\(\int e^{-2 \coth ^{-1}(a x)} x^3 \, dx\) [42]

Optimal result

Integrand size = 12, antiderivative size = 42 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \, dx=-\frac {2 x}{a^3}+\frac {x^2}{a^2}-\frac {2 x^3}{3 a}+\frac {x^4}{4}+\frac {2 \log (1+a x)}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6302, 6261, 78} \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \, dx=\frac {2 \log (a x+1)}{a^4}-\frac {2 x}{a^3}+\frac {x^2}{a^2}-\frac {2 x^3}{3 a}+\frac {x^4}{4} \]

[In]

Int[x^3/E^(2*ArcCoth[a*x]),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 78

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]))))

Rule 6261

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 6302

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]

Rubi steps \begin{align*} \text {integral}& = -\int e^{-2 \text {arctanh}(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] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \, dx=-\frac {2 x}{a^3}+\frac {x^2}{a^2}-\frac {2 x^3}{3 a}+\frac {x^4}{4}+\frac {2 \log (1+a x)}{a^4} \]

[In]

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

Time = 0.39 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93

method	result	size
norman	\(-\frac {2 x}{a^{3}}+\frac {x^{2}}{a^{2}}-\frac {2 x^{3}}{3 a}+\frac {x^{4}}{4}+\frac {2 \ln \left (a x +1\right )}{a^{4}}\)	\(39\)
risch	\(-\frac {2 x}{a^{3}}+\frac {x^{2}}{a^{2}}-\frac {2 x^{3}}{3 a}+\frac {x^{4}}{4}+\frac {2 \ln \left (a x +1\right )}{a^{4}}\)	\(39\)
default	\(\frac {2 \ln \left (a x +1\right )}{a^{4}}+\frac {\frac {1}{4} a^{3} x^{4}-\frac {2}{3} a^{2} x^{3}+a \,x^{2}-2 x}{a^{3}}\)	\(42\)
parallelrisch	\(\frac {3 a^{4} x^{4}-8 a^{3} x^{3}+12 a^{2} x^{2}-24 a x +24 \ln \left (a x +1\right )}{12 a^{4}}\)	\(43\)
meijerg	\(\frac {-\frac {a x \left (-15 a^{3} x^{3}+20 a^{2} x^{2}-30 a x +60\right )}{60}+\ln \left (a x +1\right )}{a^{4}}-\frac {\frac {a x \left (4 a^{2} x^{2}-6 a x +12\right )}{12}-\ln \left (a x +1\right )}{a^{4}}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \, dx=\frac {3 \, a^{4} x^{4} - 8 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 24 \, a x + 24 \, \log \left (a x + 1\right )}{12 \, a^{4}} \]

[In]

integrate(x^3*(a*x-1)/(a*x+1),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] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \, dx=\frac {x^{4}}{4} - \frac {2 x^{3}}{3 a} + \frac {x^{2}}{a^{2}} - \frac {2 x}{a^{3}} + \frac {2 \log {\left (a x + 1 \right )}}{a^{4}} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.02 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \, dx=\frac {3 \, a^{3} x^{4} - 8 \, a^{2} x^{3} + 12 \, a x^{2} - 24 \, x}{12 \, a^{3}} + \frac {2 \, \log \left (a x + 1\right )}{a^{4}} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.12 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \, dx=\frac {3 \, a^{4} x^{4} - 8 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 24 \, a x}{12 \, a^{4}} + \frac {2 \, \log \left ({\left | a x + 1 \right |}\right )}{a^{4}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int e^{-2 \coth ^{-1}(a x)} x^3 \, dx=\frac {2\,\ln \left (a\,x+1\right )}{a^4}-\frac {2\,x}{a^3}+\frac {x^4}{4}-\frac {2\,x^3}{3\,a}+\frac {x^2}{a^2} \]

[In]

int((x^3*(a*x - 1))/(a*x + 1),x)

[Out]

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