\(\int \frac {e^{2 \coth ^{-1}(a x)}}{x^3} \, dx\) [16]

Optimal result

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

[Out]

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

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

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

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91

method	result	size
norman	\(\frac {\frac {1}{2}+2 a x}{x^{2}}-2 a^{2} \ln \left (x \right )+2 a^{2} \ln \left (a x -1\right )\)	\(30\)
default	\(\frac {1}{2 x^{2}}+\frac {2 a}{x}-2 a^{2} \ln \left (x \right )+2 a^{2} \ln \left (a x -1\right )\)	\(31\)
risch	\(\frac {\frac {1}{2}+2 a x}{x^{2}}+2 a^{2} \ln \left (-a x +1\right )-2 a^{2} \ln \left (x \right )\)	\(31\)
parallelrisch	\(-\frac {4 a^{2} \ln \left (x \right ) x^{2}-4 a^{2} \ln \left (a x -1\right ) x^{2}-1-4 a x}{2 x^{2}}\)	\(36\)
meijerg	\(a^{2} \left (\ln \left (-a x +1\right )-\ln \left (x \right )-\ln \left (-a \right )+\frac {1}{a x}\right )-a^{2} \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )-\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}\right )\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{x^3} \, dx=\frac {4 \, a^{2} x^{2} \log \left (a x - 1\right ) - 4 \, a^{2} x^{2} \log \left (x\right ) + 4 \, a x + 1}{2 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{x^3} \, dx=2 a^{2} \left (- \log {\left (x \right )} + \log {\left (x - \frac {1}{a} \right )}\right ) + \frac {4 a x + 1}{2 x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{x^3} \, dx=2 \, a^{2} \log \left (a x - 1\right ) - 2 \, a^{2} \log \left (x\right ) + \frac {4 \, a x + 1}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{x^3} \, dx=2 \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) - 2 \, a^{2} \log \left ({\left | x \right |}\right ) + \frac {4 \, a x + 1}{2 \, x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{x^3} \, dx=\frac {2\,a\,x+\frac {1}{2}}{x^2}-4\,a^2\,\mathrm {atanh}\left (2\,a\,x-1\right ) \]

[In]

int((a*x + 1)/(x^3*(a*x - 1)),x)

[Out]

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