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

Optimal. Leaf size=91 \[ -\frac{1}{2} \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-\frac{a x}{4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac{1}{2} \tanh ^{-1}(a x)^2-\frac{1}{4} \tanh ^{-1}(a x)+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x) \]

[Out]

-(a*x)/(4*(1 - a^2*x^2)) - ArcTanh[a*x]/4 + ArcTanh[a*x]/(2*(1 - a^2*x^2)) + ArcTanh[a*x]^2/2 + ArcTanh[a*x]*L
og[2 - 2/(1 + a*x)] - PolyLog[2, -1 + 2/(1 + a*x)]/2

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Rubi [A]  time = 0.164075, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {6030, 5988, 5932, 2447, 5994, 199, 206} \[ -\frac{1}{2} \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-\frac{a x}{4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac{1}{2} \tanh ^{-1}(a x)^2-\frac{1}{4} \tanh ^{-1}(a x)+\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x)/(4*(1 - a^2*x^2)) - ArcTanh[a*x]/4 + ArcTanh[a*x]/(2*(1 - a^2*x^2)) + ArcTanh[a*x]^2/2 + ArcTanh[a*x]*L
og[2 - 2/(1 + a*x)] - PolyLog[2, -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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, 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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.168184, size = 63, normalized size = 0.69 \[ \frac{1}{8} \left (-4 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )+4 \tanh ^{-1}(a x)^2-\sinh \left (2 \tanh ^{-1}(a x)\right )+2 \tanh ^{-1}(a x) \left (4 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )+\cosh \left (2 \tanh ^{-1}(a x)\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*ArcTanh[a*x]^2 + 2*ArcTanh[a*x]*(Cosh[2*ArcTanh[a*x]] + 4*Log[1 - E^(-2*ArcTanh[a*x])]) - 4*PolyLog[2, E^(-
2*ArcTanh[a*x])] - Sinh[2*ArcTanh[a*x]])/8

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Maple [B]  time = 0.064, size = 190, normalized size = 2.1 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) }{4\,ax-4}}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) }{2}}+{\it Artanh} \left ( ax \right ) \ln \left ( ax \right ) +{\frac{{\it Artanh} \left ( ax \right ) }{4\,ax+4}}-{\frac{{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) }{2}}-{\frac{{\it dilog} \left ( ax \right ) }{2}}-{\frac{{\it dilog} \left ( ax+1 \right ) }{2}}-{\frac{\ln \left ( ax \right ) \ln \left ( ax+1 \right ) }{2}}-{\frac{ \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{8}}+{\frac{1}{2}{\it dilog} \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{\ln \left ( ax-1 \right ) }{4}\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) }+{\frac{ \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{8}}-{\frac{1}{4} \left ( \ln \left ( ax+1 \right ) -\ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) \right ) \ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) }+{\frac{1}{8\,ax-8}}+{\frac{\ln \left ( ax-1 \right ) }{8}}+{\frac{1}{8\,ax+8}}-{\frac{\ln \left ( ax+1 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*arctanh(a*x)/(a*x-1)-1/2*arctanh(a*x)*ln(a*x-1)+arctanh(a*x)*ln(a*x)+1/4*arctanh(a*x)/(a*x+1)-1/2*arctanh
(a*x)*ln(a*x+1)-1/2*dilog(a*x)-1/2*dilog(a*x+1)-1/2*ln(a*x)*ln(a*x+1)-1/8*ln(a*x-1)^2+1/2*dilog(1/2+1/2*a*x)+1
/4*ln(a*x-1)*ln(1/2+1/2*a*x)+1/8*ln(a*x+1)^2-1/4*(ln(a*x+1)-ln(1/2+1/2*a*x))*ln(-1/2*a*x+1/2)+1/8/(a*x-1)+1/8*
ln(a*x-1)+1/8/(a*x+1)-1/8*ln(a*x+1)

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Maxima [B]  time = 0.981946, size = 278, normalized size = 3.05 \begin{align*} \frac{1}{8} \, a{\left (\frac{{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) -{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} + 2 \, a x}{a^{3} x^{2} - a} + \frac{4 \,{\left (\log \left (a x - 1\right ) \log \left (\frac{1}{2} \, a x + \frac{1}{2}\right ) +{\rm Li}_2\left (-\frac{1}{2} \, a x + \frac{1}{2}\right )\right )}}{a} - \frac{4 \,{\left (\log \left (a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-a x\right )\right )}}{a} + \frac{4 \,{\left (\log \left (-a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (a x\right )\right )}}{a} - \frac{\log \left (a x + 1\right )}{a} + \frac{\log \left (a x - 1\right )}{a}\right )} - \frac{1}{2} \,{\left (\frac{1}{a^{2} x^{2} - 1} + \log \left (a^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*(((a^2*x^2 - 1)*log(a*x + 1)^2 - 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) - (a^2*x^2 - 1)*log(a*x - 1)^
2 + 2*a*x)/(a^3*x^2 - a) + 4*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/a - 4*(log(a*x + 1)*log
(x) + dilog(-a*x))/a + 4*(log(-a*x + 1)*log(x) + dilog(a*x))/a - log(a*x + 1)/a + log(a*x - 1)/a) - 1/2*(1/(a^
2*x^2 - 1) + log(a^2*x^2 - 1) - log(x^2))*arctanh(a*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 )}{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)/x/(-a^2*x^2+1)^2,x, algorithm="fricas")

[Out]

integral(arctanh(a*x)/(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}{\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)/x/(-a**2*x**2+1)**2,x)

[Out]

Integral(atanh(a*x)/(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 )}{{\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)/x/(-a^2*x^2+1)^2,x, algorithm="giac")

[Out]

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