3.270 \(\int \frac{\tanh ^{-1}(a x)^2}{x (1-a^2 x^2)^2} \, dx\)
Optimal. Leaf size=136 \[ -\frac{1}{2} \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )-\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}-\frac{a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac{1}{3} \tanh ^{-1}(a x)^3-\frac{1}{4} \tanh ^{-1}(a x)^2+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]
[Out]
1/(4*(1 - a^2*x^2)) - (a*x*ArcTanh[a*x])/(2*(1 - a^2*x^2)) - ArcTanh[a*x]^2/4 + ArcTanh[a*x]^2/(2*(1 - a^2*x^2
)) + ArcTanh[a*x]^3/3 + ArcTanh[a*x]^2*Log[2 - 2/(1 + a*x)] - ArcTanh[a*x]*PolyLog[2, -1 + 2/(1 + a*x)] - Poly
Log[3, -1 + 2/(1 + a*x)]/2
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Rubi [A] time = 0.292017, antiderivative size = 136, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used =
{6030, 5988, 5932, 5948, 6056, 6610, 5994, 5956, 261} \[ -\frac{1}{2} \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )-\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}-\frac{a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac{1}{3} \tanh ^{-1}(a x)^3-\frac{1}{4} \tanh ^{-1}(a x)^2+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]
Antiderivative was successfully verified.
[In]
Int[ArcTanh[a*x]^2/(x*(1 - a^2*x^2)^2),x]
[Out]
1/(4*(1 - a^2*x^2)) - (a*x*ArcTanh[a*x])/(2*(1 - a^2*x^2)) - ArcTanh[a*x]^2/4 + ArcTanh[a*x]^2/(2*(1 - a^2*x^2
)) + ArcTanh[a*x]^3/3 + ArcTanh[a*x]^2*Log[2 - 2/(1 + a*x)] - ArcTanh[a*x]*PolyLog[2, -1 + 2/(1 + a*x)] - Poly
Log[3, -1 + 2/(1 + a*x)]/2
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 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 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 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 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]
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 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 261
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )^2} \, dx &=a^2 \int \frac{x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac{\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )} \, dx\\ &=\frac{\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac{1}{3} \tanh ^{-1}(a x)^3-a \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac{\tanh ^{-1}(a x)^2}{x (1+a x)} \, dx\\ &=-\frac{a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}-\frac{1}{4} \tanh ^{-1}(a x)^2+\frac{\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac{1}{3} \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )-(2 a) \int \frac{\tanh ^{-1}(a x) \log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx+\frac{1}{2} a^2 \int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac{1}{4 \left (1-a^2 x^2\right )}-\frac{a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}-\frac{1}{4} \tanh ^{-1}(a x)^2+\frac{\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac{1}{3} \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )-\tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )+a \int \frac{\text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac{1}{4 \left (1-a^2 x^2\right )}-\frac{a x \tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}-\frac{1}{4} \tanh ^{-1}(a x)^2+\frac{\tanh ^{-1}(a x)^2}{2 \left (1-a^2 x^2\right )}+\frac{1}{3} \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )-\tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )-\frac{1}{2} \text{Li}_3\left (-1+\frac{2}{1+a x}\right )\\ \end{align*}
Mathematica [C] time = 0.183503, size = 106, normalized size = 0.78 \[ \frac{1}{24} \left (24 \tanh ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )-12 \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )-8 \tanh ^{-1}(a x)^3+24 \tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \sinh \left (2 \tanh ^{-1}(a x)\right )+6 \tanh ^{-1}(a x)^2 \cosh \left (2 \tanh ^{-1}(a x)\right )+3 \cosh \left (2 \tanh ^{-1}(a x)\right )+i \pi ^3\right ) \]
Warning: Unable to verify antiderivative.
[In]
Integrate[ArcTanh[a*x]^2/(x*(1 - a^2*x^2)^2),x]
[Out]
(I*Pi^3 - 8*ArcTanh[a*x]^3 + 3*Cosh[2*ArcTanh[a*x]] + 6*ArcTanh[a*x]^2*Cosh[2*ArcTanh[a*x]] + 24*ArcTanh[a*x]^
2*Log[1 - E^(2*ArcTanh[a*x])] + 24*ArcTanh[a*x]*PolyLog[2, E^(2*ArcTanh[a*x])] - 12*PolyLog[3, E^(2*ArcTanh[a*
x])] - 6*ArcTanh[a*x]*Sinh[2*ArcTanh[a*x]])/24
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Maple [C] time = 0.431, size = 1290, normalized size = 9.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(arctanh(a*x)^2/x/(-a^2*x^2+1)^2,x)
[Out]
arctanh(a*x)^2*ln(a*x)-arctanh(a*x)^2*ln((a*x+1)^2/(-a^2*x^2+1)-1)+arctanh(a*x)^2*ln(1-(a*x+1)/(-a^2*x^2+1)^(1
/2))+arctanh(a*x)^2*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+2*arctanh(a*x)*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+2*a
rctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*I*Pi*arctanh(a*x)^2+1/8*(a*x+1)*arctanh(a*x)/(a*x-1)-1/8
*(a*x-1)*arctanh(a*x)/(a*x+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)+1))*c
sgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a*x)^2-1/16*(a*x-1)/(a*x+1)+arctanh(a*x)^
2*ln(2)-1/16*(a*x+1)/(a*x-1)-1/3*arctanh(a*x)^3-1/4*arctanh(a*x)^2-1/2*I*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))
^2*arctanh(a*x)^2+1/4*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^2+1/4*I*Pi*
csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*arctanh(a*x)^2+1/2*I*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^2+1/
2*I*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^2-2*polylog(3,-(a*x+1)/(-a
^2*x^2+1)^(1/2))-2*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+1/4*I*arctanh(a*x)^2*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))^2-1/4*arctanh(a*x)^2/(a*x-1)-1/2*arctanh(a*x)^2
*ln(a*x-1)+1/4*arctanh(a*x)^2/(a*x+1)-1/2*arctanh(a*x)^2*ln(a*x+1)+arctanh(a*x)^2*ln((a*x+1)/(-a^2*x^2+1)^(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)^2+1/2*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*arctanh(a*x)^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)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^2-1/4*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))*arctanh(a*x
)^2-1/2*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)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^
2*arctanh(a*x)^2-1/2*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)-1)/((a*x+1)^2/(-a^
2*x^2+1)+1))^2*arctanh(a*x)^2
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, a^{4} \int \frac{x^{4} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} + \frac{1}{4} \, a^{3} \int \frac{x^{3} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} - \frac{1}{32} \,{\left (a{\left (\frac{2}{a^{4} x - a^{3}} - \frac{\log \left (a x + 1\right )}{a^{3}} + \frac{\log \left (a x - 1\right )}{a^{3}}\right )} + \frac{4 \, \log \left (-a x + 1\right )}{a^{4} x^{2} - a^{2}}\right )} a^{2} - \frac{1}{4} \, a^{2} \int \frac{x^{2} \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} - \frac{1}{4} \, a \int \frac{x \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} + \frac{1}{4} \, a \int \frac{x \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} - \frac{{\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )^{3} + 3 \,{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) + 1\right )} \log \left (-a x + 1\right )^{2}}{24 \,{\left (a^{2} x^{2} - 1\right )}} + \frac{1}{4} \, \int \frac{\log \left (a x + 1\right )^{2}}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} - \frac{1}{2} \, \int \frac{\log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^2/x/(-a^2*x^2+1)^2,x, algorithm="maxima")
[Out]
1/4*a^4*integrate(x^4*log(a*x + 1)*log(-a*x + 1)/(a^4*x^5 - 2*a^2*x^3 + x), x) + 1/4*a^3*integrate(x^3*log(a*x
+ 1)*log(-a*x + 1)/(a^4*x^5 - 2*a^2*x^3 + x), x) - 1/32*(a*(2/(a^4*x - a^3) - log(a*x + 1)/a^3 + log(a*x - 1)
/a^3) + 4*log(-a*x + 1)/(a^4*x^2 - a^2))*a^2 - 1/4*a^2*integrate(x^2*log(a*x + 1)*log(-a*x + 1)/(a^4*x^5 - 2*a
^2*x^3 + x), x) - 1/4*a*integrate(x*log(a*x + 1)*log(-a*x + 1)/(a^4*x^5 - 2*a^2*x^3 + x), x) + 1/4*a*integrate
(x*log(-a*x + 1)/(a^4*x^5 - 2*a^2*x^3 + x), x) - 1/24*((a^2*x^2 - 1)*log(-a*x + 1)^3 + 3*((a^2*x^2 - 1)*log(a*
x + 1) + 1)*log(-a*x + 1)^2)/(a^2*x^2 - 1) + 1/4*integrate(log(a*x + 1)^2/(a^4*x^5 - 2*a^2*x^3 + x), x) - 1/2*
integrate(log(a*x + 1)*log(-a*x + 1)/(a^4*x^5 - 2*a^2*x^3 + x), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{artanh}\left (a x\right )^{2}}{a^{4} x^{5} - 2 \, a^{2} x^{3} + x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^2/x/(-a^2*x^2+1)^2,x, algorithm="fricas")
[Out]
integral(arctanh(a*x)^2/(a^4*x^5 - 2*a^2*x^3 + x), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{x \left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(atanh(a*x)**2/x/(-a**2*x**2+1)**2,x)
[Out]
Integral(atanh(a*x)**2/(x*(a*x - 1)**2*(a*x + 1)**2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (a x\right )^{2}}{{\left (a^{2} x^{2} - 1\right )}^{2} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^2/x/(-a^2*x^2+1)^2,x, algorithm="giac")
[Out]
integrate(arctanh(a*x)^2/((a^2*x^2 - 1)^2*x), x)