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

Optimal. Leaf size=66 \[ -\frac {1}{2} \text {Li}_3\left (\frac {2}{a x+1}-1\right )-\text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)+\frac {1}{3} \tanh ^{-1}(a x)^3+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 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 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 \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*}

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Mathematica [A]  time = 0.07, size = 60, normalized size = 0.91 \[ \tanh ^{-1}(a x) \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )-\frac {1}{3} \tanh ^{-1}(a x)^3+\tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/3*ArcTanh[a*x]^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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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )^{2}}{a^{2} x^{3} - x}, x\right ) \]

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

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)

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maple [C]  time = 0.45, size = 1188, normalized size = 18.00 \[ \text {result too large to display} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{8} \, \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2} - \frac {1}{24} \, \log \left (-a x + 1\right )^{3} + \frac {1}{4} \, \int \frac {{\left (a^{2} x^{2} + a x + 2\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) - \log \left (a x + 1\right )^{2}}{a^{2} x^{3} - x}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int \frac {{\mathrm {atanh}\left (a\,x\right )}^2}{x\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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