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

Optimal. Leaf size=34 \[ -\frac{2 (1-a) \log (-a-b x+1)}{b^2}-\frac{2 x}{b}-\frac{x^2}{2} \]

[Out]

(-2*x)/b - x^2/2 - (2*(1 - a)*Log[1 - a - b*x])/b^2

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Rubi [A]  time = 0.0306465, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6163, 77} \[ -\frac{2 (1-a) \log (-a-b x+1)}{b^2}-\frac{2 x}{b}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a + b*x])*x,x]

[Out]

(-2*x)/b - x^2/2 - (2*(1 - a)*Log[1 - a - b*x])/b^2

Rule 6163

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0171643, size = 34, normalized size = 1. \[ -\frac{2 (1-a) \log (-a-b x+1)}{b^2}-\frac{2 x}{b}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a + b*x])*x,x]

[Out]

(-2*x)/b - x^2/2 - (2*(1 - a)*Log[1 - a - b*x])/b^2

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Maple [A]  time = 0.03, size = 38, normalized size = 1.1 \begin{align*} -{\frac{{x}^{2}}{2}}-2\,{\frac{x}{b}}+2\,{\frac{\ln \left ( bx+a-1 \right ) a}{{b}^{2}}}-2\,{\frac{\ln \left ( bx+a-1 \right ) }{{b}^{2}}} \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,x)

[Out]

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

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Maxima [A]  time = 0.950921, size = 41, normalized size = 1.21 \begin{align*} -\frac{b x^{2} + 4 \, x}{2 \, b} + \frac{2 \,{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2}} \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,x, algorithm="maxima")

[Out]

-1/2*(b*x^2 + 4*x)/b + 2*(a - 1)*log(b*x + a - 1)/b^2

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Fricas [A]  time = 1.71797, size = 77, normalized size = 2.26 \begin{align*} -\frac{b^{2} x^{2} + 4 \, b x - 4 \,{\left (a - 1\right )} \log \left (b x + a - 1\right )}{2 \, b^{2}} \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,x, algorithm="fricas")

[Out]

-1/2*(b^2*x^2 + 4*b*x - 4*(a - 1)*log(b*x + a - 1))/b^2

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Sympy [A]  time = 0.319083, size = 26, normalized size = 0.76 \begin{align*} - \frac{x^{2}}{2} - \frac{2 x}{b} + \frac{2 \left (a - 1\right ) \log{\left (a + b x - 1 \right )}}{b^{2}} \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,x)

[Out]

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

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Giac [A]  time = 1.18514, size = 46, normalized size = 1.35 \begin{align*} \frac{2 \,{\left (a - 1\right )} \log \left ({\left | b x + a - 1 \right |}\right )}{b^{2}} - \frac{b^{2} x^{2} + 4 \, b x}{2 \, b^{2}} \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,x, algorithm="giac")

[Out]

2*(a - 1)*log(abs(b*x + a - 1))/b^2 - 1/2*(b^2*x^2 + 4*b*x)/b^2