Integrand size = 22, antiderivative size = 105 \[ \int \frac {e^{4 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^4}+\frac {4}{5 a c^4 (1-a x)^5}-\frac {5}{a c^4 (1-a x)^4}+\frac {41}{3 a c^4 (1-a x)^3}-\frac {22}{a c^4 (1-a x)^2}+\frac {26}{a c^4 (1-a x)}+\frac {8 \log (1-a x)}{a c^4} \]
x/c^4+4/5/a/c^4/(-a*x+1)^5-5/a/c^4/(-a*x+1)^4+41/3/a/c^4/(-a*x+1)^3-22/a/c ^4/(-a*x+1)^2+26/a/c^4/(-a*x+1)+8*ln(-a*x+1)/a/c^4
Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.75 \[ \int \frac {e^{4 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {-202+890 a x-1480 a^2 x^2+1080 a^3 x^3-240 a^4 x^4-75 a^5 x^5+15 a^6 x^6+120 (-1+a x)^5 \log (1-a x)}{15 a c^4 (-1+a x)^5} \]
(-202 + 890*a*x - 1480*a^2*x^2 + 1080*a^3*x^3 - 240*a^4*x^4 - 75*a^5*x^5 + 15*a^6*x^6 + 120*(-1 + a*x)^5*Log[1 - a*x])/(15*a*c^4*(-1 + a*x)^5)
Time = 0.42 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {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.
\(\displaystyle \int \frac {e^{4 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle \frac {a^4 \int \frac {e^{4 \text {arctanh}(a x)} x^4}{(1-a x)^4}dx}{c^4}\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle \frac {a^4 \int \frac {x^4 (a x+1)^2}{(1-a x)^6}dx}{c^4}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^4 \int \left (\frac {1}{a^4}+\frac {8}{a^4 (a x-1)}+\frac {26}{a^4 (a x-1)^2}+\frac {44}{a^4 (a x-1)^3}+\frac {41}{a^4 (a x-1)^4}+\frac {20}{a^4 (a x-1)^5}+\frac {4}{a^4 (a x-1)^6}\right )dx}{c^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^4 \left (\frac {26}{a^5 (1-a x)}-\frac {22}{a^5 (1-a x)^2}+\frac {41}{3 a^5 (1-a x)^3}-\frac {5}{a^5 (1-a x)^4}+\frac {4}{5 a^5 (1-a x)^5}+\frac {8 \log (1-a x)}{a^5}+\frac {x}{a^4}\right )}{c^4}\) |
(a^4*(x/a^4 + 4/(5*a^5*(1 - a*x)^5) - 5/(a^5*(1 - a*x)^4) + 41/(3*a^5*(1 - a*x)^3) - 22/(a^5*(1 - a*x)^2) + 26/(a^5*(1 - a*x)) + (8*Log[1 - a*x])/a^ 5))/c^4
3.5.83.3.1 Defintions of rubi rules used
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]
Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {x}{c^{4}}+\frac {-26 a^{3} c^{4} x^{4}+82 a^{2} c^{4} x^{3}-\frac {311 a \,c^{4} x^{2}}{3}+\frac {181 c^{4} x}{3}-\frac {202 c^{4}}{15 a}}{c^{8} \left (a x -1\right )^{5}}+\frac {8 \ln \left (a x -1\right )}{a \,c^{4}}\) | \(78\) |
default | \(\frac {a^{4} \left (\frac {x}{a^{4}}-\frac {41}{3 a^{5} \left (a x -1\right )^{3}}-\frac {22}{a^{5} \left (a x -1\right )^{2}}-\frac {26}{a^{5} \left (a x -1\right )}-\frac {5}{a^{5} \left (a x -1\right )^{4}}-\frac {4}{5 a^{5} \left (a x -1\right )^{5}}+\frac {8 \ln \left (a x -1\right )}{a^{5}}\right )}{c^{4}}\) | \(85\) |
norman | \(\frac {\frac {a^{6} x^{7}}{c}+\frac {28 a \,x^{2}}{c}-\frac {80 a^{2} x^{3}}{3 c}-\frac {8 x}{c}-\frac {34 a^{3} x^{4}}{3 c}+\frac {493 a^{4} x^{5}}{15 c}-\frac {262 a^{5} x^{6}}{15 c}}{\left (a x -1\right )^{5} \left (a x +1\right ) c^{3}}+\frac {8 \ln \left (a x -1\right )}{a \,c^{4}}\) | \(104\) |
parallelrisch | \(\frac {15 a^{6} x^{6}+120 \ln \left (a x -1\right ) x^{5} a^{5}-277 a^{5} x^{5}-600 \ln \left (a x -1\right ) x^{4} a^{4}+770 a^{4} x^{4}+1200 a^{3} \ln \left (a x -1\right ) x^{3}-940 a^{3} x^{3}-1200 a^{2} \ln \left (a x -1\right ) x^{2}+540 a^{2} x^{2}+600 a \ln \left (a x -1\right ) x -120 a x -120 \ln \left (a x -1\right )}{15 \left (a x -1\right )^{5} c^{4} a}\) | \(135\) |
x/c^4+(-26*a^3*c^4*x^4+82*a^2*c^4*x^3-311/3*a*c^4*x^2+181/3*c^4*x-202/15*c ^4/a)/c^8/(a*x-1)^5+8/a/c^4*ln(a*x-1)
Time = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.47 \[ \int \frac {e^{4 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {15 \, a^{6} x^{6} - 75 \, a^{5} x^{5} - 240 \, a^{4} x^{4} + 1080 \, a^{3} x^{3} - 1480 \, a^{2} x^{2} + 890 \, a x + 120 \, {\left (a^{5} x^{5} - 5 \, a^{4} x^{4} + 10 \, a^{3} x^{3} - 10 \, a^{2} x^{2} + 5 \, a x - 1\right )} \log \left (a x - 1\right ) - 202}{15 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}} \]
1/15*(15*a^6*x^6 - 75*a^5*x^5 - 240*a^4*x^4 + 1080*a^3*x^3 - 1480*a^2*x^2 + 890*a*x + 120*(a^5*x^5 - 5*a^4*x^4 + 10*a^3*x^3 - 10*a^2*x^2 + 5*a*x - 1 )*log(a*x - 1) - 202)/(a^6*c^4*x^5 - 5*a^5*c^4*x^4 + 10*a^4*c^4*x^3 - 10*a ^3*c^4*x^2 + 5*a^2*c^4*x - a*c^4)
Time = 0.37 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.09 \[ \int \frac {e^{4 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {- 390 a^{4} x^{4} + 1230 a^{3} x^{3} - 1555 a^{2} x^{2} + 905 a x - 202}{15 a^{6} c^{4} x^{5} - 75 a^{5} c^{4} x^{4} + 150 a^{4} c^{4} x^{3} - 150 a^{3} c^{4} x^{2} + 75 a^{2} c^{4} x - 15 a c^{4}} + \frac {x}{c^{4}} + \frac {8 \log {\left (a x - 1 \right )}}{a c^{4}} \]
(-390*a**4*x**4 + 1230*a**3*x**3 - 1555*a**2*x**2 + 905*a*x - 202)/(15*a** 6*c**4*x**5 - 75*a**5*c**4*x**4 + 150*a**4*c**4*x**3 - 150*a**3*c**4*x**2 + 75*a**2*c**4*x - 15*a*c**4) + x/c**4 + 8*log(a*x - 1)/(a*c**4)
Time = 0.20 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.08 \[ \int \frac {e^{4 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=-\frac {390 \, a^{4} x^{4} - 1230 \, a^{3} x^{3} + 1555 \, a^{2} x^{2} - 905 \, a x + 202}{15 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}} + \frac {x}{c^{4}} + \frac {8 \, \log \left (a x - 1\right )}{a c^{4}} \]
-1/15*(390*a^4*x^4 - 1230*a^3*x^3 + 1555*a^2*x^2 - 905*a*x + 202)/(a^6*c^4 *x^5 - 5*a^5*c^4*x^4 + 10*a^4*c^4*x^3 - 10*a^3*c^4*x^2 + 5*a^2*c^4*x - a*c ^4) + x/c^4 + 8*log(a*x - 1)/(a*c^4)
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63 \[ \int \frac {e^{4 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^{4}} + \frac {8 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{4}} - \frac {390 \, a^{4} x^{4} - 1230 \, a^{3} x^{3} + 1555 \, a^{2} x^{2} - 905 \, a x + 202}{15 \, {\left (a x - 1\right )}^{5} a c^{4}} \]
x/c^4 + 8*log(abs(a*x - 1))/(a*c^4) - 1/15*(390*a^4*x^4 - 1230*a^3*x^3 + 1 555*a^2*x^2 - 905*a*x + 202)/((a*x - 1)^5*a*c^4)
Time = 3.54 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04 \[ \int \frac {e^{4 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^4}+\frac {\frac {311\,a\,x^2}{3}-\frac {181\,x}{3}+\frac {202}{15\,a}-82\,a^2\,x^3+26\,a^3\,x^4}{-a^5\,c^4\,x^5+5\,a^4\,c^4\,x^4-10\,a^3\,c^4\,x^3+10\,a^2\,c^4\,x^2-5\,a\,c^4\,x+c^4}+\frac {8\,\ln \left (a\,x-1\right )}{a\,c^4} \]