3.26 \(\int e^{4 \coth ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=47 \[ \frac{8 x}{a^2}+\frac{4}{a^3 (1-a x)}+\frac{12 \log (1-a x)}{a^3}+\frac{2 x^2}{a}+\frac{x^3}{3} \]

[Out]

(8*x)/a^2 + (2*x^2)/a + x^3/3 + 4/(a^3*(1 - a*x)) + (12*Log[1 - a*x])/a^3

________________________________________________________________________________________

Rubi [A]  time = 0.0595973, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6167, 6126, 88} \[ \frac{8 x}{a^2}+\frac{4}{a^3 (1-a x)}+\frac{12 \log (1-a x)}{a^3}+\frac{2 x^2}{a}+\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[a*x])*x^2,x]

[Out]

(8*x)/a^2 + (2*x^2)/a + x^3/3 + 4/(a^3*(1 - a*x)) + (12*Log[1 - a*x])/a^3

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{4 \coth ^{-1}(a x)} x^2 \, dx &=\int e^{4 \tanh ^{-1}(a x)} x^2 \, dx\\ &=\int \frac{x^2 (1+a x)^2}{(1-a x)^2} \, dx\\ &=\int \left (\frac{8}{a^2}+\frac{4 x}{a}+x^2+\frac{4}{a^2 (-1+a x)^2}+\frac{12}{a^2 (-1+a x)}\right ) \, dx\\ &=\frac{8 x}{a^2}+\frac{2 x^2}{a}+\frac{x^3}{3}+\frac{4}{a^3 (1-a x)}+\frac{12 \log (1-a x)}{a^3}\\ \end{align*}

Mathematica [A]  time = 0.0361972, size = 47, normalized size = 1. \[ \frac{8 x}{a^2}+\frac{4}{a^3 (1-a x)}+\frac{12 \log (1-a x)}{a^3}+\frac{2 x^2}{a}+\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[a*x])*x^2,x]

[Out]

(8*x)/a^2 + (2*x^2)/a + x^3/3 + 4/(a^3*(1 - a*x)) + (12*Log[1 - a*x])/a^3

________________________________________________________________________________________

Maple [A]  time = 0.043, size = 44, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{3}}+2\,{\frac{{x}^{2}}{a}}+8\,{\frac{x}{{a}^{2}}}-4\,{\frac{1}{{a}^{3} \left ( ax-1 \right ) }}+12\,{\frac{\ln \left ( ax-1 \right ) }{{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3+2*x^2/a+8*x/a^2-4/a^3/(a*x-1)+12/a^3*ln(a*x-1)

________________________________________________________________________________________

Maxima [A]  time = 1.07234, size = 66, normalized size = 1.4 \begin{align*} -\frac{4}{a^{4} x - a^{3}} + \frac{a^{2} x^{3} + 6 \, a x^{2} + 24 \, x}{3 \, a^{2}} + \frac{12 \, \log \left (a x - 1\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(a^4*x - a^3) + 1/3*(a^2*x^3 + 6*a*x^2 + 24*x)/a^2 + 12*log(a*x - 1)/a^3

________________________________________________________________________________________

Fricas [A]  time = 1.71211, size = 130, normalized size = 2.77 \begin{align*} \frac{a^{4} x^{4} + 5 \, a^{3} x^{3} + 18 \, a^{2} x^{2} - 24 \, a x + 36 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 12}{3 \,{\left (a^{4} x - a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a^4*x^4 + 5*a^3*x^3 + 18*a^2*x^2 - 24*a*x + 36*(a*x - 1)*log(a*x - 1) - 12)/(a^4*x - a^3)

________________________________________________________________________________________

Sympy [A]  time = 0.320149, size = 39, normalized size = 0.83 \begin{align*} \frac{x^{3}}{3} - \frac{4}{a^{4} x - a^{3}} + \frac{2 x^{2}}{a} + \frac{8 x}{a^{2}} + \frac{12 \log{\left (a x - 1 \right )}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.1178, size = 93, normalized size = 1.98 \begin{align*} \frac{{\left (a x - 1\right )}^{3}{\left (\frac{9}{a x - 1} + \frac{39}{{\left (a x - 1\right )}^{2}} + 1\right )}}{3 \, a^{3}} - \frac{12 \, \log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a^{3}} - \frac{4}{{\left (a x - 1\right )} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*x - 1)^3*(9/(a*x - 1) + 39/(a*x - 1)^2 + 1)/a^3 - 12*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a^3 - 4/((a
*x - 1)*a^3)