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

Optimal. Leaf size=168 \[ -\frac{3}{2} a^2 \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )+\frac{3}{2} a^2 \text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{a \sqrt{1-a^2 x^2}}{2 x}-a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 x^2}+a^2 \sin ^{-1}(a x)+3 a^2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(2*x) + a^2*ArcSin[a*x] - a^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - (Sqrt[1 - a^2*x^2]*ArcTa
nh[a*x])/(2*x^2) + 3*a^2*ArcTanh[a*x]*ArcTanh[Sqrt[1 - a*x]/Sqrt[1 + a*x]] - (3*a^2*PolyLog[2, -(Sqrt[1 - a*x]
/Sqrt[1 + a*x])])/2 + (3*a^2*PolyLog[2, Sqrt[1 - a*x]/Sqrt[1 + a*x]])/2

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Rubi [A]  time = 0.387339, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6014, 6010, 6026, 264, 6018, 216} \[ -\frac{3}{2} a^2 \text{PolyLog}\left (2,-\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )+\frac{3}{2} a^2 \text{PolyLog}\left (2,\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right )-\frac{a \sqrt{1-a^2 x^2}}{2 x}-a^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 x^2}+a^2 \sin ^{-1}(a x)+3 a^2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Sqrt[1 - a^2*x^2])/(2*x) + a^2*ArcSin[a*x] - a^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - (Sqrt[1 - a^2*x^2]*ArcTa
nh[a*x])/(2*x^2) + 3*a^2*ArcTanh[a*x]*ArcTanh[Sqrt[1 - a*x]/Sqrt[1 + a*x]] - (3*a^2*PolyLog[2, -(Sqrt[1 - a*x]
/Sqrt[1 + a*x])])/2 + (3*a^2*PolyLog[2, Sqrt[1 - a*x]/Sqrt[1 + 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 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 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]

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

Mathematica [A]  time = 0.942964, size = 158, normalized size = 0.94 \[ \frac{1}{8} a^2 \left (-12 \text{PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )+12 \text{PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right )-8 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+2 \tanh \left (\frac{1}{2} \tanh ^{-1}(a x)\right )-12 \tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )+12 \tanh ^{-1}(a x) \log \left (e^{-\tanh ^{-1}(a x)}+1\right )+16 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \tanh ^{-1}(a x)\right )\right )-2 \coth \left (\frac{1}{2} \tanh ^{-1}(a x)\right )-\tanh ^{-1}(a x) \text{csch}^2\left (\frac{1}{2} \tanh ^{-1}(a x)\right )-\tanh ^{-1}(a x) \text{sech}^2\left (\frac{1}{2} \tanh ^{-1}(a x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*(16*ArcTan[Tanh[ArcTanh[a*x]/2]] - 8*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 2*Coth[ArcTanh[a*x]/2] - ArcTanh[a*
x]*Csch[ArcTanh[a*x]/2]^2 - 12*ArcTanh[a*x]*Log[1 - E^(-ArcTanh[a*x])] + 12*ArcTanh[a*x]*Log[1 + E^(-ArcTanh[a
*x])] - 12*PolyLog[2, -E^(-ArcTanh[a*x])] + 12*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 = 0.204, size = 145, normalized size = 0.9 \begin{align*} -{\frac{2\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) +ax+{\it Artanh} \left ( ax \right ) }{2\,{x}^{2}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+2\,{a}^{2}\arctan \left ({\frac{ax+1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) +{\frac{3\,{a}^{2}}{2}{\it dilog} \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{3\,{a}^{2}}{2}{\it dilog} \left ( 1+{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{3\,{a}^{2}{\it Artanh} \left ( ax \right ) }{2}\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((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^3,x)

[Out]

-1/2*(-(a*x-1)*(a*x+1))^(1/2)*(2*a^2*x^2*arctanh(a*x)+a*x+arctanh(a*x))/x^2+2*a^2*arctan((a*x+1)/(-a^2*x^2+1)^
(1/2))+3/2*a^2*dilog((a*x+1)/(-a^2*x^2+1)^(1/2))+3/2*a^2*dilog(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+3/2*a^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{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} \operatorname{artanh}\left (a x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x)/x**3, 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 )}^{\frac{3}{2}} \operatorname{artanh}\left (a x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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