3.852 \(\int e^{-2 \tanh ^{-1}(a+b x)} x^3 \, dx\)

Optimal. Leaf size=57 \[ -\frac{(a+1) x^2}{b^2}+\frac{2 (a+1)^2 x}{b^3}-\frac{2 (a+1)^3 \log (a+b x+1)}{b^4}+\frac{2 x^3}{3 b}-\frac{x^4}{4} \]

[Out]

(2*(1 + a)^2*x)/b^3 - ((1 + a)*x^2)/b^2 + (2*x^3)/(3*b) - x^4/4 - (2*(1 + a)^3*Log[1 + a + b*x])/b^4

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Rubi [A]  time = 0.0559445, antiderivative size = 57, 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} \[ -\frac{(a+1) x^2}{b^2}+\frac{2 (a+1)^2 x}{b^3}-\frac{2 (a+1)^3 \log (a+b x+1)}{b^4}+\frac{2 x^3}{3 b}-\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[x^3/E^(2*ArcTanh[a + b*x]),x]

[Out]

(2*(1 + a)^2*x)/b^3 - ((1 + a)*x^2)/b^2 + (2*x^3)/(3*b) - x^4/4 - (2*(1 + a)^3*Log[1 + a + b*x])/b^4

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^3 \, dx &=\int \frac{x^3 (1-a-b x)}{1+a+b x} \, dx\\ &=\int \left (\frac{2 (1+a)^2}{b^3}-\frac{2 (1+a) x}{b^2}+\frac{2 x^2}{b}-x^3-\frac{2 (1+a)^3}{b^3 (1+a+b x)}\right ) \, dx\\ &=\frac{2 (1+a)^2 x}{b^3}-\frac{(1+a) x^2}{b^2}+\frac{2 x^3}{3 b}-\frac{x^4}{4}-\frac{2 (1+a)^3 \log (1+a+b x)}{b^4}\\ \end{align*}

Mathematica [A]  time = 0.0411919, size = 57, normalized size = 1. \[ -\frac{(a+1) x^2}{b^2}+\frac{2 (a+1)^2 x}{b^3}-\frac{2 (a+1)^3 \log (a+b x+1)}{b^4}+\frac{2 x^3}{3 b}-\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/E^(2*ArcTanh[a + b*x]),x]

[Out]

(2*(1 + a)^2*x)/b^3 - ((1 + a)*x^2)/b^2 + (2*x^3)/(3*b) - x^4/4 - (2*(1 + a)^3*Log[1 + a + b*x])/b^4

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Maple [B]  time = 0.026, size = 109, normalized size = 1.9 \begin{align*} -{\frac{{x}^{4}}{4}}+{\frac{2\,{x}^{3}}{3\,b}}-{\frac{a{x}^{2}}{{b}^{2}}}-{\frac{{x}^{2}}{{b}^{2}}}+2\,{\frac{{a}^{2}x}{{b}^{3}}}+4\,{\frac{ax}{{b}^{3}}}+2\,{\frac{x}{{b}^{3}}}-2\,{\frac{\ln \left ( bx+a+1 \right ){a}^{3}}{{b}^{4}}}-6\,{\frac{\ln \left ( bx+a+1 \right ){a}^{2}}{{b}^{4}}}-6\,{\frac{\ln \left ( bx+a+1 \right ) a}{{b}^{4}}}-2\,{\frac{\ln \left ( bx+a+1 \right ) }{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x+a+1)^2*(1-(b*x+a)^2),x)

[Out]

-1/4*x^4+2/3*x^3/b-1/b^2*x^2*a-1/b^2*x^2+2/b^3*x*a^2+4/b^3*a*x+2/b^3*x-2/b^4*ln(b*x+a+1)*a^3-6/b^4*ln(b*x+a+1)
*a^2-6/b^4*ln(b*x+a+1)*a-2/b^4*ln(b*x+a+1)

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Maxima [A]  time = 0.967719, size = 92, normalized size = 1.61 \begin{align*} -\frac{3 \, b^{3} x^{4} - 8 \, b^{2} x^{3} + 12 \,{\left (a + 1\right )} b x^{2} - 24 \,{\left (a^{2} + 2 \, a + 1\right )} x}{12 \, b^{3}} - \frac{2 \,{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x+a+1)^2*(1-(b*x+a)^2),x, algorithm="maxima")

[Out]

-1/12*(3*b^3*x^4 - 8*b^2*x^3 + 12*(a + 1)*b*x^2 - 24*(a^2 + 2*a + 1)*x)/b^3 - 2*(a^3 + 3*a^2 + 3*a + 1)*log(b*
x + a + 1)/b^4

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Fricas [A]  time = 1.41636, size = 171, normalized size = 3. \begin{align*} -\frac{3 \, b^{4} x^{4} - 8 \, b^{3} x^{3} + 12 \,{\left (a + 1\right )} b^{2} x^{2} - 24 \,{\left (a^{2} + 2 \, a + 1\right )} b x + 24 \,{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )}{12 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x+a+1)^2*(1-(b*x+a)^2),x, algorithm="fricas")

[Out]

-1/12*(3*b^4*x^4 - 8*b^3*x^3 + 12*(a + 1)*b^2*x^2 - 24*(a^2 + 2*a + 1)*b*x + 24*(a^3 + 3*a^2 + 3*a + 1)*log(b*
x + a + 1))/b^4

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Sympy [A]  time = 0.383628, size = 56, normalized size = 0.98 \begin{align*} - \frac{x^{4}}{4} + \frac{2 x^{3}}{3 b} - \frac{x^{2} \left (a + 1\right )}{b^{2}} + \frac{x \left (2 a^{2} + 4 a + 2\right )}{b^{3}} - \frac{2 \left (a + 1\right )^{3} \log{\left (a + b x + 1 \right )}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.13757, size = 200, normalized size = 3.51 \begin{align*} \frac{{\left (b x + a + 1\right )}^{4}{\left (\frac{4 \,{\left (3 \, a b + 5 \, b\right )}}{{\left (b x + a + 1\right )} b} - \frac{18 \,{\left (a^{2} b^{2} + 4 \, a b^{2} + 3 \, b^{2}\right )}}{{\left (b x + a + 1\right )}^{2} b^{2}} + \frac{12 \,{\left (a^{3} b^{3} + 9 \, a^{2} b^{3} + 15 \, a b^{3} + 7 \, b^{3}\right )}}{{\left (b x + a + 1\right )}^{3} b^{3}} - 3\right )}}{12 \, b^{4}} + \frac{2 \,{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (\frac{{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2}{\left | b \right |}}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x+a+1)^2*(1-(b*x+a)^2),x, algorithm="giac")

[Out]

1/12*(b*x + a + 1)^4*(4*(3*a*b + 5*b)/((b*x + a + 1)*b) - 18*(a^2*b^2 + 4*a*b^2 + 3*b^2)/((b*x + a + 1)^2*b^2)
+ 12*(a^3*b^3 + 9*a^2*b^3 + 15*a*b^3 + 7*b^3)/((b*x + a + 1)^3*b^3) - 3)/b^4 + 2*(a^3 + 3*a^2 + 3*a + 1)*log(
abs(b*x + a + 1)/((b*x + a + 1)^2*abs(b)))/b^4