3.38 \(\int \frac{e^{-\tanh ^{-1}(a x)}}{x} \, dx\)

Optimal. Leaf size=24 \[ -\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\sin ^{-1}(a x) \]

[Out]

-ArcSin[a*x] - ArcTanh[Sqrt[1 - a^2*x^2]]

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Rubi [A]  time = 0.0416582, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6124, 844, 216, 266, 63, 208} \[ -\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*x),x]

[Out]

-ArcSin[a*x] - ArcTanh[Sqrt[1 - a^2*x^2]]

Rule 6124

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/
2)*Sqrt[1 - a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(n - 1)/2]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[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 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]

Rubi steps

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

Mathematica [A]  time = 0.0146738, size = 28, normalized size = 1.17 \[ -\log \left (\sqrt{1-a^2 x^2}+1\right )-\sin ^{-1}(a x)+\log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcTanh[a*x]*x),x]

[Out]

-ArcSin[a*x] + Log[x] - Log[1 + Sqrt[1 - a^2*x^2]]

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Maple [B]  time = 0.041, size = 93, normalized size = 3.9 \begin{align*} \sqrt{-{a}^{2}{x}^{2}+1}-{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) -\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }-{a\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.44148, size = 57, normalized size = 2.38 \begin{align*} -a{\left (\frac{\arcsin \left (a x\right )}{a} + \frac{\log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right )}{a}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.99666, size = 103, normalized size = 4.29 \begin{align*} 2 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.21772, size = 70, normalized size = 2.92 \begin{align*} -\frac{a \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} - \frac{a \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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