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

Optimal. Leaf size=27 \[ \frac{\sqrt{1-a^2 x^2}}{a}+\frac{\sin ^{-1}(a x)}{a} \]

[Out]

Sqrt[1 - a^2*x^2]/a + ArcSin[a*x]/a

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Rubi [A]  time = 0.0108152, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6123, 641, 216} \[ \frac{\sqrt{1-a^2 x^2}}{a}+\frac{\sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/a + ArcSin[a*x]/a

Rule 6123

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

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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]

Rubi steps

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

Mathematica [A]  time = 0.0123919, size = 23, normalized size = 0.85 \[ \frac{\sqrt{1-a^2 x^2}+\sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2] + ArcSin[a*x])/a

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Maple [B]  time = 0.031, size = 66, normalized size = 2.4 \begin{align*}{\frac{1}{a}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\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)

[Out]

1/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/(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 = 1.44107, size = 34, normalized size = 1.26 \begin{align*} \frac{\arcsin \left (a x\right )}{a} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a} \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, algorithm="maxima")

[Out]

arcsin(a*x)/a + sqrt(-a^2*x^2 + 1)/a

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Fricas [A]  time = 1.9401, size = 90, normalized size = 3.33 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1} - 2 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right )}{a} \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, algorithm="fricas")

[Out]

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

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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 )}}{a x + 1}\, 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)

[Out]

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

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Giac [A]  time = 1.17369, size = 38, normalized size = 1.41 \begin{align*} \frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a} \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, algorithm="giac")

[Out]

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