Integrand size = 22, antiderivative size = 71 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {x}{c^2}+\frac {4}{3 a c^2 (1-a x)^3}-\frac {6}{a c^2 (1-a x)^2}+\frac {13}{a c^2 (1-a x)}+\frac {6 \log (1-a x)}{a c^2} \] Output:
x/c^2+4/3/a/c^2/(-a*x+1)^3-6/a/c^2/(-a*x+1)^2+13/a/c^2/(-a*x+1)+6*ln(-a*x+ 1)/a/c^2
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {-25+57 a x-30 a^2 x^2-9 a^3 x^3+3 a^4 x^4+18 (-1+a x)^3 \log (1-a x)}{3 a c^2 (-1+a x)^3} \] Input:
Integrate[E^(4*ArcCoth[a*x])/(c - c/(a*x))^2,x]
Output:
(-25 + 57*a*x - 30*a^2*x^2 - 9*a^3*x^3 + 3*a^4*x^4 + 18*(-1 + a*x)^3*Log[1 - a*x])/(3*a*c^2*(-1 + a*x)^3)
Time = 0.79 (sec) , antiderivative size = 66, 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, 99, 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.
\(\displaystyle \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle \int \frac {a^2 e^{4 \text {arctanh}(a x)}}{c^2 \left (a-\frac {1}{x}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {e^{4 \text {arctanh}(a x)}}{\left (a-\frac {1}{x}\right )^2}dx}{c^2}\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle \frac {a^2 \int \frac {e^{4 \text {arctanh}(a x)} x^2}{(1-a x)^2}dx}{c^2}\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \frac {a^2 \int \frac {x^2 (a x+1)^2}{(1-a x)^4}dx}{c^2}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^2 \int \left (\frac {1}{a^2}+\frac {6}{a^2 (a x-1)}+\frac {13}{a^2 (a x-1)^2}+\frac {12}{a^2 (a x-1)^3}+\frac {4}{a^2 (a x-1)^4}\right )dx}{c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (\frac {13}{a^3 (1-a x)}-\frac {6}{a^3 (1-a x)^2}+\frac {4}{3 a^3 (1-a x)^3}+\frac {6 \log (1-a x)}{a^3}+\frac {x}{a^2}\right )}{c^2}\) |
Input:
Int[E^(4*ArcCoth[a*x])/(c - c/(a*x))^2,x]
Output:
(a^2*(x/a^2 + 4/(3*a^3*(1 - a*x)^3) - 6/(a^3*(1 - a*x)^2) + 13/(a^3*(1 - a *x)) + (6*Log[1 - a*x])/a^3))/c^2
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {x}{c^{2}}+\frac {-13 a \,c^{2} x^{2}+20 c^{2} x -\frac {25 c^{2}}{3 a}}{c^{4} \left (a x -1\right )^{3}}+\frac {6 \ln \left (a x -1\right )}{a \,c^{2}}\) | \(56\) |
default | \(\frac {a^{2} \left (\frac {x}{a^{2}}-\frac {13}{\left (a x -1\right ) a^{3}}+\frac {6 \ln \left (a x -1\right )}{a^{3}}-\frac {6}{a^{3} \left (a x -1\right )^{2}}-\frac {4}{3 a^{3} \left (a x -1\right )^{3}}\right )}{c^{2}}\) | \(61\) |
norman | \(\frac {\frac {a^{3} x^{4}}{c}-\frac {6 x}{c}+\frac {15 a \,x^{2}}{c}-\frac {34 a^{2} x^{3}}{3 c}}{\left (a x -1\right )^{3} c}+\frac {6 \ln \left (a x -1\right )}{a \,c^{2}}\) | \(64\) |
parallelrisch | \(\frac {3 a^{4} x^{4}+18 a^{3} \ln \left (a x -1\right ) x^{3}-34 a^{3} x^{3}-54 a^{2} \ln \left (a x -1\right ) x^{2}+45 a^{2} x^{2}+54 a \ln \left (a x -1\right ) x -18 a x -18 \ln \left (a x -1\right )}{3 \left (a x -1\right )^{3} c^{2} a}\) | \(91\) |
Input:
int(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^2,x,method=_RETURNVERBOSE)
Output:
x/c^2+(-13*a*c^2*x^2+20*c^2*x-25/3*c^2/a)/c^4/(a*x-1)^3+6/a/c^2*ln(a*x-1)
Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {3 \, a^{4} x^{4} - 9 \, a^{3} x^{3} - 30 \, a^{2} x^{2} + 57 \, a x + 18 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 25}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^2,x, algorithm="fricas")
Output:
1/3*(3*a^4*x^4 - 9*a^3*x^3 - 30*a^2*x^2 + 57*a*x + 18*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x - 1) - 25)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2* x - a*c^2)
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {- 39 a^{2} x^{2} + 60 a x - 25}{3 a^{4} c^{2} x^{3} - 9 a^{3} c^{2} x^{2} + 9 a^{2} c^{2} x - 3 a c^{2}} + \frac {x}{c^{2}} + \frac {6 \log {\left (a x - 1 \right )}}{a c^{2}} \] Input:
integrate(1/(a*x-1)**2*(a*x+1)**2/(c-c/a/x)**2,x)
Output:
(-39*a**2*x**2 + 60*a*x - 25)/(3*a**4*c**2*x**3 - 9*a**3*c**2*x**2 + 9*a** 2*c**2*x - 3*a*c**2) + x/c**2 + 6*log(a*x - 1)/(a*c**2)
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=-\frac {39 \, a^{2} x^{2} - 60 \, a x + 25}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} + \frac {x}{c^{2}} + \frac {6 \, \log \left (a x - 1\right )}{a c^{2}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^2,x, algorithm="maxima")
Output:
-1/3*(39*a^2*x^2 - 60*a*x + 25)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2) + x/c^2 + 6*log(a*x - 1)/(a*c^2)
Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.32 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {a x - 1}{a c^{2}} - \frac {6 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a c^{2}} - \frac {\frac {39 \, a^{5} c^{4}}{a x - 1} + \frac {18 \, a^{5} c^{4}}{{\left (a x - 1\right )}^{2}} + \frac {4 \, a^{5} c^{4}}{{\left (a x - 1\right )}^{3}}}{3 \, a^{6} c^{6}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^2,x, algorithm="giac")
Output:
(a*x - 1)/(a*c^2) - 6*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/(a*c^2) - 1/3 *(39*a^5*c^4/(a*x - 1) + 18*a^5*c^4/(a*x - 1)^2 + 4*a^5*c^4/(a*x - 1)^3)/( a^6*c^6)
Time = 13.78 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {13\,a\,x^2-20\,x+\frac {25}{3\,a}}{-a^3\,c^2\,x^3+3\,a^2\,c^2\,x^2-3\,a\,c^2\,x+c^2}+\frac {x}{c^2}+\frac {6\,\ln \left (a\,x-1\right )}{a\,c^2} \] Input:
int((a*x + 1)^2/((c - c/(a*x))^2*(a*x - 1)^2),x)
Output:
(13*a*x^2 - 20*x + 25/(3*a))/(c^2 + 3*a^2*c^2*x^2 - a^3*c^2*x^3 - 3*a*c^2* x) + x/c^2 + (6*log(a*x - 1))/(a*c^2)
Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {18 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-54 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+54 \,\mathrm {log}\left (a x -1\right ) a x -18 \,\mathrm {log}\left (a x -1\right )+3 a^{4} x^{4}-19 a^{3} x^{3}+27 a x -15}{3 a \,c^{2} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^2,x)
Output:
(18*log(a*x - 1)*a**3*x**3 - 54*log(a*x - 1)*a**2*x**2 + 54*log(a*x - 1)*a *x - 18*log(a*x - 1) + 3*a**4*x**4 - 19*a**3*x**3 + 27*a*x - 15)/(3*a*c**2 *(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))