3.45 \(\int e^{-2 \coth ^{-1}(a x)} \, dx\)

Optimal. Leaf size=13 \[ x-\frac{2 \log (a x+1)}{a} \]

[Out]

x - (2*Log[1 + a*x])/a

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Rubi [A]  time = 0.013253, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6167, 6125, 43} \[ x-\frac{2 \log (a x+1)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x - (2*Log[1 + a*x])/a

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

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 \coth ^{-1}(a x)} \, dx &=-\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 = 0.010677, size = 13, normalized size = 1. \[ x-\frac{2 \log (a x+1)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x - (2*Log[1 + a*x])/a

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Maple [A]  time = 0.039, size = 14, normalized size = 1.1 \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)/(a*x+1),x)

[Out]

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

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Maxima [A]  time = 1.01314, size = 18, normalized size = 1.38 \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)/(a*x+1),x, algorithm="maxima")

[Out]

x - 2*log(a*x + 1)/a

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Fricas [A]  time = 1.73219, size = 35, normalized size = 2.69 \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)/(a*x+1),x, algorithm="fricas")

[Out]

(a*x - 2*log(a*x + 1))/a

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Sympy [A]  time = 0.09364, size = 10, normalized size = 0.77 \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)/(a*x+1),x)

[Out]

x - 2*log(a*x + 1)/a

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Giac [A]  time = 1.1491, size = 19, normalized size = 1.46 \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)/(a*x+1),x, algorithm="giac")

[Out]

x - 2*log(abs(a*x + 1))/a