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

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Rubi [A]  time = 0.17, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.12, 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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fricas [A]  time = 0.43, size = 107, normalized size = 1.30 \[ -\frac {2 \, a^{3} x^{3} - 2 \, a x - 2 \, {\left (a^{3} x^{3} - a x\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a x - {\left (2 \, a^{2} x^{2} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )}}{2 \, {\left (a^{2} x^{3} - x\right )}} \]

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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giac [B]  time = 0.31, size = 155, normalized size = 1.89 \[ -\frac {1}{2} \, a \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) + \frac {1}{2} \, a \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right ) + \frac {1}{4} \, {\left (\frac {a^{4} x}{{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{2} x}{a^{2} x^{2} - 1} - \frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{x {\left | a \right |}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - \frac {a}{\sqrt {-a^{2} x^{2} + 1}} \]

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)

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maple [A]  time = 0.44, size = 132, normalized size = 1.61 \[ -\frac {a \left (\arctanh \left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 \left (a x -1\right )}-\frac {\left (\arctanh \left (a x \right )+1\right ) a \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 \left (a x +1\right )}-\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )}{x}+a \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-1\right )-a \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right ) \]

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 = 0.31, size = 84, normalized size = 1.02 \[ -a {\left (\frac {1}{\sqrt {-a^{2} x^{2} + 1}} + \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )\right )} + {\left (\frac {2 \, a^{2} x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} x}\right )} \operatorname {artanh}\left (a x\right ) \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/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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