3.3.34 \(\int \frac {x^3 \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx\) [234]

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

[Out]

-x*arctanh(a*x)/a^3+1/2*arctanh(a*x)^2/a^4-1/2*x^2*arctanh(a*x)^2/a^2-1/3*arctanh(a*x)^3/a^4+arctanh(a*x)^2*ln
(2/(-a*x+1))/a^4-1/2*ln(-a^2*x^2+1)/a^4+arctanh(a*x)*polylog(2,1-2/(-a*x+1))/a^4-1/2*polylog(3,1-2/(-a*x+1))/a
^4

________________________________________________________________________________________

Rubi [A]
time = 0.21, antiderivative size = 135, normalized size of antiderivative = 1.00, 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 = {6127, 6037, 6021, 266, 6095, 6131, 6055, 6205, 6745} \begin {gather*} -\frac {\text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^4}-\frac {\tanh ^{-1}(a x)^3}{3 a^4}+\frac {\tanh ^{-1}(a x)^2}{2 a^4}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^4}-\frac {x \tanh ^{-1}(a x)}{a^3}-\frac {x^2 \tanh ^{-1}(a x)^2}{2 a^2}-\frac {\log \left (1-a^2 x^2\right )}{2 a^4} \end {gather*}

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 266

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 6021

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

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6055

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 6095

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 6127

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 6131

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 6205

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 6745

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 = 0.09, size = 112, normalized size = 0.83 \begin {gather*} -\frac {a x \tanh ^{-1}(a x)-\frac {1}{2} \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {1}{3} \tanh ^{-1}(a x)^3-\tanh ^{-1}(a x)^2 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )-\log \left (\frac {1}{\sqrt {1-a^2 x^2}}\right )+\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 )}{a^4} \end {gather*}

Antiderivative was successfully verified.

[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)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 39.98, size = 728, normalized size = 5.39 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(-1/2*a^2*x^2*arctanh(a*x)^2-1/2*arctanh(a*x)^2*ln(a*x-1)-1/2*arctanh(a*x)^2*ln(a*x+1)+arctanh(a*x)^2*ln
((a*x+1)/(-a^2*x^2+1)^(1/2))+arctanh(a*x)*polylog(2,-(a*x+1)^2/(-a^2*x^2+1))-1/2*polylog(3,-(a*x+1)^2/(-a^2*x^
2+1))+1/12*arctanh(a*x)*(3*I*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*arctanh(a*x)*Pi+6*I*csgn(I*(a*x+1)^2/(a^2*x^2-1))
^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)*Pi-3*I*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)*Pi-3*I*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))*arctanh(a*x)*Pi+3*I*csgn(I*(a*x+1)^2/(a^2
*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*arctanh(a*x)*Pi+3*I*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)*Pi+3*I*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))*arctanh(a*x)*Pi+6*I*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)*Pi-6*I*csgn(I/((a*x+1
)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)*Pi+6*I*arctanh(a*x)*Pi-4*arctanh(a*x)^2+12*arctanh(a*x)*ln(2)+6*arctanh(a*
x)-12*a*x-12)+ln((a*x+1)^2/(-a^2*x^2+1)+1))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x^{3} \operatorname {atanh}^{2}{\left (a x \right )}}{a^{2} x^{2} - 1}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {x^3\,{\mathrm {atanh}\left (a\,x\right )}^2}{a^2\,x^2-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________