\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-\frac {c}{a x})^2} \, dx\) [426]

Optimal result

Integrand size = 22, antiderivative size = 18 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^2}-\frac {\text {arctanh}(a x)}{a c^2} \]

[Out]

x/c^2-arctanh(a*x)/a/c^2

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6302, 6266, 6264, 84, 213} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^2}-\frac {\text {arctanh}(a x)}{a c^2} \]

[In]

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

[Out]

x/c^2 - ArcTanh[a*x]/(a*c^2)

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6264

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[u*(1 + d*(x/c))^
p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rule 6266

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*(1 + c*(x/d))^
p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^2}-\frac {\text {arctanh}(a x)}{a c^2} \]

[In]

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

[Out]

x/c^2 - ArcTanh[a*x]/(a*c^2)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56

method	result	size
parallelrisch	\(\frac {2 a x +\ln \left (a x -1\right )-\ln \left (a x +1\right )}{2 a \,c^{2}}\)	\(28\)
default	\(\frac {a^{2} \left (\frac {x}{a^{2}}-\frac {\ln \left (a x +1\right )}{2 a^{3}}+\frac {\ln \left (a x -1\right )}{2 a^{3}}\right )}{c^{2}}\)	\(36\)
risch	\(\frac {x}{c^{2}}-\frac {\ln \left (a x +1\right )}{2 a \,c^{2}}+\frac {\ln \left (-a x +1\right )}{2 a \,c^{2}}\)	\(36\)
norman	\(\frac {\frac {a \,x^{2}}{c}-\frac {x}{c}}{c \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{2 a \,c^{2}}-\frac {\ln \left (a x +1\right )}{2 a \,c^{2}}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {2 \, a x - \log \left (a x + 1\right ) + \log \left (a x - 1\right )}{2 \, a c^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (14) = 28\).

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.89 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=a^{2} \left (\frac {x}{a^{2} c^{2}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{2} - \frac {\log {\left (x + \frac {1}{a} \right )}}{2}}{a^{3} c^{2}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.89 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^{2}} - \frac {\log \left (a x + 1\right )}{2 \, a c^{2}} + \frac {\log \left (a x - 1\right )}{2 \, a c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^{2}} - \frac {\log \left ({\left | a x + 1 \right |}\right )}{2 \, a c^{2}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{2 \, a c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=-\frac {\mathrm {atanh}\left (a\,x\right )-a\,x}{a\,c^2} \]

[In]

int((a*x - 1)/((c - c/(a*x))^2*(a*x + 1)),x)

[Out]

-(atanh(a*x) - a*x)/(a*c^2)