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

Optimal. Leaf size=82 \[ -\frac{a}{\sqrt{1-a^2 x^2}}+\frac{a^2 x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}}-a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x} \]

[Out]

-(a/Sqrt[1 - a^2*x^2]) + (a^2*x*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x - a*ArcTa
nh[Sqrt[1 - a^2*x^2]]

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Rubi [A]  time = 0.170999, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6030, 6008, 266, 63, 208, 5958} \[ -\frac{a}{\sqrt{1-a^2 x^2}}+\frac{a^2 x \tanh ^{-1}(a x)}{\sqrt{1-a^2 x^2}}-a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a/Sqrt[1 - a^2*x^2]) + (a^2*x*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/x - a*ArcTa
nh[Sqrt[1 - a^2*x^2]]

Rule 6030

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/d, Int
[x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/d, Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTanh
[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] && ILtQ[
m, 0] && NeQ[p, -1]

Rule 6008

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

Rule 266

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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
]

Rubi steps

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

Mathematica [A]  time = 0.109833, size = 89, normalized size = 1.09 \[ \frac{a x \left (\sqrt{1-a^2 x^2} \log (x)-\sqrt{1-a^2 x^2} \log \left (\sqrt{1-a^2 x^2}+1\right )-1\right )+\left (2 a^2 x^2-1\right ) \tanh ^{-1}(a x)}{x \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.253, size = 132, normalized size = 1.6 \begin{align*} -{\frac{a \left ({\it Artanh} \left ( ax \right ) -1 \right ) }{2\,ax-2}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{ \left ({\it Artanh} \left ( ax \right ) +1 \right ) a}{2\,ax+2}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}-{\frac{{\it Artanh} \left ( ax \right ) }{x}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) }}+a\ln \left ({(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}-1 \right ) -a\ln \left ( 1+{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.957291, size = 113, normalized size = 1.38 \begin{align*} -a{\left (\frac{1}{\sqrt{-a^{2} x^{2} + 1}} + \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right )\right )} +{\left (\frac{2 \, a^{2} x}{\sqrt{-a^{2} x^{2} + 1}} - \frac{1}{\sqrt{-a^{2} x^{2} + 1} x}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(1/sqrt(-a^2*x^2 + 1) + log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x))) + (2*a^2*x/sqrt(-a^2*x^2 + 1) - 1/(sqr
t(-a^2*x^2 + 1)*x))*arctanh(a*x)

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Fricas [A]  time = 2.06518, size = 223, normalized size = 2.72 \begin{align*} -\frac{2 \, a^{3} x^{3} - 2 \, a x - 2 \,{\left (a^{3} x^{3} - a x\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt{-a^{2} x^{2} + 1}{\left (2 \, a x -{\left (2 \, a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )\right )}}{2 \,{\left (a^{2} x^{3} - x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [B]  time = 1.26088, size = 209, normalized size = 2.55 \begin{align*} -\frac{1}{2} \, a \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right ) + \frac{1}{2} \, a \log \left (-\sqrt{-a^{2} x^{2} + 1} + 1\right ) + \frac{1}{4} \,{\left (\frac{a^{4} x}{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}{\left | a \right |}} - \frac{2 \, \sqrt{-a^{2} x^{2} + 1} a^{2} x}{a^{2} x^{2} - 1} - \frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{x{\left | a \right |}}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) - \frac{a}{\sqrt{-a^{2} x^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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