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

Optimal. Leaf size=88 \[ -\frac {6}{a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5962, 5958} \[ -\frac {6}{a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5958

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 45, normalized size = 0.51 \[ \frac {a x \tanh ^{-1}(a x)^3-3 \tanh ^{-1}(a x)^2+6 a x \tanh ^{-1}(a x)-6}{a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 87, normalized size = 0.99 \[ -\frac {{\left (a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 24 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 6 \, \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 48\right )} \sqrt {-a^{2} x^{2} + 1}}{8 \, {\left (a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.27, size = 56, normalized size = 0.64 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \left (\arctanh \left (a x \right )^{3} a x +6 a x \arctanh \left (a x \right )-3 \arctanh \left (a x \right )^{2}-6\right )}{a \left (a^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 86, normalized size = 0.98 \[ \frac {x \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}} + 6 \, a {\left (\frac {x \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} a} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} a^{2}}\right )} - \frac {3 \, \operatorname {artanh}\left (a x\right )^{2}}{\sqrt {-a^{2} x^{2} + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(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 \[ \int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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