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

Optimal. Leaf size=146 \[ \frac{1}{2} \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )-\frac{1}{2} \text{PolyLog}\left (3,\frac{2}{1-a x}-1\right )-\tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )+\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{1-a x}-1\right )-\frac{1}{2} \log \left (1-a^2 x^2\right )-\frac{1}{2} a^2 x^2 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}\left (1-\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2+\frac{1}{2} \tanh ^{-1}(a x)^2-a x \tanh ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.31188, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6014, 5914, 6052, 5948, 6058, 6610, 5916, 5980, 5910, 260} \[ \frac{1}{2} \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )-\frac{1}{2} \text{PolyLog}\left (3,\frac{2}{1-a x}-1\right )-\tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )+\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{1-a x}-1\right )-\frac{1}{2} \log \left (1-a^2 x^2\right )-\frac{1}{2} a^2 x^2 \tanh ^{-1}(a x)^2+2 \tanh ^{-1}\left (1-\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2+\frac{1}{2} \tanh ^{-1}(a x)^2-a x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6014

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Dist
[d, Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(c^2*d)/f^2, Int[(f*x)^(m + 2)*(d +
e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[q
, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))

Rule 5914

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1
- c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTanh[c*x])^(p - 1)*ArcTanh[1 - 2/(1 - c*x)])/(1 - c^2*x^2), x], x]
/; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 6052

Int[(ArcTanh[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[
(Log[1 + u]*(a + b*ArcTanh[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[1 - u]*(a + b*ArcTanh[c*x])^p)/(d
+ e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[u^2 - (1 - 2/(1 - c*x
))^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 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[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 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 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, 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] && GtQ[m, 1]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

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

Rubi steps

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

Mathematica [A]  time = 0.0437522, size = 145, normalized size = 0.99 \[ -\frac{1}{2} \text{PolyLog}\left (3,\frac{-a x-1}{a x-1}\right )+\frac{1}{2} \text{PolyLog}\left (3,\frac{a x+1}{a x-1}\right )+\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{-a x-1}{a x-1}\right )-\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{a x+1}{a x-1}\right )-\frac{1}{2} \log \left (1-a^2 x^2\right )-\frac{1}{2} \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^2+2 \tanh ^{-1}\left (1-\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2-a x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.836, size = 663, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{8} \, a^{2} x^{2} \log \left (-a x + 1\right )^{2} + \frac{1}{4} \, \int -\frac{{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right )^{2} -{\left (a^{3} x^{3} + 2 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )}{a x^{2} - x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^2*x^2*log(-a*x + 1)^2 + 1/4*integrate(-((a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1)^2 - (a^3*x^3 + 2*(a^
3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1))*log(-a*x + 1))/(a*x^2 - x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )^{2}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a^2*x^2 - 1)*arctanh(a*x)^2/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}\, dx - \int a^{2} x \operatorname{atanh}^{2}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )^{2}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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