3.472 \(\int \frac{\tanh ^{-1}(a x)^2}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=139 \[ \frac{40 x}{27 \sqrt{1-a^2 x^2}}+\frac{2 x}{27 \left (1-a^2 x^2\right )^{3/2}}+\frac{2 x \tanh ^{-1}(a x)^2}{3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac{4 \tanh ^{-1}(a x)}{3 a \sqrt{1-a^2 x^2}}-\frac{2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}} \]

[Out]

(2*x)/(27*(1 - a^2*x^2)^(3/2)) + (40*x)/(27*Sqrt[1 - a^2*x^2]) - (2*ArcTanh[a*x])/(9*a*(1 - a^2*x^2)^(3/2)) -
(4*ArcTanh[a*x])/(3*a*Sqrt[1 - a^2*x^2]) + (x*ArcTanh[a*x]^2)/(3*(1 - a^2*x^2)^(3/2)) + (2*x*ArcTanh[a*x]^2)/(
3*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.0928308, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5964, 5962, 191, 192} \[ \frac{40 x}{27 \sqrt{1-a^2 x^2}}+\frac{2 x}{27 \left (1-a^2 x^2\right )^{3/2}}+\frac{2 x \tanh ^{-1}(a x)^2}{3 \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac{4 \tanh ^{-1}(a x)}{3 a \sqrt{1-a^2 x^2}}-\frac{2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^2/(1 - a^2*x^2)^(5/2),x]

[Out]

(2*x)/(27*(1 - a^2*x^2)^(3/2)) + (40*x)/(27*Sqrt[1 - a^2*x^2]) - (2*ArcTanh[a*x])/(9*a*(1 - a^2*x^2)^(3/2)) -
(4*ArcTanh[a*x])/(3*a*Sqrt[1 - a^2*x^2]) + (x*ArcTanh[a*x]^2)/(3*(1 - a^2*x^2)^(3/2)) + (2*x*ArcTanh[a*x]^2)/(
3*Sqrt[1 - a^2*x^2])

Rule 5964

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*p*(d + e*x^2)^(
q + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q
+ 1)*(a + b*ArcTanh[c*x])^p, x], x] + Dist[(b^2*p*(p - 1))/(4*(q + 1)^2), Int[(d + e*x^2)^q*(a + b*ArcTanh[c*
x])^(p - 2), x], x] - Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c
, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && GtQ[p, 1] && NeQ[q, -3/2]

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]

Rule 192

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

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{5/2}} \, dx &=-\frac{2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}}+\frac{x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac{2}{9} \int \frac{1}{\left (1-a^2 x^2\right )^{5/2}} \, dx+\frac{2}{3} \int \frac{\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=\frac{2 x}{27 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}}-\frac{4 \tanh ^{-1}(a x)}{3 a \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac{2 x \tanh ^{-1}(a x)^2}{3 \sqrt{1-a^2 x^2}}+\frac{4}{27} \int \frac{1}{\left (1-a^2 x^2\right )^{3/2}} \, dx+\frac{4}{3} \int \frac{1}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=\frac{2 x}{27 \left (1-a^2 x^2\right )^{3/2}}+\frac{40 x}{27 \sqrt{1-a^2 x^2}}-\frac{2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}}-\frac{4 \tanh ^{-1}(a x)}{3 a \sqrt{1-a^2 x^2}}+\frac{x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac{2 x \tanh ^{-1}(a x)^2}{3 \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.073697, size = 70, normalized size = 0.5 \[ \frac{-40 a^3 x^3-9 a x \left (2 a^2 x^2-3\right ) \tanh ^{-1}(a x)^2+6 \left (6 a^2 x^2-7\right ) \tanh ^{-1}(a x)+42 a x}{27 a \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]^2/(1 - a^2*x^2)^(5/2),x]

[Out]

(42*a*x - 40*a^3*x^3 + 6*(-7 + 6*a^2*x^2)*ArcTanh[a*x] - 9*a*x*(-3 + 2*a^2*x^2)*ArcTanh[a*x]^2)/(27*a*(1 - a^2
*x^2)^(3/2))

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Maple [A]  time = 0.17, size = 84, normalized size = 0.6 \begin{align*} -{\frac{18\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{x}^{3}{a}^{3}+40\,{x}^{3}{a}^{3}-36\,{a}^{2}{x}^{2}{\it Artanh} \left ( ax \right ) -27\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}xa-42\,ax+42\,{\it Artanh} \left ( ax \right ) }{27\,a \left ({a}^{2}{x}^{2}-1 \right ) ^{2}}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^2/(-a^2*x^2+1)^(5/2),x)

[Out]

-1/27/a*(-a^2*x^2+1)^(1/2)*(18*arctanh(a*x)^2*x^3*a^3+40*x^3*a^3-36*a^2*x^2*arctanh(a*x)-27*arctanh(a*x)^2*x*a
-42*a*x+42*arctanh(a*x))/(a^2*x^2-1)^2

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Maxima [B]  time = 1.6954, size = 410, normalized size = 2.95 \begin{align*} \frac{1}{3} \,{\left (\frac{2 \, x}{\sqrt{-a^{2} x^{2} + 1}} + \frac{x}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\right )} \operatorname{artanh}\left (a x\right )^{2} + \frac{1}{27} \, a{\left (\frac{\frac{2 \, x}{\sqrt{-a^{2} x^{2} + 1}} - \frac{1}{\sqrt{-a^{2} x^{2} + 1} a^{2} x + \sqrt{-a^{2} x^{2} + 1} a}}{a} + \frac{\frac{2 \, x}{\sqrt{-a^{2} x^{2} + 1}} - \frac{1}{\sqrt{-a^{2} x^{2} + 1} a^{2} x - \sqrt{-a^{2} x^{2} + 1} a}}{a} - \frac{18 \, \sqrt{-a^{2} x^{2} + 1}}{{\left (a^{2} x + a\right )} a} - \frac{18 \, \sqrt{-a^{2} x^{2} + 1}}{{\left (a^{2} x - a\right )} a} - \frac{18 \, \log \left (a x + 1\right )}{\sqrt{-a^{2} x^{2} + 1} a^{2}} + \frac{18 \, \log \left (-a x + 1\right )}{\sqrt{-a^{2} x^{2} + 1} a^{2}} - \frac{3 \, \log \left (a x + 1\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{2}} + \frac{3 \, \log \left (-a x + 1\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/(-a^2*x^2+1)^(5/2),x, algorithm="maxima")

[Out]

1/3*(2*x/sqrt(-a^2*x^2 + 1) + x/(-a^2*x^2 + 1)^(3/2))*arctanh(a*x)^2 + 1/27*a*((2*x/sqrt(-a^2*x^2 + 1) - 1/(sq
rt(-a^2*x^2 + 1)*a^2*x + sqrt(-a^2*x^2 + 1)*a))/a + (2*x/sqrt(-a^2*x^2 + 1) - 1/(sqrt(-a^2*x^2 + 1)*a^2*x - sq
rt(-a^2*x^2 + 1)*a))/a - 18*sqrt(-a^2*x^2 + 1)/((a^2*x + a)*a) - 18*sqrt(-a^2*x^2 + 1)/((a^2*x - a)*a) - 18*lo
g(a*x + 1)/(sqrt(-a^2*x^2 + 1)*a^2) + 18*log(-a*x + 1)/(sqrt(-a^2*x^2 + 1)*a^2) - 3*log(a*x + 1)/((-a^2*x^2 +
1)^(3/2)*a^2) + 3*log(-a*x + 1)/((-a^2*x^2 + 1)^(3/2)*a^2))

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Fricas [A]  time = 1.67599, size = 238, normalized size = 1.71 \begin{align*} -\frac{{\left (160 \, a^{3} x^{3} + 9 \,{\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 168 \, a x - 12 \,{\left (6 \, a^{2} x^{2} - 7\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )\right )} \sqrt{-a^{2} x^{2} + 1}}{108 \,{\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/(-a^2*x^2+1)^(5/2),x, algorithm="fricas")

[Out]

-1/108*(160*a^3*x^3 + 9*(2*a^3*x^3 - 3*a*x)*log(-(a*x + 1)/(a*x - 1))^2 - 168*a*x - 12*(6*a^2*x^2 - 7)*log(-(a
*x + 1)/(a*x - 1)))*sqrt(-a^2*x^2 + 1)/(a^5*x^4 - 2*a^3*x^2 + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{2}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**2/(-a**2*x**2+1)**(5/2),x)

[Out]

Integral(atanh(a*x)**2/(-(a*x - 1)*(a*x + 1))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (a x\right )^{2}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/(-a^2*x^2+1)^(5/2),x, algorithm="giac")

[Out]

integrate(arctanh(a*x)^2/(-a^2*x^2 + 1)^(5/2), x)