3.392 \(\int \frac{\tanh ^{-1}(a x)}{x (1-a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=112 \[ \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{a x}{\sqrt{1-a^2 x^2}}+\frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}}-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.185961, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6030, 6018, 5994, 191} \[ \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{a x}{\sqrt{1-a^2 x^2}}+\frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}}-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(x*(1 - a^2*x^2)^(3/2)),x]

[Out]

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

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 6018

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

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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.178578, size = 97, normalized size = 0.87 \[ \text{PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )-\text{PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right )-\frac{a x}{\sqrt{1-a^2 x^2}}+\frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}}+\tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x) \log \left (e^{-\tanh ^{-1}(a x)}+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]/(x*(1 - a^2*x^2)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.253, size = 157, normalized size = 1.4 \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) -1}{2\,ax-2}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{\it Artanh} \left ( ax \right ) +1}{2\,ax+2}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\it Artanh} \left ( ax \right ) \ln \left ( 1+{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) -{\it polylog} \left ( 2,-{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) +{\it Artanh} \left ( ax \right ) \ln \left ( 1-{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) +{\it polylog} \left ( 2,{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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 )}^{\frac{3}{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} x^{2} + 1} \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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)*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 (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{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)**(3/2),x)

[Out]

Integral(atanh(a*x)/(x*(-(a*x - 1)*(a*x + 1))**(3/2)), 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 )}^{\frac{3}{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)^(3/2),x, algorithm="giac")

[Out]

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