3.213 \(\int \frac {(1-a^2 x^2)^2 \tanh ^{-1}(a x)^2}{x^6} \, dx\)

Optimal. Leaf size=157 \[ -\frac {8}{15} a^5 \text {Li}_2\left (\frac {2}{a x+1}-1\right )+\frac {8}{15} a^5 \tanh ^{-1}(a x)^2-\frac {11}{30} a^5 \tanh ^{-1}(a x)+\frac {16}{15} a^5 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)+\frac {11 a^4}{30 x}-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {7 a^3 \tanh ^{-1}(a x)}{15 x^2}-\frac {a^2}{30 x^3}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}-\frac {\tanh ^{-1}(a x)^2}{5 x^5}-\frac {a \tanh ^{-1}(a x)}{10 x^4} \]

[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 = 0.59, 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 = {6012, 5916, 5982, 325, 206, 5988, 5932, 2447} \[ -\frac {8}{15} a^5 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )-\frac {a^2}{30 x^3}+\frac {7 a^3 \tanh ^{-1}(a x)}{15 x^2}+\frac {2 a^2 \tanh ^{-1}(a x)^2}{3 x^3}+\frac {11 a^4}{30 x}+\frac {8}{15} a^5 \tanh ^{-1}(a x)^2-\frac {11}{30} a^5 \tanh ^{-1}(a x)-\frac {a^4 \tanh ^{-1}(a x)^2}{x}+\frac {16}{15} a^5 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)-\frac {a \tanh ^{-1}(a x)}{10 x^4}-\frac {\tanh ^{-1}(a x)^2}{5 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a^2/(30*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 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 325

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 2447

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 5916

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

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 5982

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 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 6012

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 = 0.82, size = 118, normalized size = 0.75 \[ \frac {-16 a^5 x^5 \text {Li}_2\left (e^{-2 \tanh ^{-1}(a x)}\right )+a^2 x^2 \left (11 a^2 x^2-1\right )+2 (a x-1)^3 \left (8 a^2 x^2+9 a x+3\right ) \tanh ^{-1}(a x)^2+a x \tanh ^{-1}(a x) \left (-11 a^4 x^4+32 a^4 x^4 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )+14 a^2 x^2-3\right )}{30 x^5} \]

Warning: Unable to verify antiderivative.

[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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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {artanh}\left (a x\right )^{2}}{x^{6}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{2}}{x^{6}}\,{d x} \]

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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maple [A]  time = 0.07, size = 272, normalized size = 1.73 \[ -\frac {a^{4} \arctanh \left (a x \right )^{2}}{x}+\frac {2 a^{2} \arctanh \left (a x \right )^{2}}{3 x^{3}}-\frac {\arctanh \left (a x \right )^{2}}{5 x^{5}}-\frac {a \arctanh \left (a x \right )}{10 x^{4}}+\frac {7 a^{3} \arctanh \left (a x \right )}{15 x^{2}}+\frac {16 a^{5} \arctanh \left (a x \right ) \ln \left (a x \right )}{15}-\frac {8 a^{5} \arctanh \left (a x \right ) \ln \left (a x -1\right )}{15}-\frac {8 a^{5} \arctanh \left (a x \right ) \ln \left (a x +1\right )}{15}-\frac {a^{2}}{30 x^{3}}+\frac {11 a^{4}}{30 x}+\frac {11 a^{5} \ln \left (a x -1\right )}{60}-\frac {11 a^{5} \ln \left (a x +1\right )}{60}-\frac {8 a^{5} \dilog \left (a x \right )}{15}-\frac {8 a^{5} \dilog \left (a x +1\right )}{15}-\frac {8 a^{5} \ln \left (a x \right ) \ln \left (a x +1\right )}{15}-\frac {2 a^{5} \ln \left (a x -1\right )^{2}}{15}+\frac {8 a^{5} \dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{15}+\frac {4 a^{5} \ln \left (a x -1\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{15}+\frac {2 a^{5} \ln \left (a x +1\right )^{2}}{15}-\frac {4 a^{5} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{15}+\frac {4 a^{5} \ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{15} \]

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)

[Out]

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

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maxima [A]  time = 0.33, size = 239, normalized size = 1.52 \[ \frac {1}{60} \, {\left (32 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )} a^{3} - 32 \, {\left (\log \left (a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a x\right )\right )} a^{3} + 32 \, {\left (\log \left (-a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (a x\right )\right )} a^{3} - 11 \, a^{3} \log \left (a x + 1\right ) + 11 \, a^{3} \log \left (a x - 1\right ) + \frac {2 \, {\left (4 \, a^{3} x^{3} \log \left (a x + 1\right )^{2} - 8 \, a^{3} x^{3} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 4 \, a^{3} x^{3} \log \left (a x - 1\right )^{2} + 11 \, a^{2} x^{2} - 1\right )}}{x^{3}}\right )} a^{2} - \frac {1}{30} \, {\left (16 \, a^{4} \log \left (a^{2} x^{2} - 1\right ) - 16 \, a^{4} \log \left (x^{2}\right ) - \frac {14 \, a^{2} x^{2} - 3}{x^{4}}\right )} a \operatorname {artanh}\left (a x\right ) - \frac {{\left (15 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 3\right )} \operatorname {artanh}\left (a x\right )^{2}}{15 \, x^{5}} \]

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

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

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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