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

Optimal. Leaf size=21 \[ \frac{\sqrt{1-a^2 x^2}}{a c} \]

[Out]

Sqrt[1 - a^2*x^2]/(a*c)

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Rubi [A]  time = 0.0738308, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6131, 6128, 261} \[ \frac{\sqrt{1-a^2 x^2}}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/(a*c)

Rule 6131

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 + (c*x)/d)
^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && IntegerQ[2*p]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0127588, size = 21, normalized size = 1. \[ \frac{\sqrt{1-a^2 x^2}}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/(a*c)

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Maple [A]  time = 0.032, size = 20, normalized size = 1. \begin{align*}{\frac{1}{ac}\sqrt{-{a}^{2}{x}^{2}+1}} \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)/(c-c/a/x),x)

[Out]

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

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Maxima [A]  time = 0.974711, size = 30, normalized size = 1.43 \begin{align*} \frac{\sqrt{a x + 1} \sqrt{-a x + 1}}{a c} \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)/(c-c/a/x),x, algorithm="maxima")

[Out]

sqrt(a*x + 1)*sqrt(-a*x + 1)/(a*c)

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Fricas [A]  time = 1.9748, size = 35, normalized size = 1.67 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1}}{a c} \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)/(c-c/a/x),x, algorithm="fricas")

[Out]

sqrt(-a^2*x^2 + 1)/(a*c)

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Sympy [A]  time = 8.53791, size = 53, normalized size = 2.52 \begin{align*} a \left (\begin{cases} \tilde{\infty } \left (\begin{cases} \frac{x^{2}}{2} & \text{for}\: a^{2} = 0 \\- \frac{\left (- a^{2} x^{2} + 1\right )^{\frac{3}{2}}}{3 a^{2}} & \text{otherwise} \end{cases}\right ) & \text{for}\: c = 0 \\- \frac{x^{2}}{2 c} & \text{for}\: a^{2} = 0 \\\frac{\sqrt{- a^{2} x^{2} + 1}}{a^{2} c} & \text{otherwise} \end{cases}\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)/(c-c/a/x),x)

[Out]

a*Piecewise((zoo*Piecewise((x**2/2, Eq(a**2, 0)), (-(-a**2*x**2 + 1)**(3/2)/(3*a**2), True)), Eq(c, 0)), (-x**
2/(2*c), Eq(a**2, 0)), (sqrt(-a**2*x**2 + 1)/(a**2*c), True))

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Giac [A]  time = 1.13544, size = 26, normalized size = 1.24 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1}}{a c} \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)/(c-c/a/x),x, algorithm="giac")

[Out]

sqrt(-a^2*x^2 + 1)/(a*c)