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

Optimal. Leaf size=146 \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a^3}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a^3}-\frac{\sqrt{1-a^2 x^2}}{2 a^3}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 a^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{a^3} \]

[Out]

-Sqrt[1 - a^2*x^2]/(2*a^3) - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(2*a^2) - (ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]
*ArcTanh[a*x])/a^3 - ((I/2)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3 + ((I/2)*PolyLog[2, (I*Sqrt[1
- a*x])/Sqrt[1 + a*x]])/a^3

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Rubi [A]  time = 0.104588, antiderivative size = 146, 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 = {6016, 261, 5950} \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a^3}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{2 a^3}-\frac{\sqrt{1-a^2 x^2}}{2 a^3}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 a^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - a^2*x^2]/(2*a^3) - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(2*a^2) - (ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]
*ArcTanh[a*x])/a^3 - ((I/2)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3 + ((I/2)*PolyLog[2, (I*Sqrt[1
- a*x])/Sqrt[1 + a*x]])/a^3

Rule 6016

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

Rule 261

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

Rule 5950

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-2*(a + b*ArcTanh[c*x])*
ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(c*Sqrt[d]), x] + (-Simp[(I*b*PolyLog[2, -((I*Sqrt[1 - c*x])/Sqrt[1 + c*x
])])/(c*Sqrt[d]), x] + Simp[(I*b*PolyLog[2, (I*Sqrt[1 - c*x])/Sqrt[1 + c*x]])/(c*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{x^2 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx &=-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 a^2}+\frac{\int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{2 a^2}+\frac{\int \frac{x}{\sqrt{1-a^2 x^2}} \, dx}{2 a}\\ &=-\frac{\sqrt{1-a^2 x^2}}{2 a^3}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 a^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{a^3}-\frac{i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{2 a^3}+\frac{i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{2 a^3}\\ \end{align*}

Mathematica [A]  time = 0.218912, size = 125, normalized size = 0.86 \[ -\frac{i \text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-i \text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )+\sqrt{1-a^2 x^2}+a x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )}{2 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Sqrt[1 - a^2*x^2] + a*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] + I*ArcTanh[a*x]*Log[1 - I/E^ArcTanh[a*x]] - I*ArcTan
h[a*x]*Log[1 + I/E^ArcTanh[a*x]] + I*PolyLog[2, (-I)/E^ArcTanh[a*x]] - I*PolyLog[2, I/E^ArcTanh[a*x]])/(2*a^3)

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Maple [A]  time = 0.26, size = 154, normalized size = 1.1 \begin{align*} -{\frac{ax{\it Artanh} \left ( ax \right ) +1}{2\,{a}^{3}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{{\frac{i}{2}}{\it Artanh} \left ( ax \right ) }{{a}^{3}}\ln \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{2}}{\it Artanh} \left ( ax \right ) }{{a}^{3}}\ln \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }-{\frac{{\frac{i}{2}}}{{a}^{3}}{\it dilog} \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{2}}}{{a}^{3}}{\it dilog} \left ( 1-{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*x*arctanh(a*x)+1)*(-(a*x-1)*(a*x+1))^(1/2)/a^3-1/2*I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a
^3+1/2*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^3-1/2*I*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1
/2*I*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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