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

Optimal. Leaf size=19 \[ -\frac{2 \log (-a-b x+1)}{b}-x \]

[Out]

-x - (2*Log[1 - a - b*x])/b

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Rubi [A]  time = 0.0124833, antiderivative size = 19, 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 = 0.0099302, size = 19, 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 = 0.03, size = 17, normalized size = 0.9 \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((b*x+a+1)^2/(1-(b*x+a)^2),x)

[Out]

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

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Maxima [A]  time = 0.952147, size = 22, normalized size = 1.16 \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((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 = 1.87019, size = 42, normalized size = 2.21 \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((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 = 0.122734, size = 14, normalized size = 0.74 \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((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 [A]  time = 1.16458, size = 23, normalized size = 1.21 \begin{align*} -x - \frac{2 \, \log \left ({\left | b x + a - 1 \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*log(abs(b*x + a - 1))/b