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

Optimal. Leaf size=105 \[ -\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}+2 a \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 a \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-4 a \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x) \]

[Out]

-4*a*arctanh(a*x)*arctanh((-a*x+1)^(1/2)/(a*x+1)^(1/2))+2*a*polylog(2,-(-a*x+1)^(1/2)/(a*x+1)^(1/2))-2*a*polyl
og(2,(-a*x+1)^(1/2)/(a*x+1)^(1/2))-arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/x

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Rubi [A]  time = 0.17, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6008, 6018} \[ 2 a \text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 a \text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-4 a \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 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)^2}{x^2 \sqrt {1-a^2 x^2}} \, dx &=-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}+(2 a) \int \frac {\tanh ^{-1}(a x)}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{x}-4 a \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+2 a \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-2 a \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 89, normalized size = 0.85 \[ -\frac {\tanh ^{-1}(a x) \left (\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+2 a x \left (\log \left (e^{-\tanh ^{-1}(a x)}+1\right )-\log \left (1-e^{-\tanh ^{-1}(a x)}\right )\right )\right )}{x}+2 a \text {Li}_2\left (-e^{-\tanh ^{-1}(a x)}\right )-2 a \text {Li}_2\left (e^{-\tanh ^{-1}(a x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}}{a^{2} x^{4} - x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*arctanh(a*x)^2/(a^2*x^4 - x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{2}}{\sqrt {-a^{2} x^{2} + 1} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.41, size = 131, normalized size = 1.25 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )^{2}}{x}-2 a \arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-2 a \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+2 a \arctanh \left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+2 a \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{2}}{\sqrt {-a^{2} x^{2} + 1} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2}{x^2\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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