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

Optimal. Leaf size=49 \[ -\frac{2 (1-a) x}{b^2}-\frac{2 (1-a)^2 \log (-a-b x+1)}{b^3}-\frac{x^2}{b}-\frac{x^3}{3} \]

[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 = 0.0506131, antiderivative size = 49, 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} \[ -\frac{2 (1-a) x}{b^2}-\frac{2 (1-a)^2 \log (-a-b x+1)}{b^3}-\frac{x^2}{b}-\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a + b*x])*x^2,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 = 0.0310277, size = 44, normalized size = 0.9 \[ -\frac{b x \left (-6 a+b^2 x^2+3 b x+6\right )+6 (a-1)^2 \log (-a-b x+1)}{3 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x*(6 - 6*a + 3*b*x + b^2*x^2) + 6*(-1 + a)^2*Log[1 - a - b*x])/(3*b^3)

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Maple [A]  time = 0.028, size = 68, normalized size = 1.4 \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((b*x+a+1)^2/(1-(b*x+a)^2)*x^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 = 0.940435, size = 62, normalized size = 1.27 \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((b*x+a+1)^2/(1-(b*x+a)^2)*x^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 = 1.73783, size = 115, normalized size = 2.35 \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((b*x+a+1)^2/(1-(b*x+a)^2)*x^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 = 0.361109, size = 37, normalized size = 0.76 \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((b*x+a+1)**2/(1-(b*x+a)**2)*x**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 [A]  time = 1.15478, size = 70, normalized size = 1.43 \begin{align*} -\frac{2 \,{\left (a^{2} - 2 \, a + 1\right )} \log \left ({\left | b x + a - 1 \right |}\right )}{b^{3}} - \frac{b^{3} x^{3} + 3 \, b^{2} x^{2} - 6 \, a b x + 6 \, b x}{3 \, b^{3}} \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^2,x, algorithm="giac")

[Out]

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