3.3.73 \(\int \frac {x^3 \tanh ^{-1}(a x)^3}{(1-a^2 x^2)^2} \, dx\) [273]

Optimal. Leaf size=227 \[ -\frac {3 x}{8 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)}{8 a^4}+\frac {3 \tanh ^{-1}(a x)}{4 a^4 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^4}+\frac {\tanh ^{-1}(a x)^3}{2 a^4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{4 a^4}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}-\frac {3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \text {PolyLog}\left (4,1-\frac {2}{1-a x}\right )}{4 a^4} \]

[Out]

-3/8*x/a^3/(-a^2*x^2+1)-3/8*arctanh(a*x)/a^4+3/4*arctanh(a*x)/a^4/(-a^2*x^2+1)-3/4*x*arctanh(a*x)^2/a^3/(-a^2*
x^2+1)-1/4*arctanh(a*x)^3/a^4+1/2*arctanh(a*x)^3/a^4/(-a^2*x^2+1)+1/4*arctanh(a*x)^4/a^4-arctanh(a*x)^3*ln(2/(
-a*x+1))/a^4-3/2*arctanh(a*x)^2*polylog(2,1-2/(-a*x+1))/a^4+3/2*arctanh(a*x)*polylog(3,1-2/(-a*x+1))/a^4-3/4*p
olylog(4,1-2/(-a*x+1))/a^4

________________________________________________________________________________________

Rubi [A]
time = 0.29, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6175, 6131, 6055, 6095, 6205, 6209, 6745, 6141, 6103, 205, 212} \begin {gather*} -\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{4 a^4}-\frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{2 a^4}+\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{2 a^4}+\frac {\tanh ^{-1}(a x)^4}{4 a^4}-\frac {\tanh ^{-1}(a x)^3}{4 a^4}-\frac {3 \tanh ^{-1}(a x)}{8 a^4}-\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^4}+\frac {\tanh ^{-1}(a x)^3}{2 a^4 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 a^4 \left (1-a^2 x^2\right )}-\frac {3 x}{8 a^3 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 6103

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 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 6141

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 6175

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 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 6209

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 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)^3}{\left (1-a^2 x^2\right )^2} \, dx &=\frac {\int \frac {x \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^2} \, dx}{a^2}-\frac {\int \frac {x \tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx}{a^2}\\ &=\frac {\tanh ^{-1}(a x)^3}{2 a^4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{4 a^4}-\frac {\int \frac {\tanh ^{-1}(a x)^3}{1-a x} \, dx}{a^3}-\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx}{2 a^3}\\ &=-\frac {3 x \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^4}+\frac {\tanh ^{-1}(a x)^3}{2 a^4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{4 a^4}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}+\frac {3 \int \frac {\tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^3}+\frac {3 \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx}{2 a^2}\\ &=\frac {3 \tanh ^{-1}(a x)}{4 a^4 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^4}+\frac {\tanh ^{-1}(a x)^3}{2 a^4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{4 a^4}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx}{4 a^3}+\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^3}\\ &=-\frac {3 x}{8 a^3 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 a^4 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^4}+\frac {\tanh ^{-1}(a x)^3}{2 a^4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{4 a^4}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \int \frac {1}{1-a^2 x^2} \, dx}{8 a^3}-\frac {3 \int \frac {\text {Li}_3\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{2 a^3}\\ &=-\frac {3 x}{8 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)}{8 a^4}+\frac {3 \tanh ^{-1}(a x)}{4 a^4 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a^3 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^4}+\frac {\tanh ^{-1}(a x)^3}{2 a^4 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^4}{4 a^4}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^4}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^4}+\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^4}-\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{4 a^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.10, size = 139, normalized size = 0.61 \begin {gather*} \frac {-4 \tanh ^{-1}(a x)^4+6 \tanh ^{-1}(a x) \cosh \left (2 \tanh ^{-1}(a x)\right )+4 \tanh ^{-1}(a x)^3 \cosh \left (2 \tanh ^{-1}(a x)\right )-16 \tanh ^{-1}(a x)^3 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )+24 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+24 \tanh ^{-1}(a x) \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+12 \text {PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )-3 \sinh \left (2 \tanh ^{-1}(a x)\right )-6 \tanh ^{-1}(a x)^2 \sinh \left (2 \tanh ^{-1}(a x)\right )}{16 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*ArcTanh[a*x]^4 + 6*ArcTanh[a*x]*Cosh[2*ArcTanh[a*x]] + 4*ArcTanh[a*x]^3*Cosh[2*ArcTanh[a*x]] - 16*ArcTanh[
a*x]^3*Log[1 + E^(-2*ArcTanh[a*x])] + 24*ArcTanh[a*x]^2*PolyLog[2, -E^(-2*ArcTanh[a*x])] + 24*ArcTanh[a*x]*Pol
yLog[3, -E^(-2*ArcTanh[a*x])] + 12*PolyLog[4, -E^(-2*ArcTanh[a*x])] - 3*Sinh[2*ArcTanh[a*x]] - 6*ArcTanh[a*x]^
2*Sinh[2*ArcTanh[a*x]])/(16*a^4)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 107.56, size = 806, normalized size = 3.55 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(1/4*arctanh(a*x)^3/(a*x+1)+1/2*arctanh(a*x)^3*ln(a*x+1)-1/4*arctanh(a*x)^3/(a*x-1)+1/2*arctanh(a*x)^3*l
n(a*x-1)-arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+1/4*arctanh(a*x)^4-3/16*arctanh(a*x)^2*(a*x-1)/(a*x+1)-
3/16*arctanh(a*x)*(a*x-1)/(a*x+1)-3/32*(a*x-1)/(a*x+1)+3/16*(a*x+1)*arctanh(a*x)^2/(a*x-1)-3/16*arctanh(a*x)*(
a*x+1)/(a*x-1)+3/32*(a*x+1)/(a*x-1)-3/2*arctanh(a*x)^2*polylog(2,-(a*x+1)^2/(-a^2*x^2+1))+3/2*arctanh(a*x)*pol
ylog(3,-(a*x+1)^2/(-a^2*x^2+1))-3/4*polylog(4,-(a*x+1)^2/(-a^2*x^2+1))-1/4*(2*I*Pi+I*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-I*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))+2*I*Pi*csgn(I/((a*x+1
)^2/(-a^2*x^2+1)+1))^3+2*I*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2-2*I*Pi*csgn(I
/((a*x+1)^2/(-a^2*x^2+1)+1))^2+I*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))-I*Pi*cs
gn(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+I*Pi*csgn(I*(a*x+1)^2/(
a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3+I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3+1+4*ln(2))*arctanh(a*x)^3)

________________________________________________________________________________________

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

[Out]

-1/64*((a^2*x^2 - 1)*log(-a*x + 1)^4 + 4*((a^2*x^2 - 1)*log(a*x + 1) - 1)*log(-a*x + 1)^3)/(a^6*x^2 - a^4) + 1
/8*integrate(1/2*(2*a^3*x^3*log(a*x + 1)^3 - 6*a^3*x^3*log(a*x + 1)^2*log(-a*x + 1) - 3*(a*x - (3*a^3*x^3 + a^
2*x^2 - a*x - 1)*log(a*x + 1) + 1)*log(-a*x + 1)^2)/(a^7*x^4 - 2*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)^3/(-a^2*x^2+1)^2,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________