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

Optimal. Leaf size=193 \[ -\frac {3 a x}{8 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}-\frac {3 a x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}-\frac {3}{4} \text {Li}_4\left (\frac {2}{a x+1}-1\right )-\frac {3}{2} \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2-\frac {3}{2} \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)+\frac {1}{4} \tanh ^{-1}(a x)^4-\frac {1}{4} \tanh ^{-1}(a x)^3-\frac {3}{8} \tanh ^{-1}(a x)+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3 \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 193, 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 = {6030, 5988, 5932, 5948, 6056, 6060, 6610, 5994, 5956, 199, 206} \[ -\frac {3}{4} \text {PolyLog}\left (4,\frac {2}{a x+1}-1\right )-\frac {3}{2} \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )-\frac {3}{2} \tanh ^{-1}(a x) \text {PolyLog}\left (3,\frac {2}{a x+1}-1\right )-\frac {3 a x}{8 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}-\frac {3 a x \tanh ^{-1}(a x)^2}{4 \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac {1}{4} \tanh ^{-1}(a x)^4-\frac {1}{4} \tanh ^{-1}(a x)^3-\frac {3}{8} \tanh ^{-1}(a x)+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*x)/(8*(1 - a^2*x^2)) - (3*ArcTanh[a*x])/8 + (3*ArcTanh[a*x])/(4*(1 - a^2*x^2)) - (3*a*x*ArcTanh[a*x]^2)/
(4*(1 - a^2*x^2)) - ArcTanh[a*x]^3/4 + ArcTanh[a*x]^3/(2*(1 - a^2*x^2)) + ArcTanh[a*x]^4/4 + ArcTanh[a*x]^3*Lo
g[2 - 2/(1 + a*x)] - (3*ArcTanh[a*x]^2*PolyLog[2, -1 + 2/(1 + a*x)])/2 - (3*ArcTanh[a*x]*PolyLog[3, -1 + 2/(1
+ a*x)])/2 - (3*PolyLog[4, -1 + 2/(1 + a*x)])/4

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

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

Rule 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 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 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 5988

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

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 6030

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/d, Int
[x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/d, Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTanh
[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] && ILtQ[
m, 0] && NeQ[p, -1]

Rule 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[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 6060

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

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Mathematica [A]  time = 0.20, size = 135, normalized size = 0.70 \[ \frac {1}{64} \left (96 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )-96 \tanh ^{-1}(a x) \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )+48 \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )-16 \tanh ^{-1}(a x)^4+64 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-24 \tanh ^{-1}(a x)^2 \sinh \left (2 \tanh ^{-1}(a x)\right )-12 \sinh \left (2 \tanh ^{-1}(a x)\right )+16 \tanh ^{-1}(a x)^3 \cosh \left (2 \tanh ^{-1}(a x)\right )+24 \tanh ^{-1}(a x) \cosh \left (2 \tanh ^{-1}(a x)\right )+\pi ^4\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Pi^4 - 16*ArcTanh[a*x]^4 + 24*ArcTanh[a*x]*Cosh[2*ArcTanh[a*x]] + 16*ArcTanh[a*x]^3*Cosh[2*ArcTanh[a*x]] + 64
*ArcTanh[a*x]^3*Log[1 - E^(2*ArcTanh[a*x])] + 96*ArcTanh[a*x]^2*PolyLog[2, E^(2*ArcTanh[a*x])] - 96*ArcTanh[a*
x]*PolyLog[3, E^(2*ArcTanh[a*x])] + 48*PolyLog[4, E^(2*ArcTanh[a*x])] - 12*Sinh[2*ArcTanh[a*x]] - 24*ArcTanh[a
*x]^2*Sinh[2*ArcTanh[a*x]])/64

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (a x\right )^{3}}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctanh(a*x)^3/(a^4*x^5 - 2*a^2*x^3 + x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{3}}{{\left (a^{2} x^{2} - 1\right )}^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.74, size = 1387, normalized size = 7.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*(a*x+1)/(a*x-1)+1/4*I*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a
^2*x^2+1)))^2*arctanh(a*x)^3+3/16*(a*x+1)*arctanh(a*x)^2/(a*x-1)-3/16*arctanh(a*x)*(a*x-1)/(a*x+1)-3/16*arctan
h(a*x)^2*(a*x-1)/(a*x+1)+1/2*I*Pi*arctanh(a*x)^3-1/4*arctanh(a*x)^4-1/4*arctanh(a*x)^3-3/16*(a*x+1)*arctanh(a*
x)/(a*x-1)-1/4*arctanh(a*x)^3/(a*x-1)+1/4*arctanh(a*x)^3/(a*x+1)+arctanh(a*x)^3*ln(a*x)+6*polylog(4,-(a*x+1)/(
-a^2*x^2+1)^(1/2))+6*polylog(4,(a*x+1)/(-a^2*x^2+1)^(1/2))-6*arctanh(a*x)*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2
))-1/2*arctanh(a*x)^3*ln(a*x-1)-1/2*arctanh(a*x)^3*ln(a*x+1)+arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+1/2
*I*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*arctanh(a*x)^3-1/2*I*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*arctan
h(a*x)^3+1/4*I*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3+1/4*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x
+1)^2/(-a^2*x^2+1)))^3*arctanh(a*x)^3+1/2*I*arctanh(a*x)^3*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(
-a^2*x^2+1)))^3-6*arctanh(a*x)*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))-1/4*I*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)
))*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))*arctanh(a*x)^3+1/2*I
*arctanh(a*x)^3*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1))*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*((a*x+1)^2/(-a
^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))+3*arctanh(a*x)^2*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+3*arctanh(a*x)
^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))-arctanh(a*x)^3*ln((a*x+1)^2/(-a^2*x^2+1)-1)+ln(2)*arctanh(a*x)^3+arc
tanh(a*x)^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+arctanh(a*x)^3*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))-3/32*(a*x-1)/(a*x
+1)-1/2*I*arctanh(a*x)^3*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/
(-a^2*x^2+1)))^2+1/4*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))*arctanh(a*x)^3+1/
2*I*arctanh(a*x)^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))^2-1/4*I*Pi*csgn(I*(a*x+
1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*arctanh(a*x)^3-1/2*I*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)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )^{4} + 4 \, {\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) + 1\right )} \log \left (-a x + 1\right )^{3}}{64 \, {\left (a^{2} x^{2} - 1\right )}} - \frac {1}{8} \, \int -\frac {2 \, \log \left (a x + 1\right )^{3} - 6 \, \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right ) - 3 \, {\left (a^{2} x^{2} + a x + {\left (a^{4} x^{4} + a^{3} x^{3} - a^{2} x^{2} - a x - 2\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2}}{2 \, {\left (a^{4} x^{5} - 2 \, a^{2} x^{3} + x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/x/(-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^2*x^2 - 1) - 1/8*
integrate(-1/2*(2*log(a*x + 1)^3 - 6*log(a*x + 1)^2*log(-a*x + 1) - 3*(a^2*x^2 + a*x + (a^4*x^4 + a^3*x^3 - a^
2*x^2 - a*x - 2)*log(a*x + 1))*log(-a*x + 1)^2)/(a^4*x^5 - 2*a^2*x^3 + x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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