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

Optimal. Leaf size=74 \[ -\frac{x}{a^3 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}+\frac{\tanh ^{-1}(a x)}{a^4 \sqrt{1-a^2 x^2}}-\frac{\sin ^{-1}(a x)}{a^4} \]

[Out]

-(x/(a^3*Sqrt[1 - a^2*x^2])) - ArcSin[a*x]/a^4 + ArcTanh[a*x]/(a^4*Sqrt[1 - a^2*x^2]) + (Sqrt[1 - a^2*x^2]*Arc
Tanh[a*x])/a^4

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Rubi [A]  time = 0.169943, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6028, 5994, 216, 191} \[ -\frac{x}{a^3 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}+\frac{\tanh ^{-1}(a x)}{a^4 \sqrt{1-a^2 x^2}}-\frac{\sin ^{-1}(a x)}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/(a^3*Sqrt[1 - a^2*x^2])) - ArcSin[a*x]/a^4 + ArcTanh[a*x]/(a^4*Sqrt[1 - a^2*x^2]) + (Sqrt[1 - a^2*x^2]*Arc
Tanh[a*x])/a^4

Rule 6028

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int
[x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*A
rcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegersQ[p, 2*q] && LtQ[q, -1] &&
IGtQ[m, 1] && NeQ[p, -1]

Rule 5994

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)
^(q + 1)*(a + b*ArcTanh[c*x])^p)/(2*e*(q + 1)), x] + Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan
h[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0] && NeQ[q, -1]

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]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{x^3 \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\frac{\int \frac{x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^2}-\frac{\int \frac{x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}} \, dx}{a^2}\\ &=\frac{\tanh ^{-1}(a x)}{a^4 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}-\frac{\int \frac{1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^3}-\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{a^3}\\ &=-\frac{x}{a^3 \sqrt{1-a^2 x^2}}-\frac{\sin ^{-1}(a x)}{a^4}+\frac{\tanh ^{-1}(a x)}{a^4 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{a^4}\\ \end{align*}

Mathematica [A]  time = 0.0632908, size = 76, normalized size = 1.03 \[ \frac{a x \sqrt{1-a^2 x^2}+\left (1-a^2 x^2\right ) \sin ^{-1}(a x)+\sqrt{1-a^2 x^2} \left (a^2 x^2-2\right ) \tanh ^{-1}(a x)}{a^4 \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.249, size = 144, normalized size = 2. \begin{align*} -{\frac{{\it Artanh} \left ( ax \right ) -1}{2\,{a}^{4} \left ( ax-1 \right ) }\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{\it Artanh} \left ( ax \right ) +1}{2\,{a}^{4} \left ( ax+1 \right ) }\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{{\it Artanh} \left ( ax \right ) }{{a}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{i}{{a}^{4}}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-i \right ) }-{\frac{i}{{a}^{4}}\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(arctanh(a*x)-1)*(-(a*x-1)*(a*x+1))^(1/2)/a^4/(a*x-1)+1/2*(arctanh(a*x)+1)*(-(a*x-1)*(a*x+1))^(1/2)/a^4/(
a*x+1)+arctanh(a*x)*(-(a*x-1)*(a*x+1))^(1/2)/a^4+I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-I)/a^4-I*ln((a*x+1)/(-a^2*x^2
+1)^(1/2)+I)/a^4

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Maxima [A]  time = 1.45301, size = 146, normalized size = 1.97 \begin{align*} a{\left (\frac{\frac{x}{\sqrt{-a^{2} x^{2} + 1} a^{2}} - \frac{\arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}} a^{2}}}{a^{2}} - \frac{2 \, x}{\sqrt{-a^{2} x^{2} + 1} a^{4}}\right )} -{\left (\frac{x^{2}}{\sqrt{-a^{2} x^{2} + 1} a^{2}} - \frac{2}{\sqrt{-a^{2} x^{2} + 1} a^{4}}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.02198, size = 201, normalized size = 2.72 \begin{align*} \frac{4 \,{\left (a^{2} x^{2} - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt{-a^{2} x^{2} + 1}{\left (2 \, a x +{\left (a^{2} x^{2} - 2\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )\right )}}{2 \,{\left (a^{6} x^{2} - a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(a^2*x^2 - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^2 + 1)*(2*a*x + (a^2*x^2 - 2)*log(-(
a*x + 1)/(a*x - 1))))/(a^6*x^2 - a^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22875, size = 120, normalized size = 1.62 \begin{align*} \frac{{\left (\sqrt{-a^{2} x^{2} + 1} + \frac{1}{\sqrt{-a^{2} x^{2} + 1}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{2 \, a^{4}} - \frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{a^{3}{\left | a \right |}} + \frac{\sqrt{-a^{2} x^{2} + 1} x}{{\left (a^{2} x^{2} - 1\right )} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(-a^2*x^2 + 1) + 1/sqrt(-a^2*x^2 + 1))*log(-(a*x + 1)/(a*x - 1))/a^4 - arcsin(a*x)*sgn(a)/(a^3*abs(a)
) + sqrt(-a^2*x^2 + 1)*x/((a^2*x^2 - 1)*a^3)