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

Optimal. Leaf size=29 \[ -\frac{2 (a+1) \log (a+b x+1)}{b^2}+\frac{2 x}{b}-\frac{x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.0317985, antiderivative size = 29, 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} \[ -\frac{2 (a+1) \log (a+b x+1)}{b^2}+\frac{2 x}{b}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x/E^(2*ArcTanh[a + b*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 = 0.0172854, size = 29, normalized size = 1. \[ -\frac{2 (a+1) \log (a+b x+1)}{b^2}+\frac{2 x}{b}-\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/E^(2*ArcTanh[a + b*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 = 0.028, size = 38, normalized size = 1.3 \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(x/(b*x+a+1)^2*(1-(b*x+a)^2),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 = 0.958947, size = 41, normalized size = 1.41 \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(x/(b*x+a+1)^2*(1-(b*x+a)^2),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 = 1.48488, size = 77, normalized size = 2.66 \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(x/(b*x+a+1)^2*(1-(b*x+a)^2),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 = 0.315509, size = 26, normalized size = 0.9 \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(x/(b*x+a+1)**2*(1-(b*x+a)**2),x)

[Out]

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

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Giac [B]  time = 1.13369, size = 93, normalized size = 3.21 \begin{align*} \frac{\frac{{\left (b x + a + 1\right )}^{2}{\left (\frac{2 \,{\left (a b + 3 \, b\right )}}{{\left (b x + a + 1\right )} b} - 1\right )}}{b} + \frac{4 \,{\left (a + 1\right )} \log \left (\frac{{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2}{\left | b \right |}}\right )}{b}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*x + a + 1)^2*(2*(a*b + 3*b)/((b*x + a + 1)*b) - 1)/b + 4*(a + 1)*log(abs(b*x + a + 1)/((b*x + a + 1)^2
*abs(b)))/b)/b