\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx\) [211]

Optimal result

Integrand size = 18, antiderivative size = 14 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx=-\frac {\log (1+a x)}{a c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6302, 6264, 31} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx=-\frac {\log (a x+1)}{a c} \]

[In]

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

[Out]

-(Log[1 + a*x]/(a*c))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx=-\frac {\log (1+a x)}{a c} \]

[In]

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

[Out]

-(Log[1 + a*x]/(a*c))

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {\ln \left (a x +1\right )}{a c}\)	\(15\)
norman	\(-\frac {\ln \left (a x +1\right )}{a c}\)	\(15\)
risch	\(-\frac {\ln \left (a x +1\right )}{a c}\)	\(15\)
parallelrisch	\(-\frac {\ln \left (a x +1\right )}{a c}\)	\(15\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx=-\frac {\log \left (a x + 1\right )}{a c} \]

[In]

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

[Out]

-log(a*x + 1)/(a*c)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx=- \frac {\log {\left (a c x + c \right )}}{a c} \]

[In]

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

[Out]

-log(a*c*x + c)/(a*c)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx=-\frac {\log \left (a x + 1\right )}{a c} \]

[In]

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

[Out]

-log(a*x + 1)/(a*c)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx=-\frac {\log \left ({\left | a x + 1 \right |}\right )}{a c} \]

[In]

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

[Out]

-log(abs(a*x + 1))/(a*c)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-a c x} \, dx=-\frac {\ln \left (a\,x+1\right )}{a\,c} \]

[In]

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

[Out]

-log(a*x + 1)/(a*c)