3.238 \(\int \frac{\tanh ^{-1}(a x)^2}{x (1-a^2 x^2)} \, dx\)
Optimal. Leaf size=66 \[ -\frac{1}{2} \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )-\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{3} \tanh ^{-1}(a x)^3+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]
[Out]
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)] - PolyLog[3
, -1 + 2/(1 + a*x)]/2
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Rubi [A] time = 0.180857, antiderivative size = 66, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used =
{5988, 5932, 5948, 6056, 6610} \[ -\frac{1}{2} \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )-\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )+\frac{1}{3} \tanh ^{-1}(a x)^3+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]
Antiderivative was successfully verified.
[In]
Int[ArcTanh[a*x]^2/(x*(1 - a^2*x^2)),x]
[Out]
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)] - PolyLog[3
, -1 + 2/(1 + a*x)]/2
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]
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )} \, dx &=\frac{1}{3} \tanh ^{-1}(a x)^3+\int \frac{\tanh ^{-1}(a x)^2}{x (1+a x)} \, dx\\ &=\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}{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}{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 [A] time = 0.0626553, size = 60, normalized size = 0.91 \[ \tanh ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )-\frac{1}{3} \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right ) \]
Warning: Unable to verify antiderivative.
[In]
Integrate[ArcTanh[a*x]^2/(x*(1 - a^2*x^2)),x]
[Out]
-ArcTanh[a*x]^3/3 + ArcTanh[a*x]^2*Log[1 - E^(2*ArcTanh[a*x])] + ArcTanh[a*x]*PolyLog[2, E^(2*ArcTanh[a*x])] -
PolyLog[3, E^(2*ArcTanh[a*x])]/2
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Maple [C] time = 0.268, size = 1188, normalized size = 18. \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),x)
[Out]
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))^2*arc
tanh(a*x)^2+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*arctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+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))*csgn(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/
3*arctanh(a*x)^3+arctanh(a*x)^2*ln(2)-1/2*arctanh(a*x)^2*ln(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))+1/2*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*arcta
nh(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)/(-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)/(-a^2*x^2+1)^(1/2))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*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)^2-2*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))+1/2*I*Pi*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)/((
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)^2-1/2*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}{8} \, \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2} - \frac{1}{24} \, \log \left (-a x + 1\right )^{3} + \frac{1}{4} \, \int \frac{{\left (a^{2} x^{2} + a x + 2\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) - \log \left (a x + 1\right )^{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),x, algorithm="maxima")
[Out]
-1/8*log(a*x + 1)*log(-a*x + 1)^2 - 1/24*log(-a*x + 1)^3 + 1/4*integrate(((a^2*x^2 + a*x + 2)*log(a*x + 1)*log
(-a*x + 1) - log(a*x + 1)^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^{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),x, algorithm="fricas")
[Out]
integral(-arctanh(a*x)^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 )}}{a^{2} x^{3} - x}\, 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),x)
[Out]
-Integral(atanh(a*x)**2/(a**2*x**3 - x), 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 )} 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),x, algorithm="giac")
[Out]
integrate(-arctanh(a*x)^2/((a^2*x^2 - 1)*x), x)