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

Optimal. Leaf size=90 \[ -\frac{3}{2} a \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )-3 a \tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{4} a \tanh ^{-1}(a x)^4+a \tanh ^{-1}(a x)^3-\frac{\tanh ^{-1}(a x)^3}{x}+3 a \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]

[Out]

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

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Rubi [A]  time = 0.272736, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {5982, 5916, 5988, 5932, 5948, 6056, 6610} \[ -\frac{3}{2} a \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )-3 a \tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{4} a \tanh ^{-1}(a x)^4+a \tanh ^{-1}(a x)^3-\frac{\tanh ^{-1}(a x)^3}{x}+3 a \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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^2 \left (1-a^2 x^2\right )} \, dx &=a^2 \int \frac{\tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx+\int \frac{\tanh ^{-1}(a x)^3}{x^2} \, dx\\ &=-\frac{\tanh ^{-1}(a x)^3}{x}+\frac{1}{4} a \tanh ^{-1}(a x)^4+(3 a) \int \frac{\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )} \, dx\\ &=a \tanh ^{-1}(a x)^3-\frac{\tanh ^{-1}(a x)^3}{x}+\frac{1}{4} a \tanh ^{-1}(a x)^4+(3 a) \int \frac{\tanh ^{-1}(a x)^2}{x (1+a x)} \, dx\\ &=a \tanh ^{-1}(a x)^3-\frac{\tanh ^{-1}(a x)^3}{x}+\frac{1}{4} a \tanh ^{-1}(a x)^4+3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )-\left (6 a^2\right ) \int \frac{\tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=a \tanh ^{-1}(a x)^3-\frac{\tanh ^{-1}(a x)^3}{x}+\frac{1}{4} a \tanh ^{-1}(a x)^4+3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )-3 a \tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )+\left (3 a^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=a \tanh ^{-1}(a x)^3-\frac{\tanh ^{-1}(a x)^3}{x}+\frac{1}{4} a \tanh ^{-1}(a x)^4+3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )-3 a \tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )-\frac{3}{2} a \text{Li}_3\left (-1+\frac{2}{1+a x}\right )\\ \end{align*}

Mathematica [C]  time = 0.193665, size = 93, normalized size = 1.03 \[ -a \left (-3 \tanh ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )-\frac{1}{4} \tanh ^{-1}(a x)^4+\frac{\tanh ^{-1}(a x)^3}{a x}+\tanh ^{-1}(a x)^3-3 \tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-\frac{i \pi ^3}{8}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(a*((-I/8)*Pi^3 + ArcTanh[a*x]^3 + ArcTanh[a*x]^3/(a*x) - ArcTanh[a*x]^4/4 - 3*ArcTanh[a*x]^2*Log[1 - E^(2*Ar
cTanh[a*x])] - 3*ArcTanh[a*x]*PolyLog[2, E^(2*ArcTanh[a*x])] + (3*PolyLog[3, E^(2*ArcTanh[a*x])])/2))

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Maple [C]  time = 0.343, size = 826, normalized size = 9.2 \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^2/(-a^2*x^2+1),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a x \log \left (-a x + 1\right )^{4} - 4 \,{\left (a x \log \left (a x + 1\right ) + 2 \, a x - 2\right )} \log \left (-a x + 1\right )^{3} + 6 \,{\left (a x \log \left (a x + 1\right )^{2} - 4 \,{\left (a x + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2}}{64 \, x} - \frac{1}{8} \, \int \frac{2 \, \log \left (a x + 1\right )^{3} + 3 \,{\left ({\left (a^{3} x^{3} + a^{2} x^{2} - 2\right )} \log \left (a x + 1\right )^{2} - 4 \,{\left (a^{3} x^{3} + 2 \, a^{2} x^{2} + a x\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )}{2 \,{\left (a^{2} x^{4} - x^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(a*x*log(-a*x + 1)^4 - 4*(a*x*log(a*x + 1) + 2*a*x - 2)*log(-a*x + 1)^3 + 6*(a*x*log(a*x + 1)^2 - 4*(a*x
+ 1)*log(a*x + 1))*log(-a*x + 1)^2)/x - 1/8*integrate(1/2*(2*log(a*x + 1)^3 + 3*((a^3*x^3 + a^2*x^2 - 2)*log(a
*x + 1)^2 - 4*(a^3*x^3 + 2*a^2*x^2 + a*x)*log(a*x + 1))*log(-a*x + 1))/(a^2*x^4 - x^2), 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^{4} - x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-arctanh(a*x)^3/(a^2*x^4 - x^2), 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^{4} - x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(atanh(a*x)**3/(a**2*x**4 - x**2), 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^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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