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

Optimal. Leaf size=103 \[ -\frac{3 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^3}+\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^3}+\frac{\tanh ^{-1}(a x)^4}{4 a^3}-\frac{x \tanh ^{-1}(a x)^3}{a^2}-\frac{\tanh ^{-1}(a x)^3}{a^3}+\frac{3 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.269013, antiderivative size = 103, 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 = {5980, 5910, 5984, 5918, 5948, 6058, 6610} \[ -\frac{3 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^3}+\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^3}+\frac{\tanh ^{-1}(a x)^4}{4 a^3}-\frac{x \tanh ^{-1}(a x)^3}{a^2}-\frac{\tanh ^{-1}(a x)^3}{a^3}+\frac{3 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.242113, size = 78, normalized size = 0.76 \[ \frac{-12 \tanh ^{-1}(a x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )-6 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x)^2 \left (\tanh ^{-1}(a x)^2+(4-4 a x) \tanh ^{-1}(a x)+12 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )\right )}{4 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(ArcTanh[a*x]^2*((4 - 4*a*x)*ArcTanh[a*x] + ArcTanh[a*x]^2 + 12*Log[1 + E^(-2*ArcTanh[a*x])]) - 12*ArcTanh[a*x
]*PolyLog[2, -E^(-2*ArcTanh[a*x])] - 6*PolyLog[3, -E^(-2*ArcTanh[a*x])])/(4*a^3)

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Maple [C]  time = 0.327, size = 788, normalized size = 7.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{x^{2} \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^2*arctanh(a*x)^3/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-x^2*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^{2} \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**2*atanh(a*x)**3/(-a**2*x**2+1),x)

[Out]

-Integral(x**2*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^{2} \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^2*arctanh(a*x)^3/(-a^2*x^2+1),x, algorithm="giac")

[Out]

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