3.4.11 \(\int \frac {\text {arctanh}(a x)^2}{(1-a^2 x^2)^3} \, dx\) [311]

3.4.11.1 Optimal result
3.4.11.2 Mathematica [A] (verified)
3.4.11.3 Rubi [A] (verified)
3.4.11.4 Maple [A] (verified)
3.4.11.5 Fricas [A] (verification not implemented)
3.4.11.6 Sympy [F]
3.4.11.7 Maxima [B] (verification not implemented)
3.4.11.8 Giac [F]
3.4.11.9 Mupad [B] (verification not implemented)

3.4.11.1 Optimal result

Integrand size = 19, antiderivative size = 151 \[ \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx=\frac {x}{32 \left (1-a^2 x^2\right )^2}+\frac {15 x}{64 \left (1-a^2 x^2\right )}+\frac {15 \text {arctanh}(a x)}{64 a}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}-\frac {3 \text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )}+\frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}+\frac {3 x \text {arctanh}(a x)^2}{8 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^3}{8 a} \]

output
1/32*x/(-a^2*x^2+1)^2+15/64*x/(-a^2*x^2+1)+15/64*arctanh(a*x)/a-1/8*arctan 
h(a*x)/a/(-a^2*x^2+1)^2-3/8*arctanh(a*x)/a/(-a^2*x^2+1)+1/4*x*arctanh(a*x) 
^2/(-a^2*x^2+1)^2+3/8*x*arctanh(a*x)^2/(-a^2*x^2+1)+1/8*arctanh(a*x)^3/a
 
3.4.11.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.84 \[ \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx=\frac {1}{128} \left (\frac {4 x}{\left (-1+a^2 x^2\right )^2}-\frac {30 x}{-1+a^2 x^2}+\frac {16 \left (-4+3 a^2 x^2\right ) \text {arctanh}(a x)}{a \left (-1+a^2 x^2\right )^2}-\frac {16 x \left (-5+3 a^2 x^2\right ) \text {arctanh}(a x)^2}{\left (-1+a^2 x^2\right )^2}+\frac {16 \text {arctanh}(a x)^3}{a}-\frac {15 \log (1-a x)}{a}+\frac {15 \log (1+a x)}{a}\right ) \]

input
Integrate[ArcTanh[a*x]^2/(1 - a^2*x^2)^3,x]
 
output
((4*x)/(-1 + a^2*x^2)^2 - (30*x)/(-1 + a^2*x^2) + (16*(-4 + 3*a^2*x^2)*Arc 
Tanh[a*x])/(a*(-1 + a^2*x^2)^2) - (16*x*(-5 + 3*a^2*x^2)*ArcTanh[a*x]^2)/( 
-1 + a^2*x^2)^2 + (16*ArcTanh[a*x]^3)/a - (15*Log[1 - a*x])/a + (15*Log[1 
+ a*x])/a)/128
 
3.4.11.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.36, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6526, 215, 215, 219, 6518, 6556, 215, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx\)

\(\Big \downarrow \) 6526

\(\displaystyle \frac {3}{4} \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2}dx+\frac {1}{8} \int \frac {1}{\left (1-a^2 x^2\right )^3}dx+\frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {3}{4} \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2}dx+\frac {1}{8} \left (\frac {3}{4} \int \frac {1}{\left (1-a^2 x^2\right )^2}dx+\frac {x}{4 \left (1-a^2 x^2\right )^2}\right )+\frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {3}{4} \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2}dx+\frac {1}{8} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-a^2 x^2}dx+\frac {x}{2 \left (1-a^2 x^2\right )}\right )+\frac {x}{4 \left (1-a^2 x^2\right )^2}\right )+\frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3}{4} \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^2}dx+\frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {1}{8} \left (\frac {3}{4} \left (\frac {x}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)}{2 a}\right )+\frac {x}{4 \left (1-a^2 x^2\right )^2}\right )\)

\(\Big \downarrow \) 6518

\(\displaystyle \frac {3}{4} \left (-a \int \frac {x \text {arctanh}(a x)}{\left (1-a^2 x^2\right )^2}dx+\frac {x \text {arctanh}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^3}{6 a}\right )+\frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {1}{8} \left (\frac {3}{4} \left (\frac {x}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)}{2 a}\right )+\frac {x}{4 \left (1-a^2 x^2\right )^2}\right )\)

\(\Big \downarrow \) 6556

\(\displaystyle \frac {3}{4} \left (-a \left (\frac {\text {arctanh}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\int \frac {1}{\left (1-a^2 x^2\right )^2}dx}{2 a}\right )+\frac {x \text {arctanh}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^3}{6 a}\right )+\frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {1}{8} \left (\frac {3}{4} \left (\frac {x}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)}{2 a}\right )+\frac {x}{4 \left (1-a^2 x^2\right )^2}\right )\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {3}{4} \left (-a \left (\frac {\text {arctanh}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\frac {1}{2} \int \frac {1}{1-a^2 x^2}dx+\frac {x}{2 \left (1-a^2 x^2\right )}}{2 a}\right )+\frac {x \text {arctanh}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)^3}{6 a}\right )+\frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {1}{8} \left (\frac {3}{4} \left (\frac {x}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)}{2 a}\right )+\frac {x}{4 \left (1-a^2 x^2\right )^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {x \text {arctanh}(a x)^2}{4 \left (1-a^2 x^2\right )^2}-\frac {\text {arctanh}(a x)}{8 a \left (1-a^2 x^2\right )^2}+\frac {1}{8} \left (\frac {3}{4} \left (\frac {x}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)}{2 a}\right )+\frac {x}{4 \left (1-a^2 x^2\right )^2}\right )+\frac {3}{4} \left (\frac {x \text {arctanh}(a x)^2}{2 \left (1-a^2 x^2\right )}-a \left (\frac {\text {arctanh}(a x)}{2 a^2 \left (1-a^2 x^2\right )}-\frac {\frac {x}{2 \left (1-a^2 x^2\right )}+\frac {\text {arctanh}(a x)}{2 a}}{2 a}\right )+\frac {\text {arctanh}(a x)^3}{6 a}\right )\)

input
Int[ArcTanh[a*x]^2/(1 - a^2*x^2)^3,x]
 
output
-1/8*ArcTanh[a*x]/(a*(1 - a^2*x^2)^2) + (x*ArcTanh[a*x]^2)/(4*(1 - a^2*x^2 
)^2) + (x/(4*(1 - a^2*x^2)^2) + (3*(x/(2*(1 - a^2*x^2)) + ArcTanh[a*x]/(2* 
a)))/4)/8 + (3*((x*ArcTanh[a*x]^2)/(2*(1 - a^2*x^2)) + ArcTanh[a*x]^3/(6*a 
) - a*(ArcTanh[a*x]/(2*a^2*(1 - a^2*x^2)) - (x/(2*(1 - a^2*x^2)) + ArcTanh 
[a*x]/(2*a))/(2*a))))/4
 

3.4.11.3.1 Defintions of rubi rules used

rule 215
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 219
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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 6518
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Sy 
mbol] :> Simp[x*((a + b*ArcTanh[c*x])^p/(2*d*(d + e*x^2))), x] + (Simp[(a + 
 b*ArcTanh[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x] - Simp[b*c*(p/2)   Int[x*( 
(a + b*ArcTanh[c*x])^(p - 1)/(d + e*x^2)^2), x], x]) /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]
 

rule 6526
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_ 
Symbol] :> Simp[(-b)*p*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^(p - 1)/(4 
*c*d*(q + 1)^2)), x] + (-Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^p 
/(2*d*(q + 1))), x] + Simp[(2*q + 3)/(2*d*(q + 1))   Int[(d + e*x^2)^(q + 1 
)*(a + b*ArcTanh[c*x])^p, x], x] + Simp[b^2*p*((p - 1)/(4*(q + 1)^2))   Int 
[(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 2), x], x]) /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && GtQ[p, 1] && NeQ[q, -3/2]
 

rule 6556
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] + Simp[b*(p/(2*c*(q + 1)))   Int[(d + e*x^2)^q*(a + b*ArcTanh[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]
 
3.4.11.4 Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.79

method result size
parallelrisch \(-\frac {-8 \operatorname {arctanh}\left (a x \right )^{3} a^{4} x^{4}-15 a^{4} x^{4} \operatorname {arctanh}\left (a x \right )+24 \operatorname {arctanh}\left (a x \right )^{2} a^{3} x^{3}+16 \operatorname {arctanh}\left (a x \right )^{3} a^{2} x^{2}+15 a^{3} x^{3}+6 a^{2} x^{2} \operatorname {arctanh}\left (a x \right )-40 \operatorname {arctanh}\left (a x \right )^{2} a x -8 \operatorname {arctanh}\left (a x \right )^{3}-17 a x +17 \,\operatorname {arctanh}\left (a x \right )}{64 \left (a^{2} x^{2}-1\right )^{2} a}\) \(120\)
risch \(\frac {\ln \left (a x +1\right )^{3}}{64 a}-\frac {\left (3 a^{4} x^{4} \ln \left (-a x +1\right )+6 a^{3} x^{3}-6 x^{2} \ln \left (-a x +1\right ) a^{2}-10 a x +3 \ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )^{2}}{64 \left (a^{2} x^{2}-1\right )^{2} a}+\frac {\left (3 a^{4} x^{4} \ln \left (-a x +1\right )^{2}+12 a^{3} x^{3} \ln \left (-a x +1\right )-6 a^{2} x^{2} \ln \left (-a x +1\right )^{2}+12 a^{2} x^{2}-20 a x \ln \left (-a x +1\right )+3 \ln \left (-a x +1\right )^{2}-16\right ) \ln \left (a x +1\right )}{64 a \left (a x +1\right ) \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}-\frac {2 a^{4} x^{4} \ln \left (-a x +1\right )^{3}+15 \ln \left (a x -1\right ) a^{4} x^{4}-15 \ln \left (-a x -1\right ) a^{4} x^{4}+12 a^{3} x^{3} \ln \left (-a x +1\right )^{2}-4 a^{2} x^{2} \ln \left (-a x +1\right )^{3}+30 a^{3} x^{3}-30 \ln \left (a x -1\right ) a^{2} x^{2}+30 \ln \left (-a x -1\right ) a^{2} x^{2}+24 x^{2} \ln \left (-a x +1\right ) a^{2}-20 a \ln \left (-a x +1\right )^{2} x +2 \ln \left (-a x +1\right )^{3}-34 a x +15 \ln \left (a x -1\right )-15 \ln \left (-a x -1\right )-32 \ln \left (-a x +1\right )}{128 a \left (a x +1\right ) \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}\) \(425\)
derivativedivides \(\text {Expression too large to display}\) \(829\)
default \(\text {Expression too large to display}\) \(829\)
parts \(\text {Expression too large to display}\) \(907\)

input
int(arctanh(a*x)^2/(-a^2*x^2+1)^3,x,method=_RETURNVERBOSE)
 
output
-1/64*(-8*arctanh(a*x)^3*a^4*x^4-15*a^4*x^4*arctanh(a*x)+24*arctanh(a*x)^2 
*a^3*x^3+16*arctanh(a*x)^3*a^2*x^2+15*a^3*x^3+6*a^2*x^2*arctanh(a*x)-40*ar 
ctanh(a*x)^2*a*x-8*arctanh(a*x)^3-17*a*x+17*arctanh(a*x))/(a^2*x^2-1)^2/a
 
3.4.11.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.91 \[ \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx=-\frac {30 \, a^{3} x^{3} - 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 4 \, {\left (3 \, a^{3} x^{3} - 5 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 34 \, a x - {\left (15 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 17\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{128 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )}} \]

input
integrate(arctanh(a*x)^2/(-a^2*x^2+1)^3,x, algorithm="fricas")
 
output
-1/128*(30*a^3*x^3 - 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)) 
^3 + 4*(3*a^3*x^3 - 5*a*x)*log(-(a*x + 1)/(a*x - 1))^2 - 34*a*x - (15*a^4* 
x^4 - 6*a^2*x^2 - 17)*log(-(a*x + 1)/(a*x - 1)))/(a^5*x^4 - 2*a^3*x^2 + a)
 
3.4.11.6 Sympy [F]

\[ \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx=- \int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \]

input
integrate(atanh(a*x)**2/(-a**2*x**2+1)**3,x)
 
output
-Integral(atanh(a*x)**2/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), x)
 
3.4.11.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (129) = 258\).

Time = 0.20 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.60 \[ \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx=-\frac {1}{16} \, {\left (\frac {2 \, {\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} - \frac {3 \, \log \left (a x + 1\right )}{a} + \frac {3 \, \log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )^{2} - \frac {{\left (30 \, a^{3} x^{3} - 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{3} + 6 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) + 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{3} - 34 \, a x - 3 \, {\left (5 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} + 5\right )} \log \left (a x + 1\right ) + 15 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )\right )} a^{2}}{128 \, {\left (a^{7} x^{4} - 2 \, a^{5} x^{2} + a^{3}\right )}} + \frac {{\left (12 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 6 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 16\right )} a \operatorname {artanh}\left (a x\right )}{32 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )}} \]

input
integrate(arctanh(a*x)^2/(-a^2*x^2+1)^3,x, algorithm="maxima")
 
output
-1/16*(2*(3*a^2*x^3 - 5*x)/(a^4*x^4 - 2*a^2*x^2 + 1) - 3*log(a*x + 1)/a + 
3*log(a*x - 1)/a)*arctanh(a*x)^2 - 1/128*(30*a^3*x^3 - 2*(a^4*x^4 - 2*a^2* 
x^2 + 1)*log(a*x + 1)^3 + 6*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2*log(a 
*x - 1) + 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^3 - 34*a*x - 3*(5*a^4*x 
^4 - 10*a^2*x^2 + 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 + 5)*log(a*x 
+ 1) + 15*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1))*a^2/(a^7*x^4 - 2*a^5*x^2 
 + a^3) + 1/32*(12*a^2*x^2 - 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 
6*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)*log(a*x - 1) - 3*(a^4*x^4 - 2*a^2 
*x^2 + 1)*log(a*x - 1)^2 - 16)*a*arctanh(a*x)/(a^6*x^4 - 2*a^4*x^2 + a^2)
 
3.4.11.8 Giac [F]

\[ \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx=\int { -\frac {\operatorname {artanh}\left (a x\right )^{2}}{{\left (a^{2} x^{2} - 1\right )}^{3}} \,d x } \]

input
integrate(arctanh(a*x)^2/(-a^2*x^2+1)^3,x, algorithm="giac")
 
output
integrate(-arctanh(a*x)^2/(a^2*x^2 - 1)^3, x)
 
3.4.11.9 Mupad [B] (verification not implemented)

Time = 4.94 (sec) , antiderivative size = 358, normalized size of antiderivative = 2.37 \[ \int \frac {\text {arctanh}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx=\frac {\frac {17\,x}{8}-\frac {15\,a^2\,x^3}{8}}{8\,a^4\,x^4-16\,a^2\,x^2+8}-\ln \left (1-a\,x\right )\,\left (\frac {3\,{\ln \left (a\,x+1\right )}^2}{64\,a}-\frac {\frac {7\,x}{2}-3\,a\,x^2+\frac {4}{a}-\frac {5\,a^2\,x^3}{2}}{32\,a^4\,x^4-64\,a^2\,x^2+32}+\frac {\frac {7\,x}{2}+3\,a\,x^2-\frac {4}{a}-\frac {5\,a^2\,x^3}{2}}{32\,a^4\,x^4-64\,a^2\,x^2+32}+\frac {\ln \left (a\,x+1\right )\,\left (10\,x-6\,a^2\,x^3\right )}{32\,a^4\,x^4-64\,a^2\,x^2+32}\right )+{\ln \left (1-a\,x\right )}^2\,\left (\frac {3\,\ln \left (a\,x+1\right )}{64\,a}+\frac {\frac {5\,x}{8}-\frac {3\,a^2\,x^3}{8}}{4\,a^4\,x^4-8\,a^2\,x^2+4}\right )+\frac {{\ln \left (a\,x+1\right )}^3}{64\,a}-\frac {{\ln \left (1-a\,x\right )}^3}{64\,a}-\frac {\ln \left (a\,x+1\right )\,\left (\frac {1}{4\,a^2}-\frac {3\,x^2}{16}\right )}{\frac {1}{a}-2\,a\,x^2+a^3\,x^4}+\frac {{\ln \left (a\,x+1\right )}^2\,\left (\frac {5\,x}{32\,a}-\frac {3\,a\,x^3}{32}\right )}{\frac {1}{a}-2\,a\,x^2+a^3\,x^4}-\frac {\mathrm {atan}\left (a\,x\,1{}\mathrm {i}\right )\,15{}\mathrm {i}}{64\,a} \]

input
int(-atanh(a*x)^2/(a^2*x^2 - 1)^3,x)
 
output
((17*x)/8 - (15*a^2*x^3)/8)/(8*a^4*x^4 - 16*a^2*x^2 + 8) - log(1 - a*x)*(( 
3*log(a*x + 1)^2)/(64*a) - ((7*x)/2 - 3*a*x^2 + 4/a - (5*a^2*x^3)/2)/(32*a 
^4*x^4 - 64*a^2*x^2 + 32) + ((7*x)/2 + 3*a*x^2 - 4/a - (5*a^2*x^3)/2)/(32* 
a^4*x^4 - 64*a^2*x^2 + 32) + (log(a*x + 1)*(10*x - 6*a^2*x^3))/(32*a^4*x^4 
 - 64*a^2*x^2 + 32)) + log(1 - a*x)^2*((3*log(a*x + 1))/(64*a) + ((5*x)/8 
- (3*a^2*x^3)/8)/(4*a^4*x^4 - 8*a^2*x^2 + 4)) + log(a*x + 1)^3/(64*a) - lo 
g(1 - a*x)^3/(64*a) - (atan(a*x*1i)*15i)/(64*a) - (log(a*x + 1)*(1/(4*a^2) 
 - (3*x^2)/16))/(1/a - 2*a*x^2 + a^3*x^4) + (log(a*x + 1)^2*((5*x)/(32*a) 
- (3*a*x^3)/32))/(1/a - 2*a*x^2 + a^3*x^4)