\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx\) [212]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6302, 6264, 35, 212} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {\text {arctanh}(a x)}{a c^2} \]

[In]

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

[Out]

-(ArcTanh[a*x]/(a*c^2))

Rule 35

Int[1/(((a_) + (b_.)*(x_))*((c_) + (d_.)*(x_))), x_Symbol] :> Int[1/(a*c + b*d*x^2), x] /; FreeQ[{a, b, c, d},
x] && EqQ[b*c + a*d, 0]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 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)}}{(c-a c x)^2} \, dx \\ & = -\frac {\int \frac {1}{(1-a x) (1+a x)} \, dx}{c^2} \\ & = -\frac {\int \frac {1}{1-a^2 x^2} \, dx}{c^2} \\ & = -\frac {\text {arctanh}(a x)}{a c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {\text {arctanh}(a x)}{a c^2} \]

[In]

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

[Out]

-(ArcTanh[a*x]/(a*c^2))

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 24, normalized size of antiderivative = 2.00

method	result	size
parallelrisch	\(\frac {-\ln \left (a x +1\right )+\ln \left (a x -1\right )}{2 a \,c^{2}}\)	\(24\)
default	\(\frac {-\frac {\ln \left (a x +1\right )}{2 a}+\frac {\ln \left (a x -1\right )}{2 a}}{c^{2}}\)	\(28\)
norman	\(\frac {\ln \left (a x -1\right )}{2 a \,c^{2}}-\frac {\ln \left (a x +1\right )}{2 a \,c^{2}}\)	\(30\)
risch	\(-\frac {\ln \left (a x +1\right )}{2 a \,c^{2}}+\frac {\ln \left (-a x +1\right )}{2 a \,c^{2}}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.92 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {\log \left (a x + 1\right ) - \log \left (a x - 1\right )}{2 \, a c^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.67 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=\frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{2} - \frac {\log {\left (x + \frac {1}{a} \right )}}{2}}{a c^{2}} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (12) = 24\).

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 2.42 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {\log \left (a x + 1\right )}{2 \, a c^{2}} + \frac {\log \left (a x - 1\right )}{2 \, a c^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (12) = 24\).

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 2.08 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {\log \left ({\left | -\frac {2 \, c}{a c x - c} - 1 \right |}\right )}{2 \, a c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^2} \, dx=-\frac {\mathrm {atanh}\left (a\,x\right )}{a\,c^2} \]

[In]

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

[Out]

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