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

Optimal. Leaf size=108 \[ \frac{3 \text{PolyLog}\left (4,1-\frac{2}{1-a x}\right )}{4 a^2}+\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^2}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.206464, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5984, 5918, 5948, 6058, 6062, 6610} \[ \frac{3 \text{PolyLog}\left (4,1-\frac{2}{1-a x}\right )}{4 a^2}+\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{\tanh ^{-1}(a x)^4}{4 a^2}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 6062

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{x \tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx &=-\frac{\tanh ^{-1}(a x)^4}{4 a^2}+\frac{\int \frac{\tanh ^{-1}(a x)^3}{1-a x} \, dx}{a}\\ &=-\frac{\tanh ^{-1}(a x)^4}{4 a^2}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^2}-\frac{3 \int \frac{\tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a}\\ &=-\frac{\tanh ^{-1}(a x)^4}{4 a^2}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^2}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{3 \int \frac{\tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a}\\ &=-\frac{\tanh ^{-1}(a x)^4}{4 a^2}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^2}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1-a x}\right )}{2 a^2}+\frac{3 \int \frac{\text{Li}_3\left (1-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{2 a}\\ &=-\frac{\tanh ^{-1}(a x)^4}{4 a^2}+\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1-a x}\right )}{a^2}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{2 a^2}-\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1-a x}\right )}{2 a^2}+\frac{3 \text{Li}_4\left (1-\frac{2}{1-a x}\right )}{4 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0640732, size = 87, normalized size = 0.81 \[ -\frac{6 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text{PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x)^4-4 \tanh ^{-1}(a x)^3 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{4 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(-ArcTanh[a*x]^4 - 4*ArcTanh[a*x]^3*Log[1 + E^(-2*ArcTanh[a*x])] + 6*ArcTanh[a*x]^2*PolyLog[2, -E^(-2*ArcTanh
[a*x])] + 6*ArcTanh[a*x]*PolyLog[3, -E^(-2*ArcTanh[a*x])] + 3*PolyLog[4, -E^(-2*ArcTanh[a*x])])/(4*a^2)

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Maple [C]  time = 0.285, size = 776, normalized size = 7.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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