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

Optimal. Leaf size=28 \[ \frac{\sin ^{-1}(a x)}{a}-\frac{\sqrt{1-a^2 x^2}}{a} \]

[Out]

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

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Rubi [A]  time = 0.0109539, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6123, 641, 216} \[ \frac{\sin ^{-1}(a x)}{a}-\frac{\sqrt{1-a^2 x^2}}{a} \]

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 = 0.0103446, size = 25, normalized size = 0.89 \[ \frac{\sin ^{-1}(a x)-\sqrt{1-a^2 x^2}}{a} \]

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 [A]  time = 0.032, size = 45, normalized size = 1.6 \begin{align*} -{\frac{1}{a}\sqrt{-{a}^{2}{x}^{2}+1}}+{\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.41809, size = 47, normalized size = 1.68 \begin{align*} \frac{\arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.09674, size = 92, normalized size = 3.29 \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((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 [A]  time = 1.62957, size = 19, normalized size = 0.68 \begin{align*} \begin{cases} \frac{- \sqrt{- a^{2} x^{2} + 1} + \operatorname{asin}{\left (a x \right )}}{a} & \text{for}\: a \neq 0 \\x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-sqrt(-a**2*x**2 + 1) + asin(a*x))/a, Ne(a, 0)), (x, True))

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Giac [A]  time = 1.17587, size = 39, normalized size = 1.39 \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((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