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

Optimal. Leaf size=100 \[ \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )+\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\sin ^{-1}(a x)-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \]

[Out]

-ArcSin[a*x] + Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 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 = 0.126076, antiderivative size = 100, 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 = {6010, 6018, 216} \[ \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )+\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\sin ^{-1}(a x)-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcSin[a*x] + Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 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 6010

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((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]))/(f*(m + 2)), x] + (Dist[d/(m + 2), Int[((f*x)^m*(a + b*ArcTanh[c
*x]))/Sqrt[d + e*x^2], x], x] - Dist[(b*c*d)/(f*(m + 2)), Int[(f*x)^(m + 1)/Sqrt[d + e*x^2], x], x]) /; FreeQ[
{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && NeQ[m, -2]

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 216

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

Rubi steps

\begin{align*} \int \frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x} \, dx &=\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-a \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx+\int \frac{\tanh ^{-1}(a x)}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\sin ^{-1}(a x)+\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-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 = 0.115122, size = 91, normalized size = 0.91 \[ \text{PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )-\text{PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right )+\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x) \log \left (e^{-\tanh ^{-1}(a x)}+1\right )-2 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \tanh ^{-1}(a x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-2*ArcTan[Tanh[ArcTanh[a*x]/2]] + Sqrt[1 - a^2*x^2]*ArcTanh[a*x] + ArcTanh[a*x]*Log[1 - E^(-ArcTanh[a*x])] - A
rcTanh[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 = 0.269, size = 113, normalized size = 1.1 \begin{align*} \sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }{\it Artanh} \left ( ax \right ) -2\,\arctan \left ({\frac{ax+1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) -{\it dilog} \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) -{\it dilog} \left ( 1+{(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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(a*x-1)*(a*x+1))^(1/2)*arctanh(a*x)-2*arctan((a*x+1)/(-a^2*x^2+1)^(1/2))-dilog((a*x+1)/(-a^2*x^2+1)^(1/2))-d
ilog(1+(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-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{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname{atanh}{\left (a x \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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