\(\int e^{-2 \coth ^{-1}(a x)} x^2 \, dx\) [43]

Optimal result

Integrand size = 12, antiderivative size = 33 \[ \int e^{-2 \coth ^{-1}(a x)} x^2 \, dx=\frac {2 x}{a^2}-\frac {x^2}{a}+\frac {x^3}{3}-\frac {2 \log (1+a x)}{a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 33, 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} \[ \int e^{-2 \coth ^{-1}(a x)} x^2 \, dx=-\frac {2 \log (a x+1)}{a^3}+\frac {2 x}{a^2}-\frac {x^2}{a}+\frac {x^3}{3} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} x^2 \, dx=\frac {2 x}{a^2}-\frac {x^2}{a}+\frac {x^3}{3}-\frac {2 \log (1+a x)}{a^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97

method	result	size
norman	\(\frac {2 x}{a^{2}}-\frac {x^{2}}{a}+\frac {x^{3}}{3}-\frac {2 \ln \left (a x +1\right )}{a^{3}}\)	\(32\)
risch	\(\frac {2 x}{a^{2}}-\frac {x^{2}}{a}+\frac {x^{3}}{3}-\frac {2 \ln \left (a x +1\right )}{a^{3}}\)	\(32\)
default	\(-\frac {2 \ln \left (a x +1\right )}{a^{3}}+\frac {\frac {1}{3} a^{2} x^{3}-a \,x^{2}+2 x}{a^{2}}\)	\(35\)
parallelrisch	\(-\frac {-a^{3} x^{3}+3 a^{2} x^{2}-6 a x +6 \ln \left (a x +1\right )}{3 a^{3}}\)	\(35\)
meijerg	\(\frac {\frac {a x \left (4 a^{2} x^{2}-6 a x +12\right )}{12}-\ln \left (a x +1\right )}{a^{3}}-\frac {-\frac {a x \left (-3 a x +6\right )}{6}+\ln \left (a x +1\right )}{a^{3}}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} x^2 \, dx=\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 6 \, a x - 6 \, \log \left (a x + 1\right )}{3 \, a^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int e^{-2 \coth ^{-1}(a x)} x^2 \, dx=\frac {x^{3}}{3} - \frac {x^{2}}{a} + \frac {2 x}{a^{2}} - \frac {2 \log {\left (a x + 1 \right )}}{a^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int e^{-2 \coth ^{-1}(a x)} x^2 \, dx=\frac {a^{2} x^{3} - 3 \, a x^{2} + 6 \, x}{3 \, a^{2}} - \frac {2 \, \log \left (a x + 1\right )}{a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int e^{-2 \coth ^{-1}(a x)} x^2 \, dx=\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 6 \, a x}{3 \, a^{3}} - \frac {2 \, \log \left ({\left | a x + 1 \right |}\right )}{a^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int e^{-2 \coth ^{-1}(a x)} x^2 \, dx=\frac {2\,x}{a^2}-\frac {2\,\ln \left (a\,x+1\right )}{a^3}+\frac {x^3}{3}-\frac {x^2}{a} \]

[In]

int((x^2*(a*x - 1))/(a*x + 1),x)

[Out]

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