\(\int e^{2 \coth ^{-1}(a x)} (c-\frac {c}{a x}) \, dx\) [391]

Optimal result

Integrand size = 20, antiderivative size = 11 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=c x+\frac {c \log (x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6302, 6266, 6264, 45} \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c \log (x)}{a}+c x \]

[In]

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

[Out]

c*x + (c*Log[x])/a

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=c x+\frac {c \log (x)}{a} \]

[In]

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

[Out]

c*x + (c*Log[x])/a

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {c \left (a x +\ln \left (x \right )\right )}{a}\)	\(12\)
norman	\(c x +\frac {c \ln \left (x \right )}{a}\)	\(12\)
risch	\(c x +\frac {c \ln \left (x \right )}{a}\)	\(12\)
parallelrisch	\(\frac {a c x +c \ln \left (x \right )}{a}\)	\(14\)
meijerg	\(-\frac {c \left (-a x -\ln \left (-a x +1\right )\right )}{a}+\frac {c \left (-\ln \left (-a x +1\right )+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.18 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {a c x + c \log \left (x\right )}{a} \]

[In]

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

[Out]

(a*c*x + c*log(x))/a

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {a c x + c \log {\left (x \right )}}{a} \]

[In]

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

[Out]

(a*c*x + c*log(x))/a

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=c x + \frac {c \log \left (x\right )}{a} \]

[In]

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

[Out]

c*x + c*log(x)/a

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=c x + \frac {c \log \left ({\left | x \right |}\right )}{a} \]

[In]

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

[Out]

c*x + c*log(abs(x))/a

Mupad [B] (verification not implemented)

Time = 3.81 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int e^{2 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c\,\left (\ln \left (x\right )+a\,x\right )}{a} \]

[In]

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

[Out]

(c*(log(x) + a*x))/a