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

Optimal. Leaf size=93 \[ a \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )-a \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-a^2 x \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)^2}{x}+2 a \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)+2 a \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.217911, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6014, 5916, 5988, 5932, 2447, 5910, 5984, 5918, 2402, 2315} \[ a \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )-a \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )-a^2 x \tanh ^{-1}(a x)^2-\frac{\tanh ^{-1}(a x)^2}{x}+2 a \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)+2 a \log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 5984

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

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.132146, size = 102, normalized size = 1.1 \[ -a \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+a \left (\tanh ^{-1}(a x) \left (-\frac{\tanh ^{-1}(a x)}{a x}+\tanh ^{-1}(a x)+2 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )\right )-\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )\right )-a \tanh ^{-1}(a x) \left (a x \tanh ^{-1}(a x)-\tanh ^{-1}(a x)-2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.055, size = 170, normalized size = 1.8 \begin{align*} -{a}^{2}x \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}-{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{x}}-2\,a{\it Artanh} \left ( ax \right ) \ln \left ( ax-1 \right ) +2\,a{\it Artanh} \left ( ax \right ) \ln \left ( ax \right ) -2\,a{\it Artanh} \left ( ax \right ) \ln \left ( ax+1 \right ) -a{\it dilog} \left ( ax \right ) -a{\it dilog} \left ( ax+1 \right ) -a\ln \left ( ax \right ) \ln \left ( ax+1 \right ) -{\frac{a \left ( \ln \left ( ax-1 \right ) \right ) ^{2}}{2}}+2\,a{\it dilog} \left ( 1/2+1/2\,ax \right ) +a\ln \left ( ax-1 \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) -a\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ( ax+1 \right ) +a\ln \left ( -{\frac{ax}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{ax}{2}} \right ) +{\frac{a \left ( \ln \left ( ax+1 \right ) \right ) ^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*x*arctanh(a*x)^2-arctanh(a*x)^2/x-2*a*arctanh(a*x)*ln(a*x-1)+2*a*arctanh(a*x)*ln(a*x)-2*a*arctanh(a*x)*ln
(a*x+1)-a*dilog(a*x)-a*dilog(a*x+1)-a*ln(a*x)*ln(a*x+1)-1/2*a*ln(a*x-1)^2+2*a*dilog(1/2+1/2*a*x)+a*ln(a*x-1)*l
n(1/2+1/2*a*x)-a*ln(-1/2*a*x+1/2)*ln(a*x+1)+a*ln(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+1/2*a*ln(a*x+1)^2

________________________________________________________________________________________

Maxima [A]  time = 0.985686, size = 205, normalized size = 2.2 \begin{align*} \frac{1}{2} \, a^{2}{\left (\frac{\log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) - \log \left (a x - 1\right )^{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{2 \,{\left (\log \left (a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-a x\right )\right )}}{a} + \frac{2 \,{\left (\log \left (-a x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (a x\right )\right )}}{a}\right )} - 2 \, a{\left (\log \left (a x + 1\right ) + \log \left (a x - 1\right ) - \log \left (x\right )\right )} \operatorname{artanh}\left (a x\right ) -{\left (a^{2} x + \frac{1}{x}\right )} \operatorname{artanh}\left (a x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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^{2}}, 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^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int a^{2} \operatorname{atanh}^{2}{\left (a x \right )}\, dx - \int - \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{x^{2}}\, 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**2,x)

[Out]

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

________________________________________________________________________________________

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^{2}}\,{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^2,x, algorithm="giac")

[Out]

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