+1}-1\right )+\frac {1}{4} a^2 \tanh ^{-1}(a x)^4+\frac {1}{2} a^2 \tanh ^{-1}(a x)^3+\frac {3}{2} a^2 \tanh ^{-1}(a x)^2+a^2 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3+3 a^2 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)^3}{2 x^2}-\frac {3 a \tanh ^{-1}(a x)^2}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*ArcTanh[a*x]^2)/2 - (3*a*ArcTanh[a*x]^2)/(2*x) + (a^2*ArcTanh[a*x]^3)/2 - ArcTanh[a*x]^3/(2*x^2) + (a^2
*ArcTanh[a*x]^4)/4 + 3*a^2*ArcTanh[a*x]*Log[2 - 2/(1 + a*x)] + a^2*ArcTanh[a*x]^3*Log[2 - 2/(1 + a*x)] - (3*a^
2*PolyLog[2, -1 + 2/(1 + a*x)])/2 - (3*a^2*ArcTanh[a*x]^2*PolyLog[2, -1 + 2/(1 + a*x)])/2 - (3*a^2*ArcTanh[a*x
]*PolyLog[3, -1 + 2/(1 + a*x)])/2 - (3*a^2*PolyLog[4, -1 + 2/(1 + a*x)])/4

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

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 5982

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTanh[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -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^3 \left (1-a^2 x^2\right )} \, dx &=a^2 \int \frac {\tanh ^{-1}(a x)^3}{x \left (1-a^2 x^2\right )} \, dx+\int \frac {\tanh ^{-1}(a x)^3}{x^3} \, dx\\ &=-\frac {\tanh ^{-1}(a x)^3}{2 x^2}+\frac {1}{4} a^2 \tanh ^{-1}(a x)^4+\frac {1}{2} (3 a) \int \frac {\tanh ^{-1}(a x)^2}{x^2 \left (1-a^2 x^2\right )} \, dx+a^2 \int \frac {\tanh ^{-1}(a x)^3}{x (1+a x)} \, dx\\ &=-\frac {\tanh ^{-1}(a x)^3}{2 x^2}+\frac {1}{4} a^2 \tanh ^{-1}(a x)^4+a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )+\frac {1}{2} (3 a) \int \frac {\tanh ^{-1}(a x)^2}{x^2} \, dx+\frac {1}{2} \left (3 a^3\right ) \int \frac {\tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx-\left (3 a^3\right ) \int \frac {\tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {3 a \tanh ^{-1}(a x)^2}{2 x}+\frac {1}{2} a^2 \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{2 x^2}+\frac {1}{4} a^2 \tanh ^{-1}(a x)^4+a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )+\left (3 a^2\right ) \int \frac {\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx+\left (3 a^3\right ) \int \frac {\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac {3}{2} a^2 \tanh ^{-1}(a x)^2-\frac {3 a \tanh ^{-1}(a x)^2}{2 x}+\frac {1}{2} a^2 \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{2 x^2}+\frac {1}{4} a^2 \tanh ^{-1}(a x)^4+a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1+a x}\right )+\left (3 a^2\right ) \int \frac {\tanh ^{-1}(a x)}{x (1+a x)} \, dx+\frac {1}{2} \left (3 a^3\right ) \int \frac {\text {Li}_3\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac {3}{2} a^2 \tanh ^{-1}(a x)^2-\frac {3 a \tanh ^{-1}(a x)^2}{2 x}+\frac {1}{2} a^2 \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{2 x^2}+\frac {1}{4} a^2 \tanh ^{-1}(a x)^4+3 a^2 \tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )+a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1+a x}\right )-\frac {3}{4} a^2 \text {Li}_4\left (-1+\frac {2}{1+a x}\right )-\left (3 a^3\right ) \int \frac {\log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac {3}{2} a^2 \tanh ^{-1}(a x)^2-\frac {3 a \tanh ^{-1}(a x)^2}{2 x}+\frac {1}{2} a^2 \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{2 x^2}+\frac {1}{4} a^2 \tanh ^{-1}(a x)^4+3 a^2 \tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )+a^2 \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-\frac {3}{2} a^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {3}{2} a^2 \tanh ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1+a x}\right )-\frac {3}{4} a^2 \text {Li}_4\left (-1+\frac {2}{1+a x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 165, normalized size = 0.82 \[ -\frac {1}{64} a^2 \left (\frac {32 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}{a^2 x^2}-96 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )+96 \tanh ^{-1}(a x) \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )+96 \text {Li}_2\left (e^{-2 \tanh ^{-1}(a x)}\right )-48 \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )+16 \tanh ^{-1}(a x)^4+\frac {96 \tanh ^{-1}(a x)^2}{a x}-96 \tanh ^{-1}(a x)^2-64 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-192 \tanh ^{-1}(a x) \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )-\pi ^4\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/64*(a^2*(-Pi^4 - 96*ArcTanh[a*x]^2 + (96*ArcTanh[a*x]^2)/(a*x) + (32*(1 - a^2*x^2)*ArcTanh[a*x]^3)/(a^2*x^2
) + 16*ArcTanh[a*x]^4 - 192*ArcTanh[a*x]*Log[1 - E^(-2*ArcTanh[a*x])] - 64*ArcTanh[a*x]^3*Log[1 - E^(2*ArcTanh
[a*x])] + 96*PolyLog[2, E^(-2*ArcTanh[a*x])] - 96*ArcTanh[a*x]^2*PolyLog[2, E^(2*ArcTanh[a*x])] + 96*ArcTanh[a
*x]*PolyLog[3, E^(2*ArcTanh[a*x])] - 48*PolyLog[4, E^(2*ArcTanh[a*x])]))

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )^{3}}{a^{2} x^{5} - x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-arctanh(a*x)^3/(a^2*x^5 - x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 4.73, size = 406, normalized size = 2.03 \[ -\frac {a^{2} \arctanh \left (a x \right )^{4}}{4}+\frac {a^{2} \arctanh \left (a x \right )^{3}}{2}-\frac {3 a^{2} \arctanh \left (a x \right )^{2}}{2}-\frac {3 a \arctanh \left (a x \right )^{2}}{2 x}-\frac {\arctanh \left (a x \right )^{3}}{2 x^{2}}+a^{2} \arctanh \left (a x \right )^{3} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 a^{2} \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-6 a^{2} \arctanh \left (a x \right ) \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+6 a^{2} \polylog \left (4, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+a^{2} \arctanh \left (a x \right )^{3} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 a^{2} \arctanh \left (a x \right )^{2} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-6 a^{2} \arctanh \left (a x \right ) \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+6 a^{2} \polylog \left (4, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 a^{2} \arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 a^{2} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 a^{2} \arctanh \left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+3 a^{2} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*arctanh(a*x)^4+1/2*a^2*arctanh(a*x)^3-3/2*a^2*arctanh(a*x)^2-3/2*a*arctanh(a*x)^2/x-1/2*arctanh(a*x)^
3/x^2+a^2*arctanh(a*x)^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2*arctanh(a*x)^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)
^(1/2))-6*a^2*arctanh(a*x)*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+6*a^2*polylog(4,-(a*x+1)/(-a^2*x^2+1)^(1/2))
+a^2*arctanh(a*x)^3*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2*arctanh(a*x)^2*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2)
)-6*a^2*arctanh(a*x)*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+6*a^2*polylog(4,(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2*a
rctanh(a*x)*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^2*arctanh(a*x)*l
n(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a^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 \[ \frac {a^{2} x^{2} \log \left (-a x + 1\right )^{4} + 4 \, {\left (a^{2} x^{2} \log \left (a x + 1\right ) + 1\right )} \log \left (-a x + 1\right )^{3}}{64 \, x^{2}} - \frac {1}{8} \, \int \frac {2 \, \log \left (a x + 1\right )^{3} - 6 \, \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right ) + 3 \, {\left (a^{2} x^{2} + a x + {\left (a^{4} x^{4} + a^{3} x^{3} + 2\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2}}{2 \, {\left (a^{2} x^{5} - x^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(a^2*x^2*log(-a*x + 1)^4 + 4*(a^2*x^2*log(a*x + 1) + 1)*log(-a*x + 1)^3)/x^2 - 1/8*integrate(1/2*(2*log(a
*x + 1)^3 - 6*log(a*x + 1)^2*log(-a*x + 1) + 3*(a^2*x^2 + a*x + (a^4*x^4 + a^3*x^3 + 2)*log(a*x + 1))*log(-a*x
+ 1)^2)/(a^2*x^5 - x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{a^{2} x^{5} - x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(atanh(a*x)**3/(a**2*x**5 - x**3), x)

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