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

Optimal. Leaf size=194 \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a^3}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a^3}-\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a^3}+\frac{\sqrt{1-a^2 x^2}}{8 a^3}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{4 a^3} \]

[Out]

Sqrt[1 - a^2*x^2]/(8*a^3) - (1 - a^2*x^2)^(3/2)/(12*a^3) - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(8*a^2) + (x^3*S
qrt[1 - a^2*x^2]*ArcTanh[a*x])/4 - (ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(4*a^3) - ((I/8)*PolyLog
[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3 + ((I/8)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3

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Rubi [A]  time = 0.193429, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6010, 6016, 261, 5950, 266, 43} \[ -\frac{i \text{PolyLog}\left (2,-\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a^3}+\frac{i \text{PolyLog}\left (2,\frac{i \sqrt{1-a x}}{\sqrt{a x+1}}\right )}{8 a^3}-\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a^3}+\frac{\sqrt{1-a^2 x^2}}{8 a^3}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}\right ) \tanh ^{-1}(a x)}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/(8*a^3) - (1 - a^2*x^2)^(3/2)/(12*a^3) - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(8*a^2) + (x^3*S
qrt[1 - a^2*x^2]*ArcTanh[a*x])/4 - (ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(4*a^3) - ((I/8)*PolyLog
[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3 + ((I/8)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3

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 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]

Rule 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x) \, dx &=\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{1}{4} \int \frac{x^2 \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx-\frac{1}{4} a \int \frac{x^3}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)+\frac{\int \frac{\tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{8 a^2}+\frac{\int \frac{x}{\sqrt{1-a^2 x^2}} \, dx}{8 a}-\frac{1}{8} a \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1-a^2 x^2}}{8 a^3}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{4 a^3}-\frac{i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a^3}+\frac{i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a^3}-\frac{1}{8} a \operatorname{Subst}\left (\int \left (\frac{1}{a^2 \sqrt{1-a^2 x}}-\frac{\sqrt{1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )\\ &=\frac{\sqrt{1-a^2 x^2}}{8 a^3}-\frac{\left (1-a^2 x^2\right )^{3/2}}{12 a^3}-\frac{x \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2}+\frac{1}{4} x^3 \sqrt{1-a^2 x^2} \tanh ^{-1}(a x)-\frac{\tan ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{1+a x}}\right ) \tanh ^{-1}(a x)}{4 a^3}-\frac{i \text{Li}_2\left (-\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a^3}+\frac{i \text{Li}_2\left (\frac{i \sqrt{1-a x}}{\sqrt{1+a x}}\right )}{8 a^3}\\ \end{align*}

Mathematica [A]  time = 0.437281, size = 160, normalized size = 0.82 \[ \frac{\sqrt{1-a^2 x^2} \left (-\frac{3 i \left (\text{PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-\text{PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt{1-a^2 x^2}}+2 a^2 x^2+6 a x \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)-\frac{3 i \tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt{1-a^2 x^2}}+3 a x \tanh ^{-1}(a x)+1\right )}{24 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(1 + 2*a^2*x^2 + 3*a*x*ArcTanh[a*x] + 6*a*x*(-1 + a^2*x^2)*ArcTanh[a*x] - ((3*I)*ArcTanh[a*
x]*(Log[1 - I/E^ArcTanh[a*x]] - Log[1 + I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2] - ((3*I)*(PolyLog[2, (-I)/E^ArcT
anh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/(24*a^3)

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Maple [A]  time = 0.26, size = 175, normalized size = 0.9 \begin{align*}{\frac{6\,{a}^{3}{x}^{3}{\it Artanh} \left ( ax \right ) +2\,{a}^{2}{x}^{2}-3\,ax{\it Artanh} \left ( ax \right ) +1}{24\,{a}^{3}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{{\frac{i}{8}}{\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}{8}}{\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}{8}}}{{a}^{3}}{\it dilog} \left ( 1+{i \left ( ax+1 \right ){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) }+{\frac{{\frac{i}{8}}}{{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]

1/24/a^3*(-(a*x-1)*(a*x+1))^(1/2)*(6*a^3*x^3*arctanh(a*x)+2*a^2*x^2-3*a*x*arctanh(a*x)+1)-1/8*I*ln(1+I*(a*x+1)
/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^3+1/8*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^3-1/8*I*dilog(1+
I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1/8*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 \sqrt{-a^{2} x^{2} + 1} x^{2} \operatorname{artanh}\left (a x\right )\,{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(sqrt(-a^2*x^2 + 1)*x^2*arctanh(a*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-a^{2} x^{2} + 1} x^{2} \operatorname{artanh}\left (a x\right ), 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), x)

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

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