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

Optimal. Leaf size=161 \[ \frac{\text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^4}+\frac{1}{4 a^4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^2}{2 a^4 \left (1-a^2 x^2\right )}-\frac{x \tanh ^{-1}(a x)}{2 a^3 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{3 a^4}-\frac{\tanh ^{-1}(a x)^2}{4 a^4}-\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^4} \]

[Out]

1/(4*a^4*(1 - a^2*x^2)) - (x*ArcTanh[a*x])/(2*a^3*(1 - a^2*x^2)) - ArcTanh[a*x]^2/(4*a^4) + ArcTanh[a*x]^2/(2*
a^4*(1 - a^2*x^2)) + ArcTanh[a*x]^3/(3*a^4) - (ArcTanh[a*x]^2*Log[2/(1 - a*x)])/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 = 0.291178, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {6028, 5984, 5918, 5948, 6058, 6610, 5994, 5956, 261} \[ \frac{\text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{2 a^4}-\frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^4}+\frac{1}{4 a^4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^2}{2 a^4 \left (1-a^2 x^2\right )}-\frac{x \tanh ^{-1}(a x)}{2 a^3 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^3}{3 a^4}-\frac{\tanh ^{-1}(a x)^2}{4 a^4}-\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(4*a^4*(1 - a^2*x^2)) - (x*ArcTanh[a*x])/(2*a^3*(1 - a^2*x^2)) - ArcTanh[a*x]^2/(4*a^4) + ArcTanh[a*x]^2/(2*
a^4*(1 - a^2*x^2)) + ArcTanh[a*x]^3/(3*a^4) - (ArcTanh[a*x]^2*Log[2/(1 - a*x)])/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 6028

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

Rule 5994

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

Rule 5956

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

Rule 261

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

Rubi steps

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

Mathematica [A]  time = 0.178218, size = 103, normalized size = 0.64 \[ \frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )-\frac{1}{3} \tanh ^{-1}(a x)^3-\tanh ^{-1}(a x)^2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-\frac{1}{4} \tanh ^{-1}(a x) \sinh \left (2 \tanh ^{-1}(a x)\right )+\frac{1}{8} \left (2 \tanh ^{-1}(a x)^2+1\right ) \cosh \left (2 \tanh ^{-1}(a x)\right )}{a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-ArcTanh[a*x]^3/3 + ((1 + 2*ArcTanh[a*x]^2)*Cosh[2*ArcTanh[a*x]])/8 - ArcTanh[a*x]^2*Log[1 + E^(-2*ArcTanh[a*
x])] + ArcTanh[a*x]*PolyLog[2, -E^(-2*ArcTanh[a*x])] + PolyLog[3, -E^(-2*ArcTanh[a*x])]/2 - (ArcTanh[a*x]*Sinh
[2*ArcTanh[a*x]])/4)/a^4

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Maple [C]  time = 0.348, size = 907, normalized size = 5.6 \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)^2,x)

[Out]

-1/16/a^3/(a*x+1)*x-1/16/a^3*x/(a*x-1)-1/4/a^4*arctanh(a*x)^2/(a*x-1)+1/4/a^4*arctanh(a*x)^2/(a*x+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)^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*arctanh(a*x)^2/a^4+1/2/a^4*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))-1/4*I/a^4*arc
tanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3-1/4*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))^3-1/2*I/a^4*arctanh(a*x)^2*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3+1/2*I/a^4*arctanh(a*x
)^2*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2-1/8/a^3*arctanh(a*x)/(a*x+1)*x-1/2*I/a^4*arctanh(a*x)^2*Pi+1/8/a^3
*arctanh(a*x)/(a*x-1)*x-1/4*I/a^4*arctanh(a*x)^2*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))-1/2*I/a^4*arctanh(a*x)^2*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+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*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))-1/16/a^4/(a*x-1)+1/16/a^4/(a*x+1)+1/3*arctanh(a*x)^3/a^4-1/a^4*arctanh(a*x)^2*ln(2)+1/8/a^4*arcta
nh(a*x)/(a*x-1)+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))+1/8/a^4*arctanh(a*x)/(a*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{3}{4} \, a^{3} \int \frac{x^{3} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}}\,{d x} - \frac{1}{4} \, a^{2} \int \frac{x^{2} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}}\,{d x} - \frac{1}{32} \,{\left (a{\left (\frac{2}{a^{7} x - a^{6}} - \frac{\log \left (a x + 1\right )}{a^{6}} + \frac{\log \left (a x - 1\right )}{a^{6}}\right )} + \frac{4 \, \log \left (-a x + 1\right )}{a^{7} x^{2} - a^{5}}\right )} a + \frac{1}{4} \, a \int \frac{x \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}}\,{d x} + \frac{{\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )^{3} + 3 \,{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) - 1\right )} \log \left (-a x + 1\right )^{2}}{24 \,{\left (a^{6} x^{2} - a^{4}\right )}} + \frac{1}{4} \, \int \frac{a^{3} x^{3} \log \left (a x + 1\right )^{2}}{a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}}\,{d x} + \frac{1}{4} \, \int \frac{\log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}}\,{d x} + \frac{1}{4} \, \int \frac{\log \left (-a x + 1\right )}{a^{7} x^{4} - 2 \, 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)^2,x, algorithm="maxima")

[Out]

-3/4*a^3*integrate(x^3*log(a*x + 1)*log(-a*x + 1)/(a^7*x^4 - 2*a^5*x^2 + a^3), x) - 1/4*a^2*integrate(x^2*log(
a*x + 1)*log(-a*x + 1)/(a^7*x^4 - 2*a^5*x^2 + a^3), x) - 1/32*(a*(2/(a^7*x - a^6) - log(a*x + 1)/a^6 + log(a*x
- 1)/a^6) + 4*log(-a*x + 1)/(a^7*x^2 - a^5))*a + 1/4*a*integrate(x*log(a*x + 1)*log(-a*x + 1)/(a^7*x^4 - 2*a^
5*x^2 + a^3), x) + 1/24*((a^2*x^2 - 1)*log(-a*x + 1)^3 + 3*((a^2*x^2 - 1)*log(a*x + 1) - 1)*log(-a*x + 1)^2)/(
a^6*x^2 - a^4) + 1/4*integrate(a^3*x^3*log(a*x + 1)^2/(a^7*x^4 - 2*a^5*x^2 + a^3), x) + 1/4*integrate(log(a*x
+ 1)*log(-a*x + 1)/(a^7*x^4 - 2*a^5*x^2 + a^3), x) + 1/4*integrate(log(-a*x + 1)/(a^7*x^4 - 2*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^{4} x^{4} - 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)^2,x, algorithm="fricas")

[Out]

integral(x^3*arctanh(a*x)^2/(a^4*x^4 - 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 )}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, 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)**2,x)

[Out]

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

[Out]

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