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

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Rubi [A]  time = 0.12, antiderivative size = 94, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.42, size = 91, normalized size = 0.97 \[ \frac {\sqrt {-a^{2} x^{2} + 1} {\left (6 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 48 \, a x - 24 \, \log \left (-\frac {a x + 1}{a x - 1}\right )\right )}}{8 \, {\left (a^{4} x^{2} - a^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \operatorname {artanh}\left (a x\right )^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.33, size = 98, normalized size = 1.04 \[ -\frac {\left (\arctanh \left (a x \right )^{3}-3 \arctanh \left (a x \right )^{2}+6 \arctanh \left (a x \right )-6\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{2} \left (a x -1\right )}+\frac {\left (\arctanh \left (a x \right )^{3}+3 \arctanh \left (a x \right )^{2}+6 \arctanh \left (a x \right )+6\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{2} \left (a x +1\right )} \]

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 = 0.32, size = 88, normalized size = 0.94 \[ -\frac {3 \, x \operatorname {artanh}\left (a x\right )^{2}}{\sqrt {-a^{2} x^{2} + 1} a} + \frac {\operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {6 \, {\left (\frac {x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {\operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} a}\right )}}{a} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\mathrm {atanh}\left (a\,x\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \operatorname {atanh}^{3}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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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