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

Optimal. Leaf size=135 \[ -\frac{\text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^4}-\frac{\log \left (1-a^2 x^2\right )}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^2}{2 a^2}-\frac{\tanh ^{-1}(a x)^3}{3 a^4}+\frac{\tanh ^{-1}(a x)^2}{2 a^4}-\frac{x \tanh ^{-1}(a x)}{a^3}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^4} \]

[Out]

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

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Rubi [A]  time = 0.302269, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {5980, 5916, 5910, 260, 5948, 5984, 5918, 6058, 6610} \[ -\frac{\text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^4}+\frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^4}-\frac{\log \left (1-a^2 x^2\right )}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^2}{2 a^2}-\frac{\tanh ^{-1}(a x)^3}{3 a^4}+\frac{\tanh ^{-1}(a x)^2}{2 a^4}-\frac{x \tanh ^{-1}(a x)}{a^3}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, 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] && GtQ[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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 5984

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

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[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 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[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{x^3 \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx &=-\frac{\int x \tanh ^{-1}(a x)^2 \, dx}{a^2}+\frac{\int \frac{x \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{a^2}\\ &=-\frac{x^2 \tanh ^{-1}(a x)^2}{2 a^2}-\frac{\tanh ^{-1}(a x)^3}{3 a^4}+\frac{\int \frac{\tanh ^{-1}(a x)^2}{1-a x} \, dx}{a^3}+\frac{\int \frac{x^2 \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{a}\\ &=-\frac{x^2 \tanh ^{-1}(a x)^2}{2 a^2}-\frac{\tanh ^{-1}(a x)^3}{3 a^4}+\frac{\tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a^4}-\frac{\int \tanh ^{-1}(a x) \, dx}{a^3}+\frac{\int \frac{\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{a^3}-\frac{2 \int \frac{\tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}\\ &=-\frac{x \tanh ^{-1}(a x)}{a^3}+\frac{\tanh ^{-1}(a x)^2}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^2}{2 a^2}-\frac{\tanh ^{-1}(a x)^3}{3 a^4}+\frac{\tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a^4}+\frac{\tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{a^4}-\frac{\int \frac{\text{Li}_2\left (1-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}+\frac{\int \frac{x}{1-a^2 x^2} \, dx}{a^2}\\ &=-\frac{x \tanh ^{-1}(a x)}{a^3}+\frac{\tanh ^{-1}(a x)^2}{2 a^4}-\frac{x^2 \tanh ^{-1}(a x)^2}{2 a^2}-\frac{\tanh ^{-1}(a x)^3}{3 a^4}+\frac{\tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a^4}-\frac{\log \left (1-a^2 x^2\right )}{2 a^4}+\frac{\tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{a^4}-\frac{\text{Li}_3\left (1-\frac{2}{1-a x}\right )}{2 a^4}\\ \end{align*}

Mathematica [A]  time = 0.121508, size = 112, normalized size = 0.83 \[ -\frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )-\log \left (\frac{1}{\sqrt{1-a^2 x^2}}\right )-\frac{1}{2} \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac{1}{3} \tanh ^{-1}(a x)^3+a x \tanh ^{-1}(a x)-\tanh ^{-1}(a x)^2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.292, size = 812, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{3 \,{\left (a^{2} x^{2} + \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2} + \log \left (-a x + 1\right )^{3}}{24 \, a^{4}} + \frac{1}{4} \, \int -\frac{a^{3} x^{3} \log \left (a x + 1\right )^{2} -{\left (a^{3} x^{3} + a^{2} x^{2} +{\left (2 \, a^{3} x^{3} + a x + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )}{a^{5} x^{2} - a^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(a^2*x^2 + log(a*x + 1))*log(-a*x + 1)^2 + log(-a*x + 1)^3)/a^4 + 1/4*integrate(-(a^3*x^3*log(a*x + 1
)^2 - (a^3*x^3 + a^2*x^2 + (2*a^3*x^3 + a*x + 1)*log(a*x + 1))*log(-a*x + 1))/(a^5*x^2 - a^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{x^{3} \operatorname{artanh}\left (a x\right )^{2}}{a^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{3} \operatorname{atanh}^{2}{\left (a x \right )}}{a^{2} x^{2} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{3} \operatorname{artanh}\left (a x\right )^{2}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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