\(\int e^{4 \coth ^{-1}(a x)} (c-\frac {c}{a x}) \, dx\) [408]

Optimal result

Integrand size = 20, antiderivative size = 25 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=c x-\frac {c \log (x)}{a}+\frac {4 c \log (1-a x)}{a} \]

[Out]

c*x-c*ln(x)/a+4*c*ln(-a*x+1)/a

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, 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, 84} \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=-\frac {c \log (x)}{a}+\frac {4 c \log (1-a x)}{a}+c x \]

[In]

Int[E^(4*ArcCoth[a*x])*(c - c/(a*x)),x]

[Out]

c*x - (c*Log[x])/a + (4*c*Log[1 - a*x])/a

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 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^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right ) \, dx \\ & = -\frac {c \int \frac {e^{4 \text {arctanh}(a x)} (1-a x)}{x} \, dx}{a} \\ & = -\frac {c \int \frac {(1+a x)^2}{x (1-a x)} \, dx}{a} \\ & = -\frac {c \int \left (-a+\frac {1}{x}-\frac {4 a}{-1+a x}\right ) \, dx}{a} \\ & = c x-\frac {c \log (x)}{a}+\frac {4 c \log (1-a x)}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {c (a x-\log (x)+4 \log (1-a x))}{a} \]

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - c/(a*x)),x]

[Out]

(c*(a*x - Log[x] + 4*Log[1 - a*x]))/a

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {c \left (a x -\ln \left (x \right )+4 \ln \left (a x -1\right )\right )}{a}\)	\(22\)
parallelrisch	\(-\frac {-a c x +c \ln \left (x \right )-4 c \ln \left (a x -1\right )}{a}\)	\(25\)
risch	\(c x -\frac {c \ln \left (x \right )}{a}+\frac {4 c \ln \left (-a x +1\right )}{a}\)	\(26\)
norman	\(\frac {a c \,x^{2}-c x}{a x -1}-\frac {c \ln \left (x \right )}{a}+\frac {4 c \ln \left (a x -1\right )}{a}\)	\(41\)
meijerg	\(-\frac {c \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}+\frac {c \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}-\frac {c x}{-a x +1}-\frac {c \left (\frac {2 a x}{-2 a x +2}-\ln \left (-a x +1\right )+1+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}\)	\(107\)

[In]

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

[Out]

c/a*(a*x-ln(x)+4*ln(a*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {a c x + 4 \, c \log \left (a x - 1\right ) - c \log \left (x\right )}{a} \]

[In]

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

[Out]

(a*c*x + 4*c*log(a*x - 1) - c*log(x))/a

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=c x + \frac {c \left (- \log {\left (x \right )} + 4 \log {\left (x - \frac {1}{a} \right )}\right )}{a} \]

[In]

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

[Out]

c*x + c*(-log(x) + 4*log(x - 1/a))/a

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=c x + \frac {4 \, c \log \left (a x - 1\right )}{a} - \frac {c \log \left (x\right )}{a} \]

[In]

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

[Out]

c*x + 4*c*log(a*x - 1)/a - c*log(x)/a

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=\frac {{\left (a x - 1\right )} c}{a} - \frac {3 \, c \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {c \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right )}{a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right ) \, dx=c\,x-\frac {c\,\ln \left (x\right )}{a}+\frac {4\,c\,\ln \left (a\,x-1\right )}{a} \]

[In]

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

[Out]

c*x - (c*log(x))/a + (4*c*log(a*x - 1))/a