3.14 \(\int e^{2 \tanh ^{-1}(a x)} \, dx\)

Optimal. Leaf size=16 \[ -\frac{2 \log (1-a x)}{a}-x \]

[Out]

-x - (2*Log[1 - a*x])/a

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Rubi [A]  time = 0.0120185, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6125, 43} \[ -\frac{2 \log (1-a x)}{a}-x \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x]),x]

[Out]

-x - (2*Log[1 - a*x])/a

Rule 6125

Int[E^(ArcTanh[(a_.)*(x_)]*(n_)), x_Symbol] :> Int[(1 + a*x)^(n/2)/(1 - a*x)^(n/2), x] /; FreeQ[{a, n}, x] &&
!IntegerQ[(n - 1)/2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{2 \tanh ^{-1}(a x)} \, dx &=\int \frac{1+a x}{1-a x} \, dx\\ &=\int \left (-1-\frac{2}{-1+a x}\right ) \, dx\\ &=-x-\frac{2 \log (1-a x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0094114, size = 16, normalized size = 1. \[ -\frac{2 \log (1-a x)}{a}-x \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x]),x]

[Out]

-x - (2*Log[1 - a*x])/a

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Maple [A]  time = 0.027, size = 16, normalized size = 1. \begin{align*} -x-2\,{\frac{\ln \left ( ax-1 \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x-2/a*ln(a*x-1)

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Maxima [A]  time = 0.952724, size = 20, normalized size = 1.25 \begin{align*} -x - \frac{2 \, \log \left (a x - 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(a*x - 1)/a

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Fricas [A]  time = 1.74374, size = 36, normalized size = 2.25 \begin{align*} -\frac{a x + 2 \, \log \left (a x - 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x + 2*log(a*x - 1))/a

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Sympy [A]  time = 0.187183, size = 12, normalized size = 0.75 \begin{align*} - x - \frac{2 \log{\left (a x - 1 \right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(a*x - 1)/a

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Giac [A]  time = 1.17178, size = 22, normalized size = 1.38 \begin{align*} -x - \frac{2 \, \log \left ({\left | a x - 1 \right |}\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(abs(a*x - 1))/a