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

Optimal. Leaf size=30 \[ \frac{2 \sqrt{1-a^2 x^2}}{a \sqrt{c-a c x}} \]

[Out]

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

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Rubi [A]  time = 0.0473599, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {6127, 649} \[ \frac{2 \sqrt{1-a^2 x^2}}{a \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6127

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

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rubi steps

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

Mathematica [A]  time = 0.015685, size = 30, normalized size = 1. \[ \frac{2 \sqrt{1-a^2 x^2}}{a \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.028, size = 27, normalized size = 0.9 \begin{align*} 2\,{\frac{\sqrt{-{a}^{2}{x}^{2}+1}}{a\sqrt{-acx+c}}} \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)/(-a*c*x+c)^(1/2),x)

[Out]

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

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Maxima [A]  time = 0.980754, size = 20, normalized size = 0.67 \begin{align*} \frac{2 \, \sqrt{a x + 1}}{a \sqrt{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)/(-a*c*x+c)^(1/2),x, algorithm="maxima")

[Out]

2*sqrt(a*x + 1)/(a*sqrt(c))

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Fricas [A]  time = 2.1293, size = 76, normalized size = 2.53 \begin{align*} -\frac{2 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{a^{2} c x - 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)/(-a*c*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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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 )}}{\sqrt{- c \left (a x - 1\right )} \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)/(-a*c*x+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.23812, size = 42, normalized size = 1.4 \begin{align*} -\frac{2 \,{\left (\frac{\sqrt{2} \sqrt{c}}{a} - \frac{\sqrt{a c x + c}}{a}\right )}{\left | c \right |}}{c^{2}} \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)/(-a*c*x+c)^(1/2),x, algorithm="giac")

[Out]

-2*(sqrt(2)*sqrt(c)/a - sqrt(a*c*x + c)/a)*abs(c)/c^2