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

Optimal. Leaf size=11 \[ \frac{\sin ^{-1}(a x)}{a c} \]

[Out]

ArcSin[a*x]/(a*c)

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Rubi [A]  time = 0.0279087, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6127, 216} \[ \frac{\sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - a*c*x)),x]

[Out]

ArcSin[a*x]/(a*c)

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

Mathematica [A]  time = 0.0068999, size = 11, normalized size = 1. \[ \frac{\sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcSin[a*x]/(a*c)

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Maple [B]  time = 0.041, size = 154, normalized size = 14. \begin{align*}{\frac{1}{2\,ac}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{1}{2\,c}\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}}}}}-{\frac{1}{2\,ac}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}+{\frac{1}{2\,c}\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)/(-a*c*x+c),x)

[Out]

1/2/c/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/2/c/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))
^(1/2))-1/2/c/a*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+1/2/c/(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 [B]  time = 1.47068, size = 42, normalized size = 3.82 \begin{align*} \frac{\arcsin \left (\frac{x}{c \sqrt{\frac{1}{a^{2} c^{2}}}}\right )}{a^{2} c^{2} \sqrt{\frac{1}{a^{2} 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),x, algorithm="maxima")

[Out]

arcsin(x/(c*sqrt(1/(a^2*c^2))))/(a^2*c^2*sqrt(1/(a^2*c^2)))

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Fricas [B]  time = 1.58867, size = 66, normalized size = 6. \begin{align*} -\frac{2 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right )}{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),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.22909, size = 44, normalized size = 4. \begin{align*} \frac{\begin{cases} \sqrt{\frac{1}{a^{2}}} \operatorname{asin}{\left (x \sqrt{a^{2}} \right )} & \text{for}\: a^{2} > 0 \\\sqrt{- \frac{1}{a^{2}}} \operatorname{asinh}{\left (x \sqrt{- a^{2}} \right )} & \text{for}\: a^{2} < 0 \end{cases}}{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),x)

[Out]

Piecewise((sqrt(a**(-2))*asin(x*sqrt(a**2)), a**2 > 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0))/c

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Giac [A]  time = 1.19563, size = 19, normalized size = 1.73 \begin{align*} \frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c{\left | a \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)/(-a*c*x+c),x, algorithm="giac")

[Out]

arcsin(a*x)*sgn(a)/(c*abs(a))