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

Optimal. Leaf size=15 \[ \frac {2 \log (a x+1)}{a}-x \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6125, 43} \[ \frac {2 \log (a x+1)}{a}-x \]

Antiderivative was successfully verified.

[In]

Int[E^(-2*ArcTanh[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]

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*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \[ \frac {2 \log (a x+1)}{a}-x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x + (2*Log[1 + a*x])/a

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fricas [A]  time = 0.49, size = 17, normalized size = 1.13 \[ -\frac {a x - 2 \, \log \left (a x + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x - 2*log(a*x + 1))/a

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giac [B]  time = 0.16, size = 64, normalized size = 4.27 \[ -a^{2} {\left (\frac {a x + 1}{a^{3}} + \frac {2 \, \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a^{3}} - \frac {1}{{\left (a x + 1\right )} a^{3}}\right )} - \frac {1}{{\left (a x + 1\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*((a*x + 1)/a^3 + 2*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/a^3 - 1/((a*x + 1)*a^3)) - 1/((a*x + 1)*a)

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maple [A]  time = 0.02, size = 16, normalized size = 1.07 \[ -x +\frac {2 \ln \left (a x +1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 15, normalized size = 1.00 \[ -x + \frac {2 \, \log \left (a x + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 2*log(a*x + 1)/a

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mupad [B]  time = 0.03, size = 15, normalized size = 1.00 \[ \frac {2\,\ln \left (a\,x+1\right )}{a}-x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a^2*x^2 - 1)/(a*x + 1)^2,x)

[Out]

(2*log(a*x + 1))/a - x

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sympy [A]  time = 0.09, size = 10, normalized size = 0.67 \[ - x + \frac {2 \log {\left (a x + 1 \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 2*log(a*x + 1)/a

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