3.10 \(\int e^{2 \coth ^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=43 \[ \frac{x^2}{a^2}+\frac{2 x}{a^3}+\frac{2 \log (1-a x)}{a^4}+\frac{2 x^3}{3 a}+\frac{x^4}{4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0526951, antiderivative size = 43, 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, 77} \[ \frac{x^2}{a^2}+\frac{2 x}{a^3}+\frac{2 \log (1-a x)}{a^4}+\frac{2 x^3}{3 a}+\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])*x^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0189926, size = 43, normalized size = 1. \[ \frac{x^2}{a^2}+\frac{2 x}{a^3}+\frac{2 \log (1-a x)}{a^4}+\frac{2 x^3}{3 a}+\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCoth[a*x])*x^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 39, normalized size = 0.9 \begin{align*}{\frac{{x}^{4}}{4}}+{\frac{2\,{x}^{3}}{3\,a}}+{\frac{{x}^{2}}{{a}^{2}}}+2\,{\frac{x}{{a}^{3}}}+2\,{\frac{\ln \left ( ax-1 \right ) }{{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.01774, size = 58, normalized size = 1.35 \begin{align*} \frac{3 \, a^{3} x^{4} + 8 \, a^{2} x^{3} + 12 \, a x^{2} + 24 \, x}{12 \, a^{3}} + \frac{2 \, \log \left (a x - 1\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.8717, size = 100, normalized size = 2.33 \begin{align*} \frac{3 \, a^{4} x^{4} + 8 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 24 \, a x + 24 \, \log \left (a x - 1\right )}{12 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.90545, size = 37, normalized size = 0.86 \begin{align*} \frac{x^{4}}{4} + \frac{2 x^{3}}{3 a} + \frac{x^{2}}{a^{2}} + \frac{2 x}{a^{3}} + \frac{2 \log{\left (a x - 1 \right )}}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.16499, size = 63, normalized size = 1.47 \begin{align*} \frac{3 \, a^{4} x^{4} + 8 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 24 \, a x}{12 \, a^{4}} + \frac{2 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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