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

Optimal. Leaf size=14 \[ \frac {2 \log (1-a x)}{a}+x \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, 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} \[ \frac {2 \log (1-a x)}{a}+x \]

Antiderivative was successfully verified.

[In]

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

[Out]

x + (2*Log[1 - a*x])/a

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])

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 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]

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*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \[ \frac {2 \log (1-a x)}{a}+x \]

Antiderivative was successfully verified.

[In]

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

[Out]

x + (2*Log[1 - a*x])/a

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fricas [A]  time = 0.45, size = 16, normalized size = 1.14 \[ \frac {a x + 2 \, \log \left (a x - 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x + 2*log(a*x - 1))/a

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giac [A]  time = 0.11, size = 14, normalized size = 1.00 \[ x + \frac {2 \, \log \left ({\left | a x - 1 \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*log(abs(a*x - 1))/a

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maple [A]  time = 0.03, size = 14, normalized size = 1.00 \[ x +\frac {2 \ln \left (a x -1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 13, normalized size = 0.93 \[ x + \frac {2 \, \log \left (a x - 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*log(a*x - 1)/a

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mupad [B]  time = 1.17, size = 13, normalized size = 0.93 \[ x+\frac {2\,\ln \left (a\,x-1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)/(a*x - 1),x)

[Out]

x + (2*log(a*x - 1))/a

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sympy [A]  time = 0.07, size = 10, normalized size = 0.71 \[ x + \frac {2 \log {\left (a x - 1 \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*log(a*x - 1)/a

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