Integrand size = 22, antiderivative size = 87 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^4}-\frac {1}{2 a c^4 (1-a x)^4}+\frac {3}{a c^4 (1-a x)^3}-\frac {8}{a c^4 (1-a x)^2}+\frac {14}{a c^4 (1-a x)}+\frac {6 \log (1-a x)}{a c^4} \] Output:
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
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {17-56 a x+60 a^2 x^2-16 a^3 x^3-8 a^4 x^4+2 a^5 x^5+12 (-1+a x)^4 \log (1-a x)}{2 a c^4 (-1+a x)^4} \] Input:
Integrate[E^(2*ArcCoth[a*x])/(c - c/(a*x))^4,x]
Output:
(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)
Time = 0.80 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6717, 27, 6681, 6679, 86, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {a^4 e^{2 \text {arctanh}(a x)}}{c^4 \left (a-\frac {1}{x}\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^4 \int \frac {e^{2 \text {arctanh}(a x)}}{\left (a-\frac {1}{x}\right )^4}dx}{c^4}\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle -\frac {a^4 \int \frac {e^{2 \text {arctanh}(a x)} x^4}{(1-a x)^4}dx}{c^4}\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle -\frac {a^4 \int \frac {x^4 (a x+1)}{(1-a x)^5}dx}{c^4}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {a^4 \int \left (-\frac {1}{a^4}-\frac {6}{a^4 (a x-1)}-\frac {14}{a^4 (a x-1)^2}-\frac {16}{a^4 (a x-1)^3}-\frac {9}{a^4 (a x-1)^4}-\frac {2}{a^4 (a x-1)^5}\right )dx}{c^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 \left (-\frac {14}{a^5 (1-a x)}+\frac {8}{a^5 (1-a x)^2}-\frac {3}{a^5 (1-a x)^3}+\frac {1}{2 a^5 (1-a x)^4}-\frac {6 \log (1-a x)}{a^5}-\frac {x}{a^4}\right )}{c^4}\) |
Input:
Int[E^(2*ArcCoth[a*x])/(c - c/(a*x))^4,x]
Output:
-((a^4*(-(x/a^4) + 1/(2*a^5*(1 - a*x)^4) - 3/(a^5*(1 - a*x)^3) + 8/(a^5*(1 - a*x)^2) - 14/(a^5*(1 - a*x)) - (6*Log[1 - a*x])/a^5))/c^4)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((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]))))
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Simp[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])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol ] :> Simp[d^p Int[u*(1 + c*(x/d))^p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; F reeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2) Int[ u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
Time = 0.13 (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 | \(\frac {a^{4} \left (\frac {x}{a^{4}}-\frac {14}{a^{5} \left (a x -1\right )}+\frac {6 \ln \left (a x -1\right )}{a^{5}}-\frac {1}{2 a^{5} \left (a x -1\right )^{4}}-\frac {8}{a^{5} \left (a x -1\right )^{2}}-\frac {3}{a^{5} \left (a x -1\right )^{3}}\right )}{c^{4}}\) | \(73\) |
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 | \(\frac {2 a^{5} x^{5}+12 \ln \left (a x -1\right ) x^{4} a^{4}-25 a^{4} x^{4}-48 a^{3} \ln \left (a x -1\right ) x^{3}+52 a^{3} x^{3}+72 a^{2} \ln \left (a x -1\right ) x^{2}-42 a^{2} x^{2}-48 a \ln \left (a x -1\right ) x +12 a x +12 \ln \left (a x -1\right )}{2 c^{4} \left (a x -1\right )^{4} a}\) | \(113\) |
Input:
int(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x,method=_RETURNVERBOSE)
Output:
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)
Time = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.45 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {2 \, a^{5} x^{5} - 8 \, a^{4} x^{4} - 16 \, a^{3} x^{3} + 60 \, a^{2} x^{2} - 56 \, a x + 12 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x - 1\right ) + 17}{2 \, {\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x, algorithm="fricas")
Output:
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)
Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.08 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {- 28 a^{3} x^{3} + 68 a^{2} x^{2} - 58 a x + 17}{2 a^{5} c^{4} x^{4} - 8 a^{4} c^{4} x^{3} + 12 a^{3} c^{4} x^{2} - 8 a^{2} c^{4} x + 2 a c^{4}} + \frac {x}{c^{4}} + \frac {6 \log {\left (a x - 1 \right )}}{a c^{4}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)**4,x)
Output:
(-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)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=-\frac {28 \, a^{3} x^{3} - 68 \, a^{2} x^{2} + 58 \, a x - 17}{2 \, {\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} + \frac {x}{c^{4}} + \frac {6 \, \log \left (a x - 1\right )}{a c^{4}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x, algorithm="maxima")
Output:
-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)
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^{4}} + \frac {6 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{4}} - \frac {28 \, a^{3} x^{3} - 68 \, a^{2} x^{2} + 58 \, a x - 17}{2 \, {\left (a x - 1\right )}^{4} a c^{4}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x, algorithm="giac")
Output:
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)
Time = 13.65 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^4}-\frac {29\,x-34\,a\,x^2-\frac {17}{2\,a}+14\,a^2\,x^3}{a^4\,c^4\,x^4-4\,a^3\,c^4\,x^3+6\,a^2\,c^4\,x^2-4\,a\,c^4\,x+c^4}+\frac {6\,\ln \left (a\,x-1\right )}{a\,c^4} \] Input:
int((a*x + 1)/((c - c/(a*x))^4*(a*x - 1)),x)
Output:
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)
Time = 0.18 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.48 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {12 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-48 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+72 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-48 \,\mathrm {log}\left (a x -1\right ) a x +12 \,\mathrm {log}\left (a x -1\right )+2 a^{5} x^{5}-12 a^{4} x^{4}+36 a^{2} x^{2}-40 a x +13}{2 a \,c^{4} \left (a^{4} x^{4}-4 a^{3} x^{3}+6 a^{2} x^{2}-4 a x +1\right )} \] Input:
int(1/(a*x-1)*(a*x+1)/(c-c/a/x)^4,x)
Output:
(12*log(a*x - 1)*a**4*x**4 - 48*log(a*x - 1)*a**3*x**3 + 72*log(a*x - 1)*a **2*x**2 - 48*log(a*x - 1)*a*x + 12*log(a*x - 1) + 2*a**5*x**5 - 12*a**4*x **4 + 36*a**2*x**2 - 40*a*x + 13)/(2*a*c**4*(a**4*x**4 - 4*a**3*x**3 + 6*a **2*x**2 - 4*a*x + 1))