\(\int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx\) [425]

Optimal result

Integrand size = 22, antiderivative size = 20 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {x}{c}-\frac {\log (1+a x)}{a c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6302, 6266, 6264, 45} \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {x}{c}-\frac {\log (a x+1)}{a c} \]

[In]

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

[Out]

x/c - Log[1 + a*x]/(a*c)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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)}}{c-\frac {c}{a x}} \, dx \\ & = \frac {a \int \frac {e^{-2 \text {arctanh}(a x)} x}{1-a x} \, dx}{c} \\ & = \frac {a \int \frac {x}{1+a x} \, dx}{c} \\ & = \frac {a \int \left (\frac {1}{a}-\frac {1}{a (1+a x)}\right ) \, dx}{c} \\ & = \frac {x}{c}-\frac {\log (1+a x)}{a c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {a \left (\frac {x}{a}-\frac {\log (1+a x)}{a^2}\right )}{c} \]

[In]

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

[Out]

(a*(x/a - Log[1 + a*x]/a^2))/c

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {a x -\ln \left (a x +1\right )}{a c}\)	\(20\)
norman	\(\frac {x}{c}-\frac {\ln \left (a x +1\right )}{a c}\)	\(21\)
risch	\(\frac {x}{c}-\frac {\ln \left (a x +1\right )}{a c}\)	\(21\)
default	\(\frac {a \left (\frac {x}{a}-\frac {\ln \left (a x +1\right )}{a^{2}}\right )}{c}\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {a x - \log \left (a x + 1\right )}{a c} \]

[In]

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

[Out]

(a*x - log(a*x + 1))/(a*c)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=a \left (\frac {x}{a c} - \frac {\log {\left (a x + 1 \right )}}{a^{2} c}\right ) \]

[In]

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

[Out]

a*(x/(a*c) - log(a*x + 1)/(a**2*c))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {x}{c} - \frac {\log \left (a x + 1\right )}{a c} \]

[In]

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

[Out]

x/c - log(a*x + 1)/(a*c)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {x}{c} - \frac {\log \left ({\left | a x + 1 \right |}\right )}{a c} \]

[In]

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

[Out]

x/c - log(abs(a*x + 1))/(a*c)

Mupad [B] (verification not implemented)

Time = 3.80 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=-\frac {\ln \left (a\,x+1\right )-a\,x}{a\,c} \]

[In]

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

[Out]

-(log(a*x + 1) - a*x)/(a*c)