3.371 \(\int \frac {\tanh ^{-1}(a x)}{x^2 \sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=42 \[ -\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

-a*arctanh((-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 = 0.08, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6008, 266, 63, 208} \[ -\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)}{x^2 \sqrt {1-a^2 x^2}} \, dx &=-\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 {\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 {\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 {\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 = 0.05, size = 48, normalized size = 1.14 \[ -a \log \left (\sqrt {1-a^2 x^2}+1\right )-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x}+a \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 58, normalized size = 1.38 \[ \frac {2 \, a x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} \log \left (-\frac {a x + 1}{a x - 1}\right )}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.27, size = 111, normalized size = 2.64 \[ -\frac {1}{2} \, a {\left (\log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right ) - \log \left (-\sqrt {-a^{2} x^{2} + 1} + 1\right )\right )} + \frac {1}{4} \, {\left (\frac {a^{4} x}{{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{x {\left | a \right |}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.40, size = 72, normalized size = 1.71 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )}{x}-a \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )+a \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 51, normalized size = 1.21 \[ -a \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )}{x^2\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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