\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-\frac {c}{a x})^2} \, dx\) [393]

Optimal result

Integrand size = 22, antiderivative size = 53 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^2}-\frac {1}{a c^2 (1-a x)^2}+\frac {5}{a c^2 (1-a x)}+\frac {4 \log (1-a x)}{a c^2} \]

[Out]

x/c^2-1/a/c^2/(-a*x+1)^2+5/a/c^2/(-a*x+1)+4*ln(-a*x+1)/a/c^2

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 53, 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, 78} \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {5}{a c^2 (1-a x)}-\frac {1}{a c^2 (1-a x)^2}+\frac {4 \log (1-a x)}{a c^2}+\frac {x}{c^2} \]

[In]

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

[Out]

x/c^2 - 1/(a*c^2*(1 - a*x)^2) + 5/(a*c^2*(1 - a*x)) + (4*Log[1 - a*x])/(a*c^2)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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)}}{\left (c-\frac {c}{a x}\right )^2} \, dx \\ & = -\frac {a^2 \int \frac {e^{2 \text {arctanh}(a x)} x^2}{(1-a x)^2} \, dx}{c^2} \\ & = -\frac {a^2 \int \frac {x^2 (1+a x)}{(1-a x)^3} \, dx}{c^2} \\ & = -\frac {a^2 \int \left (-\frac {1}{a^2}-\frac {2}{a^2 (-1+a x)^3}-\frac {5}{a^2 (-1+a x)^2}-\frac {4}{a^2 (-1+a x)}\right ) \, dx}{c^2} \\ & = \frac {x}{c^2}-\frac {1}{a c^2 (1-a x)^2}+\frac {5}{a c^2 (1-a x)}+\frac {4 \log (1-a x)}{a c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=-\frac {a^2 \left (-\frac {x}{a^2}+\frac {1}{a^3 (1-a x)^2}-\frac {5}{a^3 (1-a x)}-\frac {4 \log (1-a x)}{a^3}\right )}{c^2} \]

[In]

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

[Out]

-((a^2*(-(x/a^2) + 1/(a^3*(1 - a*x)^2) - 5/(a^3*(1 - a*x)) - (4*Log[1 - a*x])/a^3))/c^2)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {x}{c^{2}}+\frac {-5 c^{2} x +\frac {4 c^{2}}{a}}{c^{4} \left (a x -1\right )^{2}}+\frac {4 \ln \left (a x -1\right )}{a \,c^{2}}\)	\(47\)
default	\(\frac {a^{2} \left (\frac {x}{a^{2}}-\frac {1}{a^{3} \left (a x -1\right )^{2}}-\frac {5}{a^{3} \left (a x -1\right )}+\frac {4 \ln \left (a x -1\right )}{a^{3}}\right )}{c^{2}}\)	\(49\)
norman	\(\frac {\frac {a^{2} x^{3}}{c}-\frac {6 a \,x^{2}}{c}+\frac {4 x}{c}}{c \left (a x -1\right )^{2}}+\frac {4 \ln \left (a x -1\right )}{a \,c^{2}}\)	\(53\)
parallelrisch	\(\frac {a^{3} x^{3}+4 a^{2} \ln \left (a x -1\right ) x^{2}-6 a^{2} x^{2}-8 a \ln \left (a x -1\right ) x +4 a x +4 \ln \left (a x -1\right )}{c^{2} \left (a x -1\right )^{2} a}\)	\(67\)

[In]

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

[Out]

x/c^2+(-5*c^2*x+4*c^2/a)/c^4/(a*x-1)^2+4/a/c^2*ln(a*x-1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.32 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {a^{3} x^{3} - 2 \, a^{2} x^{2} - 4 \, a x + 4 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 4}{a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}} \]

[In]

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

[Out]

(a^3*x^3 - 2*a^2*x^2 - 4*a*x + 4*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) + 4)/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {- 5 a x + 4}{a^{3} c^{2} x^{2} - 2 a^{2} c^{2} x + a c^{2}} + \frac {x}{c^{2}} + \frac {4 \log {\left (a x - 1 \right )}}{a c^{2}} \]

[In]

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

[Out]

(-5*a*x + 4)/(a**3*c**2*x**2 - 2*a**2*c**2*x + a*c**2) + x/c**2 + 4*log(a*x - 1)/(a*c**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=-\frac {5 \, a x - 4}{a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}} + \frac {x}{c^{2}} + \frac {4 \, \log \left (a x - 1\right )}{a c^{2}} \]

[In]

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

[Out]

-(5*a*x - 4)/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2) + x/c^2 + 4*log(a*x - 1)/(a*c^2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^{2}} + \frac {4 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{2}} - \frac {5 \, a x - 4}{{\left (a x - 1\right )}^{2} a c^{2}} \]

[In]

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

[Out]

x/c^2 + 4*log(abs(a*x - 1))/(a*c^2) - (5*a*x - 4)/((a*x - 1)^2*a*c^2)

Mupad [B] (verification not implemented)

Time = 3.82 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^2}-\frac {5\,x-\frac {4}{a}}{a^2\,c^2\,x^2-2\,a\,c^2\,x+c^2}+\frac {4\,\ln \left (a\,x-1\right )}{a\,c^2} \]

[In]

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

[Out]

x/c^2 - (5*x - 4/a)/(c^2 + a^2*c^2*x^2 - 2*a*c^2*x) + (4*log(a*x - 1))/(a*c^2)