\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx\) [47]

Optimal result

Integrand size = 12, antiderivative size = 18 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx=\frac {1}{x}+2 a \log (x)-2 a \log (1+a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, 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} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx=2 a \log (x)-2 a \log (a x+1)+\frac {1}{x} \]

[In]

Int[1/(E^(2*ArcCoth[a*x])*x^2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx=\frac {1}{x}+2 a \log (x)-2 a \log (1+a x) \]

[In]

Integrate[1/(E^(2*ArcCoth[a*x])*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {1}{x}+2 a \ln \left (x \right )-2 a \ln \left (a x +1\right )\)	\(19\)
norman	\(\frac {1}{x}+2 a \ln \left (x \right )-2 a \ln \left (a x +1\right )\)	\(19\)
risch	\(\frac {1}{x}+2 a \ln \left (-x \right )-2 a \ln \left (a x +1\right )\)	\(21\)
parallelrisch	\(\frac {2 a \ln \left (x \right ) x -2 a \ln \left (a x +1\right ) x +1}{x}\)	\(23\)
meijerg	\(a \left (-\ln \left (a x +1\right )+\ln \left (x \right )+\ln \left (a \right )\right )-a \left (\ln \left (a x +1\right )-\ln \left (x \right )-\ln \left (a \right )-\frac {1}{a x}\right )\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx=-\frac {2 \, a x \log \left (a x + 1\right ) - 2 \, a x \log \left (x\right ) - 1}{x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx=2 a \left (\log {\left (x \right )} - \log {\left (x + \frac {1}{a} \right )}\right ) + \frac {1}{x} \]

[In]

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

[Out]

2*a*(log(x) - log(x + 1/a)) + 1/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx=-2 \, a \log \left (a x + 1\right ) + 2 \, a \log \left (x\right ) + \frac {1}{x} \]

[In]

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

[Out]

-2*a*log(a*x + 1) + 2*a*log(x) + 1/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx=-2 \, a \log \left ({\left | a x + 1 \right |}\right ) + 2 \, a \log \left ({\left | x \right |}\right ) + \frac {1}{x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{x^2} \, dx=\frac {1}{x}-4\,a\,\mathrm {atanh}\left (2\,a\,x+1\right ) \]

[In]

int((a*x - 1)/(x^2*(a*x + 1)),x)

[Out]

1/x - 4*a*atanh(2*a*x + 1)