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

Optimal. Leaf size=41 \[ -\frac{2 (a+1) x}{b^2}+\frac{2 (a+1)^2 \log (a+b x+1)}{b^3}+\frac{x^2}{b}-\frac{x^3}{3} \]

[Out]

(-2*(1 + a)*x)/b^2 + x^2/b - x^3/3 + (2*(1 + a)^2*Log[1 + a + b*x])/b^3

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Rubi [A]  time = 0.0456562, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {6163, 77} \[ -\frac{2 (a+1) x}{b^2}+\frac{2 (a+1)^2 \log (a+b x+1)}{b^3}+\frac{x^2}{b}-\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/E^(2*ArcTanh[a + b*x]),x]

[Out]

(-2*(1 + a)*x)/b^2 + x^2/b - x^3/3 + (2*(1 + a)^2*Log[1 + a + b*x])/b^3

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^2 \, dx &=\int \frac{x^2 (1-a-b x)}{1+a+b x} \, dx\\ &=\int \left (-\frac{2 (1+a)}{b^2}+\frac{2 x}{b}-x^2+\frac{2 (1+a)^2}{b^2 (1+a+b x)}\right ) \, dx\\ &=-\frac{2 (1+a) x}{b^2}+\frac{x^2}{b}-\frac{x^3}{3}+\frac{2 (1+a)^2 \log (1+a+b x)}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.0296369, size = 41, normalized size = 1. \[ -\frac{2 (a+1) x}{b^2}+\frac{2 (a+1)^2 \log (a+b x+1)}{b^3}+\frac{x^2}{b}-\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/E^(2*ArcTanh[a + b*x]),x]

[Out]

(-2*(1 + a)*x)/b^2 + x^2/b - x^3/3 + (2*(1 + a)^2*Log[1 + a + b*x])/b^3

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Maple [A]  time = 0.028, size = 67, normalized size = 1.6 \begin{align*} -{\frac{{x}^{3}}{3}}+{\frac{{x}^{2}}{b}}-2\,{\frac{ax}{{b}^{2}}}-2\,{\frac{x}{{b}^{2}}}+2\,{\frac{\ln \left ( bx+a+1 \right ){a}^{2}}{{b}^{3}}}+4\,{\frac{\ln \left ( bx+a+1 \right ) a}{{b}^{3}}}+2\,{\frac{\ln \left ( bx+a+1 \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.966852, size = 62, normalized size = 1.51 \begin{align*} -\frac{b^{2} x^{3} - 3 \, b x^{2} + 6 \,{\left (a + 1\right )} x}{3 \, b^{2}} + \frac{2 \,{\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^2*x^3 - 3*b*x^2 + 6*(a + 1)*x)/b^2 + 2*(a^2 + 2*a + 1)*log(b*x + a + 1)/b^3

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Fricas [A]  time = 1.44515, size = 115, normalized size = 2.8 \begin{align*} -\frac{b^{3} x^{3} - 3 \, b^{2} x^{2} + 6 \,{\left (a + 1\right )} b x - 6 \,{\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 - 3*b^2*x^2 + 6*(a + 1)*b*x - 6*(a^2 + 2*a + 1)*log(b*x + a + 1))/b^3

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Sympy [A]  time = 0.353981, size = 37, normalized size = 0.9 \begin{align*} - \frac{x^{3}}{3} + \frac{x^{2}}{b} - \frac{x \left (2 a + 2\right )}{b^{2}} + \frac{2 \left (a + 1\right )^{2} \log{\left (a + b x + 1 \right )}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.19284, size = 138, normalized size = 3.37 \begin{align*} \frac{{\left (b x + a + 1\right )}^{3}{\left (\frac{3 \,{\left (a b + 2 \, b\right )}}{{\left (b x + a + 1\right )} b} - \frac{3 \,{\left (a^{2} b^{2} + 6 \, a b^{2} + 5 \, b^{2}\right )}}{{\left (b x + a + 1\right )}^{2} b^{2}} - 1\right )}}{3 \, b^{3}} - \frac{2 \,{\left (a^{2} + 2 \, a + 1\right )} \log \left (\frac{{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2}{\left | b \right |}}\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x + a + 1)^3*(3*(a*b + 2*b)/((b*x + a + 1)*b) - 3*(a^2*b^2 + 6*a*b^2 + 5*b^2)/((b*x + a + 1)^2*b^2) - 1
)/b^3 - 2*(a^2 + 2*a + 1)*log(abs(b*x + a + 1)/((b*x + a + 1)^2*abs(b)))/b^3