3.3.13 \(\int \frac {(1-a^2 x^2)^2 \tanh ^{-1}(a x)^2}{x^6} \, dx\) [213]

Optimal. Leaf size=157 \[ -\frac {a^2}{30 x^3}+\frac {11 a^4}{30 x}-\frac {11}{30} a^5 \tanh ^{-1}(a x)-\frac {a \tanh ^{-1}(a x)}{10 x^4}+\frac {7 a^3 \tanh ^{-1}(a x)}{15 x^2}+\frac {8}{15} a^5 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {16}{15} a^5 \tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )-\frac {8}{15} a^5 \text {PolyLog}\left (2,-1+\frac {2}{1+a x}\right ) \]

[Out]

-1/30*a^2/x^3+11/30*a^4/x-11/30*a^5*arctanh(a*x)-1/10*a*arctanh(a*x)/x^4+7/15*a^3*arctanh(a*x)/x^2+8/15*a^5*ar
ctanh(a*x)^2-1/5*arctanh(a*x)^2/x^5+2/3*a^2*arctanh(a*x)^2/x^3-a^4*arctanh(a*x)^2/x+16/15*a^5*arctanh(a*x)*ln(
2-2/(a*x+1))-8/15*a^5*polylog(2,-1+2/(a*x+1))

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Rubi [A]
time = 0.43, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6159, 6037, 6129, 331, 212, 6135, 6079, 2497} \begin {gather*} -\frac {8}{15} a^5 \text {Li}_2\left (\frac {2}{a x+1}-1\right )+\frac {8}{15} a^5 \tanh ^{-1}(a x)^2-\frac {11}{30} a^5 \tanh ^{-1}(a x)+\frac {16}{15} a^5 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)+\frac {11 a^4}{30 x}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {7 a^3 \tanh ^{-1}(a x)}{15 x^2}-\frac {a^2}{30 x^3}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {\tanh ^{-1}(a x)^2}{5 x^5}-\frac {a \tanh ^{-1}(a x)}{10 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)^2*ArcTanh[a*x]^2)/x^6,x]

[Out]

-1/30*a^2/x^3 + (11*a^4)/(30*x) - (11*a^5*ArcTanh[a*x])/30 - (a*ArcTanh[a*x])/(10*x^4) + (7*a^3*ArcTanh[a*x])/
(15*x^2) + (8*a^5*ArcTanh[a*x]^2)/15 - ArcTanh[a*x]^2/(5*x^5) + (2*a^2*ArcTanh[a*x]^2)/(3*x^3) - (a^4*ArcTanh[
a*x]^2)/x + (16*a^5*ArcTanh[a*x]*Log[2 - 2/(1 + a*x)])/15 - (8*a^5*PolyLog[2, -1 + 2/(1 + a*x)])/15

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 6079

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 6129

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

Rule 6135

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 6159

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

Rubi steps

\begin {align*} \int \frac {\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{x^6} \, dx &=\int \left (\frac {\tanh ^{-1}(a x)^2}{x^6}-\frac {2 a^2 \tanh ^{-1}(a x)^2}{x^4}+\frac {a^4 \tanh ^{-1}(a x)^2}{x^2}\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int \frac {\tanh ^{-1}(a x)^2}{x^4} \, dx\right )+a^4 \int \frac {\tanh ^{-1}(a x)^2}{x^2} \, dx+\int \frac {\tanh ^{-1}(a x)^2}{x^6} \, dx\\ &=-\frac {\tanh ^{-1}(a x)^2}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {1}{5} (2 a) \int \frac {\tanh ^{-1}(a x)}{x^5 \left (1-a^2 x^2\right )} \, dx-\frac {1}{3} \left (4 a^3\right ) \int \frac {\tanh ^{-1}(a x)}{x^3 \left (1-a^2 x^2\right )} \, dx+\left (2 a^5\right ) \int \frac {\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx\\ &=a^5 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {1}{5} (2 a) \int \frac {\tanh ^{-1}(a x)}{x^5} \, dx+\frac {1}{5} \left (2 a^3\right ) \int \frac {\tanh ^{-1}(a x)}{x^3 \left (1-a^2 x^2\right )} \, dx-\frac {1}{3} \left (4 a^3\right ) \int \frac {\tanh ^{-1}(a x)}{x^3} \, dx-\frac {1}{3} \left (4 a^5\right ) \int \frac {\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx+\left (2 a^5\right ) \int \frac {\tanh ^{-1}(a x)}{x (1+a x)} \, dx\\ &=-\frac {a \tanh ^{-1}(a x)}{10 x^4}+\frac {2 a^3 \tanh ^{-1}(a x)}{3 x^2}+\frac {1}{3} a^5 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+2 a^5 \tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )+\frac {1}{10} a^2 \int \frac {1}{x^4 \left (1-a^2 x^2\right )} \, dx+\frac {1}{5} \left (2 a^3\right ) \int \frac {\tanh ^{-1}(a x)}{x^3} \, dx-\frac {1}{3} \left (2 a^4\right ) \int \frac {1}{x^2 \left (1-a^2 x^2\right )} \, dx+\frac {1}{5} \left (2 a^5\right ) \int \frac {\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx-\frac {1}{3} \left (4 a^5\right ) \int \frac {\tanh ^{-1}(a x)}{x (1+a x)} \, dx-\left (2 a^6\right ) \int \frac {\log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {a^2}{30 x^3}+\frac {2 a^4}{3 x}-\frac {a \tanh ^{-1}(a x)}{10 x^4}+\frac {7 a^3 \tanh ^{-1}(a x)}{15 x^2}+\frac {8}{15} a^5 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {2}{3} a^5 \tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )-a^5 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )+\frac {1}{10} a^4 \int \frac {1}{x^2 \left (1-a^2 x^2\right )} \, dx+\frac {1}{5} a^4 \int \frac {1}{x^2 \left (1-a^2 x^2\right )} \, dx+\frac {1}{5} \left (2 a^5\right ) \int \frac {\tanh ^{-1}(a x)}{x (1+a x)} \, dx-\frac {1}{3} \left (2 a^6\right ) \int \frac {1}{1-a^2 x^2} \, dx+\frac {1}{3} \left (4 a^6\right ) \int \frac {\log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {a^2}{30 x^3}+\frac {11 a^4}{30 x}-\frac {2}{3} a^5 \tanh ^{-1}(a x)-\frac {a \tanh ^{-1}(a x)}{10 x^4}+\frac {7 a^3 \tanh ^{-1}(a x)}{15 x^2}+\frac {8}{15} a^5 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {16}{15} a^5 \tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )-\frac {1}{3} a^5 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )+\frac {1}{10} a^6 \int \frac {1}{1-a^2 x^2} \, dx+\frac {1}{5} a^6 \int \frac {1}{1-a^2 x^2} \, dx-\frac {1}{5} \left (2 a^6\right ) \int \frac {\log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {a^2}{30 x^3}+\frac {11 a^4}{30 x}-\frac {11}{30} a^5 \tanh ^{-1}(a x)-\frac {a \tanh ^{-1}(a x)}{10 x^4}+\frac {7 a^3 \tanh ^{-1}(a x)}{15 x^2}+\frac {8}{15} a^5 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{5 x^5}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {16}{15} a^5 \tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )-\frac {8}{15} a^5 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 118, normalized size = 0.75 \begin {gather*} \frac {a^2 x^2 \left (-1+11 a^2 x^2\right )+2 (-1+a x)^3 \left (3+9 a x+8 a^2 x^2\right ) \tanh ^{-1}(a x)^2+a x \tanh ^{-1}(a x) \left (-3+14 a^2 x^2-11 a^4 x^4+32 a^4 x^4 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )\right )-16 a^5 x^5 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )}{30 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)^2*ArcTanh[a*x]^2)/x^6,x]

[Out]

(a^2*x^2*(-1 + 11*a^2*x^2) + 2*(-1 + a*x)^3*(3 + 9*a*x + 8*a^2*x^2)*ArcTanh[a*x]^2 + a*x*ArcTanh[a*x]*(-3 + 14
*a^2*x^2 - 11*a^4*x^4 + 32*a^4*x^4*Log[1 - E^(-2*ArcTanh[a*x])]) - 16*a^5*x^5*PolyLog[2, E^(-2*ArcTanh[a*x])])
/(30*x^5)

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Maple [A]
time = 0.57, size = 233, normalized size = 1.48

method	result	size
derivativedivides	\(a^{5} \left (-\frac {\arctanh \left (a x \right )^{2}}{a x}-\frac {\arctanh \left (a x \right )^{2}}{5 a^{5} x^{5}}+\frac {2 \arctanh \left (a x \right )^{2}}{3 a^{3} x^{3}}-\frac {\arctanh \left (a x \right )}{10 a^{4} x^{4}}+\frac {7 \arctanh \left (a x \right )}{15 a^{2} x^{2}}+\frac {16 \arctanh \left (a x \right ) \ln \left (a x \right )}{15}-\frac {8 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{15}-\frac {8 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{15}-\frac {1}{30 a^{3} x^{3}}+\frac {11}{30 a x}-\frac {11 \ln \left (a x +1\right )}{60}+\frac {11 \ln \left (a x -1\right )}{60}-\frac {8 \dilog \left (a x +1\right )}{15}-\frac {8 \ln \left (a x \right ) \ln \left (a x +1\right )}{15}-\frac {8 \dilog \left (a x \right )}{15}-\frac {2 \ln \left (a x -1\right )^{2}}{15}+\frac {8 \dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{15}+\frac {4 \ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{15}+\frac {2 \ln \left (a x +1\right )^{2}}{15}-\frac {4 \left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{15}\right )\)	\(233\)
default	\(a^{5} \left (-\frac {\arctanh \left (a x \right )^{2}}{a x}-\frac {\arctanh \left (a x \right )^{2}}{5 a^{5} x^{5}}+\frac {2 \arctanh \left (a x \right )^{2}}{3 a^{3} x^{3}}-\frac {\arctanh \left (a x \right )}{10 a^{4} x^{4}}+\frac {7 \arctanh \left (a x \right )}{15 a^{2} x^{2}}+\frac {16 \arctanh \left (a x \right ) \ln \left (a x \right )}{15}-\frac {8 \arctanh \left (a x \right ) \ln \left (a x +1\right )}{15}-\frac {8 \arctanh \left (a x \right ) \ln \left (a x -1\right )}{15}-\frac {1}{30 a^{3} x^{3}}+\frac {11}{30 a x}-\frac {11 \ln \left (a x +1\right )}{60}+\frac {11 \ln \left (a x -1\right )}{60}-\frac {8 \dilog \left (a x +1\right )}{15}-\frac {8 \ln \left (a x \right ) \ln \left (a x +1\right )}{15}-\frac {8 \dilog \left (a x \right )}{15}-\frac {2 \ln \left (a x -1\right )^{2}}{15}+\frac {8 \dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{15}+\frac {4 \ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{15}+\frac {2 \ln \left (a x +1\right )^{2}}{15}-\frac {4 \left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{15}\right )\)	\(233\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)^2*arctanh(a*x)^2/x^6,x,method=_RETURNVERBOSE)

[Out]

a^5*(-arctanh(a*x)^2/a/x-1/5*arctanh(a*x)^2/a^5/x^5+2/3*arctanh(a*x)^2/a^3/x^3-1/10*arctanh(a*x)/a^4/x^4+7/15*
arctanh(a*x)/a^2/x^2+16/15*arctanh(a*x)*ln(a*x)-8/15*arctanh(a*x)*ln(a*x+1)-8/15*arctanh(a*x)*ln(a*x-1)-1/30/a
^3/x^3+11/30/a/x-11/60*ln(a*x+1)+11/60*ln(a*x-1)-8/15*dilog(a*x+1)-8/15*ln(a*x)*ln(a*x+1)-8/15*dilog(a*x)-2/15
*ln(a*x-1)^2+8/15*dilog(1/2*a*x+1/2)+4/15*ln(a*x-1)*ln(1/2*a*x+1/2)+2/15*ln(a*x+1)^2-4/15*(ln(a*x+1)-ln(1/2*a*
x+1/2))*ln(-1/2*a*x+1/2))

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Maxima [A]
time = 0.27, size = 239, normalized size = 1.52 \begin {gather*} \frac {1}{60} \, {\left (32 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )} a^{3} - 32 \, {\left (\log \left (a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-a x\right )\right )} a^{3} + 32 \, {\left (\log \left (-a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (a x\right )\right )} a^{3} - 11 \, a^{3} \log \left (a x + 1\right ) + 11 \, a^{3} \log \left (a x - 1\right ) + \frac {2 \, {\left (4 \, a^{3} x^{3} \log \left (a x + 1\right )^{2} - 8 \, a^{3} x^{3} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 4 \, a^{3} x^{3} \log \left (a x - 1\right )^{2} + 11 \, a^{2} x^{2} - 1\right )}}{x^{3}}\right )} a^{2} - \frac {1}{30} \, {\left (16 \, a^{4} \log \left (a^{2} x^{2} - 1\right ) - 16 \, a^{4} \log \left (x^{2}\right ) - \frac {14 \, a^{2} x^{2} - 3}{x^{4}}\right )} a \operatorname {artanh}\left (a x\right ) - \frac {{\left (15 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 3\right )} \operatorname {artanh}\left (a x\right )^{2}}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*x^2+1)^2*arctanh(a*x)^2/x^6,x, algorithm="maxima")

[Out]

1/60*(32*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))*a^3 - 32*(log(a*x + 1)*log(x) + dilog(-a*x)
)*a^3 + 32*(log(-a*x + 1)*log(x) + dilog(a*x))*a^3 - 11*a^3*log(a*x + 1) + 11*a^3*log(a*x - 1) + 2*(4*a^3*x^3*
log(a*x + 1)^2 - 8*a^3*x^3*log(a*x + 1)*log(a*x - 1) - 4*a^3*x^3*log(a*x - 1)^2 + 11*a^2*x^2 - 1)/x^3)*a^2 - 1
/30*(16*a^4*log(a^2*x^2 - 1) - 16*a^4*log(x^2) - (14*a^2*x^2 - 3)/x^4)*a*arctanh(a*x) - 1/15*(15*a^4*x^4 - 10*
a^2*x^2 + 3)*arctanh(a*x)^2/x^5

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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((-a^2*x^2+1)^2*arctanh(a*x)^2/x^6,x, algorithm="fricas")

[Out]

integral((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)^2/x^6, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{2}{\left (a x \right )}}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)**2*atanh(a*x)**2/x**6,x)

[Out]

Integral((a*x - 1)**2*(a*x + 1)**2*atanh(a*x)**2/x**6, x)

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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((-a^2*x^2+1)^2*arctanh(a*x)^2/x^6,x, algorithm="giac")

[Out]

integrate((a^2*x^2 - 1)^2*arctanh(a*x)^2/x^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^2}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atanh(a*x)^2*(a^2*x^2 - 1)^2)/x^6,x)

[Out]

int((atanh(a*x)^2*(a^2*x^2 - 1)^2)/x^6, x)

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