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

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

[Out]

-x + (2*Log[1 + a + b*x])/b

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Rubi [A]  time = 0.0121516, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6161, 43} \[ \frac{2 \log (a+b x+1)}{b}-x \]

Antiderivative was successfully verified.

[In]

Int[E^(-2*ArcTanh[a + b*x]),x]

[Out]

-x + (2*Log[1 + a + b*x])/b

Rule 6161

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

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

Mathematica [A]  time = 0.0111211, size = 16, normalized size = 1. \[ \frac{2 \log (a+b x+1)}{b}-x \]

Antiderivative was successfully verified.

[In]

Integrate[E^(-2*ArcTanh[a + b*x]),x]

[Out]

-x + (2*Log[1 + a + b*x])/b

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Maple [A]  time = 0.029, size = 17, normalized size = 1.1 \begin{align*} -x+2\,{\frac{\ln \left ( bx+a+1 \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.959308, size = 22, normalized size = 1.38 \begin{align*} -x + \frac{2 \, \log \left (b x + a + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 2*log(b*x + a + 1)/b

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Fricas [A]  time = 1.44847, size = 42, normalized size = 2.62 \begin{align*} -\frac{b x - 2 \, \log \left (b x + a + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x - 2*log(b*x + a + 1))/b

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Sympy [A]  time = 0.123091, size = 12, normalized size = 0.75 \begin{align*} - x + \frac{2 \log{\left (a + b x + 1 \right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 2*log(a + b*x + 1)/b

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Giac [B]  time = 1.17688, size = 51, normalized size = 3.19 \begin{align*} -\frac{b x + a + 1}{b} - \frac{2 \, \log \left (\frac{{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2}{\left | b \right |}}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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