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

Optimal. Leaf size=94 \[ -\frac{6 x}{a \sqrt{1-a^2 x^2}}+\frac{\tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 x \tanh ^{-1}(a x)^2}{a \sqrt{1-a^2 x^2}}+\frac{6 \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}} \]

[Out]

(-6*x)/(a*Sqrt[1 - a^2*x^2]) + (6*ArcTanh[a*x])/(a^2*Sqrt[1 - a^2*x^2]) - (3*x*ArcTanh[a*x]^2)/(a*Sqrt[1 - a^2
*x^2]) + ArcTanh[a*x]^3/(a^2*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.124535, antiderivative size = 94, 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 = {5994, 5962, 191} \[ -\frac{6 x}{a \sqrt{1-a^2 x^2}}+\frac{\tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 x \tanh ^{-1}(a x)^2}{a \sqrt{1-a^2 x^2}}+\frac{6 \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*x)/(a*Sqrt[1 - a^2*x^2]) + (6*ArcTanh[a*x])/(a^2*Sqrt[1 - a^2*x^2]) - (3*x*ArcTanh[a*x]^2)/(a*Sqrt[1 - a^2
*x^2]) + ArcTanh[a*x]^3/(a^2*Sqrt[1 - a^2*x^2])

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 5962

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> -Simp[(b*p*(a + b*ArcTa
nh[c*x])^(p - 1))/(c*d*Sqrt[d + e*x^2]), x] + (Dist[b^2*p*(p - 1), Int[(a + b*ArcTanh[c*x])^(p - 2)/(d + e*x^2
)^(3/2), x], x] + Simp[(x*(a + b*ArcTanh[c*x])^p)/(d*Sqrt[d + e*x^2]), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[c^2*d + e, 0] && GtQ[p, 1]

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 \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\frac{\tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 \int \frac{\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a}\\ &=\frac{6 \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 x \tanh ^{-1}(a x)^2}{a \sqrt{1-a^2 x^2}}+\frac{\tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}-\frac{6 \int \frac{1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a}\\ &=-\frac{6 x}{a \sqrt{1-a^2 x^2}}+\frac{6 \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}}-\frac{3 x \tanh ^{-1}(a x)^2}{a \sqrt{1-a^2 x^2}}+\frac{\tanh ^{-1}(a x)^3}{a^2 \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0618625, size = 45, normalized size = 0.48 \[ \frac{-6 a x+\tanh ^{-1}(a x)^3-3 a x \tanh ^{-1}(a x)^2+6 \tanh ^{-1}(a x)}{a^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*x + 6*ArcTanh[a*x] - 3*a*x*ArcTanh[a*x]^2 + ArcTanh[a*x]^3)/(a^2*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.212, size = 98, normalized size = 1. \begin{align*} -{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}-3\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}+6\,{\it Artanh} \left ( ax \right ) -6}{2\,{a}^{2} \left ( ax-1 \right ) }\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}+3\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}+6\,{\it Artanh} \left ( ax \right ) +6}{2\,{a}^{2} \left ( ax+1 \right ) }\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(arctanh(a*x)^3-3*arctanh(a*x)^2+6*arctanh(a*x)-6)*(-(a*x-1)*(a*x+1))^(1/2)/a^2/(a*x-1)+1/2*(arctanh(a*x)
^3+3*arctanh(a*x)^2+6*arctanh(a*x)+6)*(-(a*x-1)*(a*x+1))^(1/2)/a^2/(a*x+1)

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Maxima [A]  time = 0.960278, size = 119, normalized size = 1.27 \begin{align*} -\frac{3 \, x \operatorname{artanh}\left (a x\right )^{2}}{\sqrt{-a^{2} x^{2} + 1} a} + \frac{\operatorname{artanh}\left (a x\right )^{3}}{\sqrt{-a^{2} x^{2} + 1} a^{2}} - \frac{6 \,{\left (\frac{x}{\sqrt{-a^{2} x^{2} + 1}} - \frac{\operatorname{artanh}\left (a x\right )}{\sqrt{-a^{2} x^{2} + 1} a}\right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x*arctanh(a*x)^2/(sqrt(-a^2*x^2 + 1)*a) + arctanh(a*x)^3/(sqrt(-a^2*x^2 + 1)*a^2) - 6*(x/sqrt(-a^2*x^2 + 1)
- arctanh(a*x)/(sqrt(-a^2*x^2 + 1)*a))/a

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Fricas [A]  time = 2.28271, size = 197, normalized size = 2.1 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1}{\left (6 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - \log \left (-\frac{a x + 1}{a x - 1}\right )^{3} + 48 \, a x - 24 \, \log \left (-\frac{a x + 1}{a x - 1}\right )\right )}}{8 \,{\left (a^{4} x^{2} - a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(-a^2*x^2 + 1)*(6*a*x*log(-(a*x + 1)/(a*x - 1))^2 - log(-(a*x + 1)/(a*x - 1))^3 + 48*a*x - 24*log(-(a*
x + 1)/(a*x - 1)))/(a^4*x^2 - a^2)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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