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

Optimal. Leaf size=48 \[ -\frac{1}{2} \text{PolyLog}(2,-a x)+\frac{1}{2} \text{PolyLog}(2,a x)-\frac{1}{2} a^2 x^2 \tanh ^{-1}(a x)-\frac{a x}{2}+\frac{1}{2} \tanh ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.0473091, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {6014, 5912, 5916, 321, 206} \[ -\frac{1}{2} \text{PolyLog}(2,-a x)+\frac{1}{2} \text{PolyLog}(2,a x)-\frac{1}{2} a^2 x^2 \tanh ^{-1}(a x)-\frac{a x}{2}+\frac{1}{2} \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)*ArcTanh[a*x])/x,x]

[Out]

-(a*x)/2 + ArcTanh[a*x]/2 - (a^2*x^2*ArcTanh[a*x])/2 - PolyLog[2, -(a*x)]/2 + PolyLog[2, 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 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

Mathematica [A]  time = 0.0177563, size = 60, normalized size = 1.25 \[ \frac{1}{2} (\text{PolyLog}(2,a x)-\text{PolyLog}(2,-a x))-\frac{1}{2} a^2 x^2 \tanh ^{-1}(a x)-\frac{a x}{2}-\frac{1}{4} \log (1-a x)+\frac{1}{4} \log (a x+1) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 69, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) }{2}}+{\it Artanh} \left ( ax \right ) \ln \left ( ax \right ) -{\frac{ax}{2}}-{\frac{\ln \left ( ax-1 \right ) }{4}}+{\frac{\ln \left ( ax+1 \right ) }{4}}-{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.952142, size = 120, normalized size = 2.5 \begin{align*} -\frac{1}{4} \, a{\left (2 \, x + \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} - \frac{\log \left (a x + 1\right )}{a} + \frac{\log \left (a x - 1\right )}{a}\right )} - \frac{1}{2} \,{\left (a^{2} x^{2} - \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((-a^2*x^2+1)*arctanh(a*x)/x,x, algorithm="maxima")

[Out]

-1/4*a*(2*x + 2*(log(a*x + 1)*log(x) + dilog(-a*x))/a - 2*(log(-a*x + 1)*log(x) + dilog(a*x))/a - log(a*x + 1)
/a + log(a*x - 1)/a) - 1/2*(a^2*x^2 - 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{{\left (a^{2} x^{2} - 1\right )} \operatorname{artanh}\left (a x\right )}{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)/x,x, algorithm="fricas")

[Out]

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

[Out]

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

[Out]

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