Integrand size = 22, antiderivative size = 73 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {x}{c^3}+\frac {2}{3 a c^3 (1-a x)^3}-\frac {7}{2 a c^3 (1-a x)^2}+\frac {9}{a c^3 (1-a x)}+\frac {5 \log (1-a x)}{a c^3} \] Output:
x/c^3+2/3/a/c^3/(-a*x+1)^3-7/2/a/c^3/(-a*x+1)^2+9/a/c^3/(-a*x+1)+5*ln(-a*x +1)/a/c^3
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {-37+81 a x-36 a^2 x^2-18 a^3 x^3+6 a^4 x^4+30 (-1+a x)^3 \log (1-a x)}{6 a c^3 (-1+a x)^3} \] Input:
Integrate[E^(2*ArcCoth[a*x])/(c - c/(a*x))^3,x]
Output:
(-37 + 81*a*x - 36*a^2*x^2 - 18*a^3*x^3 + 6*a^4*x^4 + 30*(-1 + a*x)^3*Log[ 1 - a*x])/(6*a*c^3*(-1 + a*x)^3)
Time = 0.78 (sec) , antiderivative size = 68, 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.
\(\displaystyle \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {a^3 e^{2 \text {arctanh}(a x)}}{c^3 \left (a-\frac {1}{x}\right )^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^3 \int \frac {e^{2 \text {arctanh}(a x)}}{\left (a-\frac {1}{x}\right )^3}dx}{c^3}\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle \frac {a^3 \int \frac {e^{2 \text {arctanh}(a x)} x^3}{(1-a x)^3}dx}{c^3}\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \frac {a^3 \int \frac {x^3 (a x+1)}{(1-a x)^4}dx}{c^3}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {a^3 \int \left (\frac {1}{a^3}+\frac {5}{a^3 (a x-1)}+\frac {9}{a^3 (a x-1)^2}+\frac {7}{a^3 (a x-1)^3}+\frac {2}{a^3 (a x-1)^4}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \left (\frac {9}{a^4 (1-a x)}-\frac {7}{2 a^4 (1-a x)^2}+\frac {2}{3 a^4 (1-a x)^3}+\frac {5 \log (1-a x)}{a^4}+\frac {x}{a^3}\right )}{c^3}\) |
Input:
Int[E^(2*ArcCoth[a*x])/(c - c/(a*x))^3,x]
Output:
(a^3*(x/a^3 + 2/(3*a^4*(1 - a*x)^3) - 7/(2*a^4*(1 - a*x)^2) + 9/(a^4*(1 - a*x)) + (5*Log[1 - a*x])/a^4))/c^3
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 = 56, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {x}{c^{3}}+\frac {-9 a \,c^{3} x^{2}+\frac {29 c^{3} x}{2}-\frac {37 c^{3}}{6 a}}{c^{6} \left (a x -1\right )^{3}}+\frac {5 \ln \left (a x -1\right )}{a \,c^{3}}\) | \(56\) |
default | \(\frac {a^{3} \left (\frac {x}{a^{3}}-\frac {9}{a^{4} \left (a x -1\right )}+\frac {5 \ln \left (a x -1\right )}{a^{4}}-\frac {7}{2 a^{4} \left (a x -1\right )^{2}}-\frac {2}{3 a^{4} \left (a x -1\right )^{3}}\right )}{c^{3}}\) | \(61\) |
norman | \(\frac {\frac {a^{3} x^{4}}{c}-\frac {5 x}{c}+\frac {25 a \,x^{2}}{2 c}-\frac {55 a^{2} x^{3}}{6 c}}{\left (a x -1\right )^{3} c^{2}}+\frac {5 \ln \left (a x -1\right )}{a \,c^{3}}\) | \(64\) |
parallelrisch | \(\frac {6 a^{4} x^{4}+30 a^{3} \ln \left (a x -1\right ) x^{3}-55 a^{3} x^{3}-90 a^{2} \ln \left (a x -1\right ) x^{2}+75 a^{2} x^{2}+90 a \ln \left (a x -1\right ) x -30 a x -30 \ln \left (a x -1\right )}{6 c^{3} \left (a x -1\right )^{3} a}\) | \(91\) |
Input:
int(1/(a*x-1)*(a*x+1)/(c-c/a/x)^3,x,method=_RETURNVERBOSE)
Output:
x/c^3+(-9*a*c^3*x^2+29/2*c^3*x-37/6*c^3/a)/c^6/(a*x-1)^3+5/a/c^3*ln(a*x-1)
Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.37 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {6 \, a^{4} x^{4} - 18 \, a^{3} x^{3} - 36 \, a^{2} x^{2} + 81 \, a x + 30 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 37}{6 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^3,x, algorithm="fricas")
Output:
1/6*(6*a^4*x^4 - 18*a^3*x^3 - 36*a^2*x^2 + 81*a*x + 30*(a^3*x^3 - 3*a^2*x^ 2 + 3*a*x - 1)*log(a*x - 1) - 37)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3 *x - a*c^3)
Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {- 54 a^{2} x^{2} + 87 a x - 37}{6 a^{4} c^{3} x^{3} - 18 a^{3} c^{3} x^{2} + 18 a^{2} c^{3} x - 6 a c^{3}} + \frac {x}{c^{3}} + \frac {5 \log {\left (a x - 1 \right )}}{a c^{3}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)**3,x)
Output:
(-54*a**2*x**2 + 87*a*x - 37)/(6*a**4*c**3*x**3 - 18*a**3*c**3*x**2 + 18*a **2*c**3*x - 6*a*c**3) + x/c**3 + 5*log(a*x - 1)/(a*c**3)
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=-\frac {54 \, a^{2} x^{2} - 87 \, a x + 37}{6 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} + \frac {x}{c^{3}} + \frac {5 \, \log \left (a x - 1\right )}{a c^{3}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^3,x, algorithm="maxima")
Output:
-1/6*(54*a^2*x^2 - 87*a*x + 37)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3) + x/c^3 + 5*log(a*x - 1)/(a*c^3)
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {x}{c^{3}} + \frac {5 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{3}} - \frac {54 \, a^{2} x^{2} - 87 \, a x + 37}{6 \, {\left (a x - 1\right )}^{3} a c^{3}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a/x)^3,x, algorithm="giac")
Output:
x/c^3 + 5*log(abs(a*x - 1))/(a*c^3) - 1/6*(54*a^2*x^2 - 87*a*x + 37)/((a*x - 1)^3*a*c^3)
Time = 13.67 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {9\,a\,x^2-\frac {29\,x}{2}+\frac {37}{6\,a}}{-a^3\,c^3\,x^3+3\,a^2\,c^3\,x^2-3\,a\,c^3\,x+c^3}+\frac {x}{c^3}+\frac {5\,\ln \left (a\,x-1\right )}{a\,c^3} \] Input:
int((a*x + 1)/((c - c/(a*x))^3*(a*x - 1)),x)
Output:
(9*a*x^2 - (29*x)/2 + 37/(6*a))/(c^3 + 3*a^2*c^3*x^2 - a^3*c^3*x^3 - 3*a*c ^3*x) + x/c^3 + (5*log(a*x - 1))/(a*c^3)
Time = 0.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.36 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {30 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-90 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+90 \,\mathrm {log}\left (a x -1\right ) a x -30 \,\mathrm {log}\left (a x -1\right )+6 a^{4} x^{4}-30 a^{3} x^{3}+45 a x -25}{6 a \,c^{3} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int(1/(a*x-1)*(a*x+1)/(c-c/a/x)^3,x)
Output:
(30*log(a*x - 1)*a**3*x**3 - 90*log(a*x - 1)*a**2*x**2 + 90*log(a*x - 1)*a *x - 30*log(a*x - 1) + 6*a**4*x**4 - 30*a**3*x**3 + 45*a*x - 25)/(6*a*c**3 *(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))