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

Optimal. Leaf size=66 \[ -\frac{(1-a) x^2}{b^2}-\frac{2 (1-a)^2 x}{b^3}-\frac{2 (1-a)^3 \log (-a-b x+1)}{b^4}-\frac{2 x^3}{3 b}-\frac{x^4}{4} \]

[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 = 0.0600754, antiderivative size = 66, 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} \[ -\frac{(1-a) x^2}{b^2}-\frac{2 (1-a)^2 x}{b^3}-\frac{2 (1-a)^3 \log (-a-b x+1)}{b^4}-\frac{2 x^3}{3 b}-\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a + b*x])*x^3,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 = 0.0426213, size = 66, normalized size = 1. \[ -\frac{(1-a) x^2}{b^2}-\frac{2 (1-a)^2 x}{b^3}-\frac{2 (1-a)^3 \log (-a-b x+1)}{b^4}-\frac{2 x^3}{3 b}-\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a + b*x])*x^3,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 [A]  time = 0.027, size = 108, normalized size = 1.6 \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((b*x+a+1)^2/(1-(b*x+a)^2)*x^3,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 = 0.966067, size = 92, normalized size = 1.39 \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((b*x+a+1)^2/(1-(b*x+a)^2)*x^3,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 = 1.83136, size = 171, normalized size = 2.59 \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((b*x+a+1)^2/(1-(b*x+a)^2)*x^3,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 = 0.40406, size = 56, normalized size = 0.85 \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((b*x+a+1)**2/(1-(b*x+a)**2)*x**3,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 [A]  time = 1.21377, size = 111, normalized size = 1.68 \begin{align*} \frac{2 \,{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left ({\left | b x + a - 1 \right |}\right )}{b^{4}} - \frac{3 \, b^{4} x^{4} + 8 \, b^{3} x^{3} - 12 \, a b^{2} x^{2} + 24 \, a^{2} b x + 12 \, b^{2} x^{2} - 48 \, a b x + 24 \, b x}{12 \, b^{4}} \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^3,x, algorithm="giac")

[Out]

2*(a^3 - 3*a^2 + 3*a - 1)*log(abs(b*x + a - 1))/b^4 - 1/12*(3*b^4*x^4 + 8*b^3*x^3 - 12*a*b^2*x^2 + 24*a^2*b*x
+ 12*b^2*x^2 - 48*a*b*x + 24*b*x)/b^4