\(\int e^{2 \coth ^{-1}(a x)} x \, dx\) [12]

Optimal result

Integrand size = 10, antiderivative size = 26 \[ \int e^{2 \coth ^{-1}(a x)} x \, dx=\frac {2 x}{a}+\frac {x^2}{2}+\frac {2 \log (1-a x)}{a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6302, 6261, 78} \[ \int e^{2 \coth ^{-1}(a x)} x \, dx=\frac {2 \log (1-a x)}{a^2}+\frac {2 x}{a}+\frac {x^2}{2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{2 \coth ^{-1}(a x)} x \, dx=\frac {2 x}{a}+\frac {x^2}{2}+\frac {2 \log (1-a x)}{a^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
norman	\(\frac {x^{2}}{2}+\frac {2 x}{a}+\frac {2 \ln \left (a x -1\right )}{a^{2}}\)	\(24\)
risch	\(\frac {x^{2}}{2}+\frac {2 x}{a}+\frac {2 \ln \left (a x -1\right )}{a^{2}}\)	\(24\)
parallelrisch	\(\frac {a^{2} x^{2}+4 a x +4 \ln \left (a x -1\right )}{2 a^{2}}\)	\(26\)
default	\(\frac {\frac {1}{2} a \,x^{2}+2 x}{a}+\frac {2 \ln \left (a x -1\right )}{a^{2}}\)	\(27\)
meijerg	\(\frac {\frac {a x \left (3 a x +6\right )}{6}+\ln \left (-a x +1\right )}{a^{2}}-\frac {-a x -\ln \left (-a x +1\right )}{a^{2}}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int e^{2 \coth ^{-1}(a x)} x \, dx=\frac {a^{2} x^{2} + 4 \, a x + 4 \, \log \left (a x - 1\right )}{2 \, a^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int e^{2 \coth ^{-1}(a x)} x \, dx=\frac {x^{2}}{2} + \frac {2 x}{a} + \frac {2 \log {\left (a x - 1 \right )}}{a^{2}} \]

[In]

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

[Out]

x**2/2 + 2*x/a + 2*log(a*x - 1)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{2 \coth ^{-1}(a x)} x \, dx=\frac {a x^{2} + 4 \, x}{2 \, a} + \frac {2 \, \log \left (a x - 1\right )}{a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int e^{2 \coth ^{-1}(a x)} x \, dx=\frac {a^{2} x^{2} + 4 \, a x}{2 \, a^{2}} + \frac {2 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int e^{2 \coth ^{-1}(a x)} x \, dx=\frac {2\,\ln \left (a\,x-1\right )}{a^2}+\frac {2\,x}{a}+\frac {x^2}{2} \]

[In]

int((x*(a*x + 1))/(a*x - 1),x)

[Out]

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