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

Optimal. Leaf size=137 \[ \frac{1}{2} a^2 \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{1}{2} a^2 \text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{a \sqrt{1-a^2 x^2}}{2 x}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 x^2}+a^2 \left (-\tanh ^{-1}(a x)\right ) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.135654, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6026, 264, 6018} \[ \frac{1}{2} a^2 \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{1}{2} a^2 \text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{a \sqrt{1-a^2 x^2}}{2 x}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 x^2}+a^2 \left (-\tanh ^{-1}(a x)\right ) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6026

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

Rule 264

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

Rubi steps

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

Mathematica [A]  time = 0.689746, size = 126, normalized size = 0.92 \[ \frac{1}{8} a^2 \left (4 \text{PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )-4 \text{PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right )+2 \tanh \left (\frac{1}{2} \tanh ^{-1}(a x)\right )+4 \tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-4 \tanh ^{-1}(a x) \log \left (e^{-\tanh ^{-1}(a x)}+1\right )-2 \coth \left (\frac{1}{2} \tanh ^{-1}(a x)\right )-\tanh ^{-1}(a x) \text{csch}^2\left (\frac{1}{2} \tanh ^{-1}(a x)\right )-\tanh ^{-1}(a x) \text{sech}^2\left (\frac{1}{2} \tanh ^{-1}(a x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*(-2*Coth[ArcTanh[a*x]/2] - ArcTanh[a*x]*Csch[ArcTanh[a*x]/2]^2 + 4*ArcTanh[a*x]*Log[1 - E^(-ArcTanh[a*x])
] - 4*ArcTanh[a*x]*Log[1 + E^(-ArcTanh[a*x])] + 4*PolyLog[2, -E^(-ArcTanh[a*x])] - 4*PolyLog[2, E^(-ArcTanh[a*
x])] - ArcTanh[a*x]*Sech[ArcTanh[a*x]/2]^2 + 2*Tanh[ArcTanh[a*x]/2]))/8

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Maple [A]  time = 0.254, size = 141, normalized size = 1. \begin{align*} -{\frac{ax+{\it Artanh} \left ( ax \right ) }{2\,{x}^{2}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) }{2}\ln \left ( 1+{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{a}^{2}}{2}{\it polylog} \left ( 2,-{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{a}^{2}{\it Artanh} \left ( ax \right ) }{2}\ln \left ( 1-{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{a}^{2}}{2}{\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^3/(-a^2*x^2+1)^(1/2),x)

[Out]

-1/2*(-(a*x-1)*(a*x+1))^(1/2)*(a*x+arctanh(a*x))/x^2-1/2*a^2*arctanh(a*x)*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-1/2
*a^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*a^2*arctanh(a*x)*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*a^2*poly
log(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 )}{\sqrt{-a^{2} x^{2} + 1} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(arctanh(a*x)/(sqrt(-a^2*x^2 + 1)*x^3), 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^{2} x^{5} - x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*arctanh(a*x)/(a^2*x^5 - x^3), 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^{3} \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atanh(a*x)/(x**3*sqrt(-(a*x - 1)*(a*x + 1))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

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