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]

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

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.57, size = 41, normalized size = 1.52 \[ \frac {\sqrt {-a^{2} x^{2} + 1} - 2 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{a} \]

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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giac [A]  time = 0.59, size = 28, normalized size = 1.04 \[ \frac {\arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a} \]

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

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maple [B]  time = 0.03, size = 66, normalized size = 2.44 \[ \frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}} \]

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 = 0.42, size = 25, normalized size = 0.93 \[ \frac {\arcsin \left (a x\right )}{a} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a} \]

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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mupad [B]  time = 0.03, size = 35, normalized size = 1.30 \[ \frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {\sqrt {1-a^2\,x^2}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - a^2*x^2)^(1/2)/(a*x + 1),x)

[Out]

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

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

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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