\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-\frac {c}{a x})^4} \, dx\) [395]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 87 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^4}-\frac {1}{2 a c^4 (1-a x)^4}+\frac {3}{a c^4 (1-a x)^3}-\frac {8}{a c^4 (1-a x)^2}+\frac {14}{a c^4 (1-a x)}+\frac {6 \log (1-a x)}{a c^4} \]

[Out]

x/c^4-1/2/a/c^4/(-a*x+1)^4+3/a/c^4/(-a*x+1)^3-8/a/c^4/(-a*x+1)^2+14/a/c^4/(-a*x+1)+6*ln(-a*x+1)/a/c^4

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 87, 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} \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {14}{a c^4 (1-a x)}-\frac {8}{a c^4 (1-a x)^2}+\frac {3}{a c^4 (1-a x)^3}-\frac {1}{2 a c^4 (1-a x)^4}+\frac {6 \log (1-a x)}{a c^4}+\frac {x}{c^4} \]

[In]

Int[E^(2*ArcCoth[a*x])/(c - c/(a*x))^4,x]

[Out]

x/c^4 - 1/(2*a*c^4*(1 - a*x)^4) + 3/(a*c^4*(1 - a*x)^3) - 8/(a*c^4*(1 - a*x)^2) + 14/(a*c^4*(1 - a*x)) + (6*Lo
g[1 - a*x])/(a*c^4)

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 )^4} \, dx \\ & = -\frac {a^4 \int \frac {e^{2 \text {arctanh}(a x)} x^4}{(1-a x)^4} \, dx}{c^4} \\ & = -\frac {a^4 \int \frac {x^4 (1+a x)}{(1-a x)^5} \, dx}{c^4} \\ & = -\frac {a^4 \int \left (-\frac {1}{a^4}-\frac {2}{a^4 (-1+a x)^5}-\frac {9}{a^4 (-1+a x)^4}-\frac {16}{a^4 (-1+a x)^3}-\frac {14}{a^4 (-1+a x)^2}-\frac {6}{a^4 (-1+a x)}\right ) \, dx}{c^4} \\ & = \frac {x}{c^4}-\frac {1}{2 a c^4 (1-a x)^4}+\frac {3}{a c^4 (1-a x)^3}-\frac {8}{a c^4 (1-a x)^2}+\frac {14}{a c^4 (1-a x)}+\frac {6 \log (1-a x)}{a c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {17-56 a x+60 a^2 x^2-16 a^3 x^3-8 a^4 x^4+2 a^5 x^5+12 (-1+a x)^4 \log (1-a x)}{2 a c^4 (-1+a x)^4} \]

[In]

Integrate[E^(2*ArcCoth[a*x])/(c - c/(a*x))^4,x]

[Out]

(17 - 56*a*x + 60*a^2*x^2 - 16*a^3*x^3 - 8*a^4*x^4 + 2*a^5*x^5 + 12*(-1 + a*x)^4*Log[1 - a*x])/(2*a*c^4*(-1 +
a*x)^4)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77

method result size
risch \(\frac {x}{c^{4}}+\frac {-14 a^{2} c^{4} x^{3}+34 a \,c^{4} x^{2}-29 c^{4} x +\frac {17 c^{4}}{2 a}}{c^{8} \left (a x -1\right )^{4}}+\frac {6 \ln \left (a x -1\right )}{a \,c^{4}}\) \(67\)
default \(\frac {a^{4} \left (\frac {x}{a^{4}}-\frac {8}{a^{5} \left (a x -1\right )^{2}}-\frac {1}{2 a^{5} \left (a x -1\right )^{4}}-\frac {14}{a^{5} \left (a x -1\right )}-\frac {3}{a^{5} \left (a x -1\right )^{3}}+\frac {6 \ln \left (a x -1\right )}{a^{5}}\right )}{c^{4}}\) \(73\)
norman \(\frac {\frac {a^{4} x^{5}}{c}+\frac {6 x}{c}-\frac {21 a \,x^{2}}{c}+\frac {26 a^{2} x^{3}}{c}-\frac {25 a^{3} x^{4}}{2 c}}{\left (a x -1\right )^{4} c^{3}}+\frac {6 \ln \left (a x -1\right )}{a \,c^{4}}\) \(75\)
parallelrisch \(\frac {2 a^{5} x^{5}+12 \ln \left (a x -1\right ) x^{4} a^{4}-25 a^{4} x^{4}-48 a^{3} \ln \left (a x -1\right ) x^{3}+52 a^{3} x^{3}+72 a^{2} \ln \left (a x -1\right ) x^{2}-42 a^{2} x^{2}-48 a \ln \left (a x -1\right ) x +12 a x +12 \ln \left (a x -1\right )}{2 c^{4} \left (a x -1\right )^{4} a}\) \(113\)

[In]

int(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x,method=_RETURNVERBOSE)

[Out]

x/c^4+(-14*a^2*c^4*x^3+34*a*c^4*x^2-29*c^4*x+17/2*c^4/a)/c^8/(a*x-1)^4+6/a/c^4*ln(a*x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.45 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {2 \, a^{5} x^{5} - 8 \, a^{4} x^{4} - 16 \, a^{3} x^{3} + 60 \, a^{2} x^{2} - 56 \, a x + 12 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x - 1\right ) + 17}{2 \, {\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x, algorithm="fricas")

[Out]

1/2*(2*a^5*x^5 - 8*a^4*x^4 - 16*a^3*x^3 + 60*a^2*x^2 - 56*a*x + 12*(a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x +
1)*log(a*x - 1) + 17)/(a^5*c^4*x^4 - 4*a^4*c^4*x^3 + 6*a^3*c^4*x^2 - 4*a^2*c^4*x + a*c^4)

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.08 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {- 28 a^{3} x^{3} + 68 a^{2} x^{2} - 58 a x + 17}{2 a^{5} c^{4} x^{4} - 8 a^{4} c^{4} x^{3} + 12 a^{3} c^{4} x^{2} - 8 a^{2} c^{4} x + 2 a c^{4}} + \frac {x}{c^{4}} + \frac {6 \log {\left (a x - 1 \right )}}{a c^{4}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)**4,x)

[Out]

(-28*a**3*x**3 + 68*a**2*x**2 - 58*a*x + 17)/(2*a**5*c**4*x**4 - 8*a**4*c**4*x**3 + 12*a**3*c**4*x**2 - 8*a**2
*c**4*x + 2*a*c**4) + x/c**4 + 6*log(a*x - 1)/(a*c**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=-\frac {28 \, a^{3} x^{3} - 68 \, a^{2} x^{2} + 58 \, a x - 17}{2 \, {\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} + \frac {x}{c^{4}} + \frac {6 \, \log \left (a x - 1\right )}{a c^{4}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x, algorithm="maxima")

[Out]

-1/2*(28*a^3*x^3 - 68*a^2*x^2 + 58*a*x - 17)/(a^5*c^4*x^4 - 4*a^4*c^4*x^3 + 6*a^3*c^4*x^2 - 4*a^2*c^4*x + a*c^
4) + x/c^4 + 6*log(a*x - 1)/(a*c^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^{4}} + \frac {6 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{4}} - \frac {28 \, a^{3} x^{3} - 68 \, a^{2} x^{2} + 58 \, a x - 17}{2 \, {\left (a x - 1\right )}^{4} a c^{4}} \]

[In]

integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x, algorithm="giac")

[Out]

x/c^4 + 6*log(abs(a*x - 1))/(a*c^4) - 1/2*(28*a^3*x^3 - 68*a^2*x^2 + 58*a*x - 17)/((a*x - 1)^4*a*c^4)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^4}-\frac {29\,x-34\,a\,x^2-\frac {17}{2\,a}+14\,a^2\,x^3}{a^4\,c^4\,x^4-4\,a^3\,c^4\,x^3+6\,a^2\,c^4\,x^2-4\,a\,c^4\,x+c^4}+\frac {6\,\ln \left (a\,x-1\right )}{a\,c^4} \]

[In]

int((a*x + 1)/((c - c/(a*x))^4*(a*x - 1)),x)

[Out]

x/c^4 - (29*x - 34*a*x^2 - 17/(2*a) + 14*a^2*x^3)/(c^4 + 6*a^2*c^4*x^2 - 4*a^3*c^4*x^3 + a^4*c^4*x^4 - 4*a*c^4
*x) + (6*log(a*x - 1))/(a*c^4)