3.240 \(\int \frac {\tanh ^{-1}(a x)^2}{x^3 (1-a^2 x^2)} \, dx\)

Optimal. Leaf size=138 \[ -\frac {1}{2} a^2 \text {Li}_3\left (\frac {2}{a x+1}-1\right )-a^2 \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)-\frac {1}{2} a^2 \log \left (1-a^2 x^2\right )+a^2 \log (x)+\frac {1}{3} a^2 \tanh ^{-1}(a x)^3+\frac {1}{2} a^2 \tanh ^{-1}(a x)^2+a^2 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{2 x^2}-\frac {a \tanh ^{-1}(a x)}{x} \]

[Out]

-a*arctanh(a*x)/x+1/2*a^2*arctanh(a*x)^2-1/2*arctanh(a*x)^2/x^2+1/3*a^2*arctanh(a*x)^3+a^2*ln(x)-1/2*a^2*ln(-a
^2*x^2+1)+a^2*arctanh(a*x)^2*ln(2-2/(a*x+1))-a^2*arctanh(a*x)*polylog(2,-1+2/(a*x+1))-1/2*a^2*polylog(3,-1+2/(
a*x+1))

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Rubi [A]  time = 0.36, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5982, 5916, 266, 36, 29, 31, 5948, 5988, 5932, 6056, 6610} \[ -\frac {1}{2} a^2 \text {PolyLog}\left (3,\frac {2}{a x+1}-1\right )-a^2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )-\frac {1}{2} a^2 \log \left (1-a^2 x^2\right )+a^2 \log (x)+\frac {1}{3} a^2 \tanh ^{-1}(a x)^3+\frac {1}{2} a^2 \tanh ^{-1}(a x)^2+a^2 \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{2 x^2}-\frac {a \tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^2/(x^3*(1 - a^2*x^2)),x]

[Out]

-((a*ArcTanh[a*x])/x) + (a^2*ArcTanh[a*x]^2)/2 - ArcTanh[a*x]^2/(2*x^2) + (a^2*ArcTanh[a*x]^3)/3 + a^2*Log[x]
- (a^2*Log[1 - a^2*x^2])/2 + a^2*ArcTanh[a*x]^2*Log[2 - 2/(1 + a*x)] - a^2*ArcTanh[a*x]*PolyLog[2, -1 + 2/(1 +
a*x)] - (a^2*PolyLog[3, -1 + 2/(1 + a*x)])/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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 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 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^3 \left (1-a^2 x^2\right )} \, dx &=a^2 \int \frac {\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )} \, dx+\int \frac {\tanh ^{-1}(a x)^2}{x^3} \, dx\\ &=-\frac {\tanh ^{-1}(a x)^2}{2 x^2}+\frac {1}{3} a^2 \tanh ^{-1}(a x)^3+a \int \frac {\tanh ^{-1}(a x)}{x^2 \left (1-a^2 x^2\right )} \, dx+a^2 \int \frac {\tanh ^{-1}(a x)^2}{x (1+a x)} \, dx\\ &=-\frac {\tanh ^{-1}(a x)^2}{2 x^2}+\frac {1}{3} a^2 \tanh ^{-1}(a x)^3+a^2 \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )+a \int \frac {\tanh ^{-1}(a x)}{x^2} \, dx+a^3 \int \frac {\tanh ^{-1}(a x)}{1-a^2 x^2} \, dx-\left (2 a^3\right ) \int \frac {\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {a \tanh ^{-1}(a x)}{x}+\frac {1}{2} a^2 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{2 x^2}+\frac {1}{3} a^2 \tanh ^{-1}(a x)^3+a^2 \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-a^2 \tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )+a^2 \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx+a^3 \int \frac {\text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {a \tanh ^{-1}(a x)}{x}+\frac {1}{2} a^2 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{2 x^2}+\frac {1}{3} a^2 \tanh ^{-1}(a x)^3+a^2 \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-a^2 \tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {1}{2} a^2 \text {Li}_3\left (-1+\frac {2}{1+a x}\right )+\frac {1}{2} a^2 \operatorname {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {a \tanh ^{-1}(a x)}{x}+\frac {1}{2} a^2 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{2 x^2}+\frac {1}{3} a^2 \tanh ^{-1}(a x)^3+a^2 \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-a^2 \tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {1}{2} a^2 \text {Li}_3\left (-1+\frac {2}{1+a x}\right )+\frac {1}{2} a^2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} a^4 \operatorname {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a \tanh ^{-1}(a x)}{x}+\frac {1}{2} a^2 \tanh ^{-1}(a x)^2-\frac {\tanh ^{-1}(a x)^2}{2 x^2}+\frac {1}{3} a^2 \tanh ^{-1}(a x)^3+a^2 \log (x)-\frac {1}{2} a^2 \log \left (1-a^2 x^2\right )+a^2 \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-a^2 \tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {1}{2} a^2 \text {Li}_3\left (-1+\frac {2}{1+a x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.37, size = 133, normalized size = 0.96 \[ -a^2 \left (-\log \left (\frac {a x}{\sqrt {1-a^2 x^2}}\right )+\frac {\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a^2 x^2}-\tanh ^{-1}(a x) \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {1}{2} \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {1}{3} \tanh ^{-1}(a x)^3+\frac {\tanh ^{-1}(a x)}{a x}-\tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-\frac {i \pi ^3}{24}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]^2/(x^3*(1 - a^2*x^2)),x]

[Out]

-(a^2*((-1/24*I)*Pi^3 + ArcTanh[a*x]/(a*x) + ((1 - a^2*x^2)*ArcTanh[a*x]^2)/(2*a^2*x^2) + ArcTanh[a*x]^3/3 - A
rcTanh[a*x]^2*Log[1 - E^(2*ArcTanh[a*x])] - Log[(a*x)/Sqrt[1 - a^2*x^2]] - ArcTanh[a*x]*PolyLog[2, E^(2*ArcTan
h[a*x])] + PolyLog[3, E^(2*ArcTanh[a*x])]/2))

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )^{2}}{a^{2} x^{5} - x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/x^3/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-arctanh(a*x)^2/(a^2*x^5 - x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\operatorname {artanh}\left (a x\right )^{2}}{{\left (a^{2} x^{2} - 1\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/x^3/(-a^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-arctanh(a*x)^2/((a^2*x^2 - 1)*x^3), x)

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maple [C]  time = 0.89, size = 1360, normalized size = 9.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^2/x^3/(-a^2*x^2+1),x)

[Out]

2*a^2*arctanh(a*x)*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+2*a^2*arctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1
/2))-a*arctanh(a*x)/x+a^2*arctanh(a*x)^2*ln(2)+a^2*arctanh(a*x)^2*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+a^2*arctanh
(a*x)^2*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-1/3*a^2*arctanh(a*x)^3+1/2*I*a^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)/(-a
^2*x^2+1)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2-1/4*I*a^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csg
n(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2+1/4*I*a^2*arctanh(a*x)^2*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))+1/4*I*a^2*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I
*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2-1/2*I*a^2*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2
+1)))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2-1/2*I*a^2*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)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2+1/2*a^2*arctanh(a*x)^2-
1/2*arctanh(a*x)^2/x^2-a^2*arctanh(a*x)-2*a^2*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))-2*a^2*polylog(3,(a*x+1)/(
-a^2*x^2+1)^(1/2))+a^2*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*I*a^2*arctanh(a*x)^2*Pi+a^2*ln((a*x+1)/(-a^2*x^2+1
)^(1/2)-1)+a^2*arctanh(a*x)^2*ln(a*x)-1/2*a^2*arctanh(a*x)^2*ln(a*x-1)-a^2*arctanh(a*x)^2*ln((a*x+1)^2/(-a^2*x
^2+1)-1)-1/2*a^2*arctanh(a*x)^2*ln(a*x+1)+a^2*arctanh(a*x)^2*ln((a*x+1)/(-a^2*x^2+1)^(1/2))-1/4*I*a^2*arctanh(
a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*
x+1)^2/(-a^2*x^2+1)))+1/2*I*a^2*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+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)/(1+(a*x+1)^2/(-a^2*x^2+1)))-1/2*I*a^2*arctanh(a*x)^2*Pi*csgn(I/(1+
(a*x+1)^2/(-a^2*x^2+1)))^2+1/4*I*a^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3+1/2*I*a^2*arctanh(a*x)^
2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3+1/4*I*a^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^
2/(-a^2*x^2+1)))^3+1/2*I*a^2*arctanh(a*x)^2*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {a^{2} x^{2} \log \left (-a x + 1\right )^{3} + 3 \, {\left (a^{2} x^{2} \log \left (a x + 1\right ) + 1\right )} \log \left (-a x + 1\right )^{2}}{24 \, x^{2}} + \frac {1}{4} \, \int -\frac {\log \left (a x + 1\right )^{2} - {\left (a^{2} x^{2} + a x + {\left (a^{4} x^{4} + a^{3} x^{3} + 2\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )}{a^{2} x^{5} - x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/x^3/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-1/24*(a^2*x^2*log(-a*x + 1)^3 + 3*(a^2*x^2*log(a*x + 1) + 1)*log(-a*x + 1)^2)/x^2 + 1/4*integrate(-(log(a*x +
1)^2 - (a^2*x^2 + a*x + (a^4*x^4 + a^3*x^3 + 2)*log(a*x + 1))*log(-a*x + 1))/(a^2*x^5 - x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\mathrm {atanh}\left (a\,x\right )}^2}{x^3\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)^2/(x^3*(a^2*x^2 - 1)),x)

[Out]

-int(atanh(a*x)^2/(x^3*(a^2*x^2 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{a^{2} x^{5} - x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**2/x**3/(-a**2*x**2+1),x)

[Out]

-Integral(atanh(a*x)**2/(a**2*x**5 - x**3), x)

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