Integrand size = 22, antiderivative size = 75 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^4}-\frac {1}{4 a c^4 (1-a x)^2}+\frac {7}{4 a c^4 (1-a x)}+\frac {17 \log (1-a x)}{8 a c^4}-\frac {\log (1+a x)}{8 a c^4} \] Output:
x/c^4-1/4/a/c^4/(-a*x+1)^2+7/4/a/c^4/(-a*x+1)+17/8*ln(-a*x+1)/a/c^4-1/8*ln (a*x+1)/a/c^4
Time = 0.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=-\frac {a^4 \left (-\frac {x}{a^4}+\frac {1}{4 a^5 (1-a x)^2}-\frac {7}{4 a^5 (1-a x)}-\frac {17 \log (1-a x)}{8 a^5}+\frac {\log (1+a x)}{8 a^5}\right )}{c^4} \] Input:
Integrate[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^4),x]
Output:
-((a^4*(-(x/a^4) + 1/(4*a^5*(1 - a*x)^2) - 7/(4*a^5*(1 - a*x)) - (17*Log[1 - a*x])/(8*a^5) + Log[1 + a*x]/(8*a^5)))/c^4)
Time = 0.77 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.96, 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.
\(\displaystyle \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle -\frac {a^4 \int \frac {x^4}{(1-a x)^3 (a x+1)}dx}{c^4}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a^4 \int \left (\frac {1}{8 a^4 (a x+1)}-\frac {1}{a^4}-\frac {17}{8 a^4 (a x-1)}-\frac {7}{4 a^4 (a x-1)^2}-\frac {1}{2 a^4 (a x-1)^3}\right )dx}{c^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 \left (-\frac {7}{4 a^5 (1-a x)}+\frac {1}{4 a^5 (1-a x)^2}-\frac {17 \log (1-a x)}{8 a^5}+\frac {\log (a x+1)}{8 a^5}-\frac {x}{a^4}\right )}{c^4}\) |
Input:
Int[1/(E^(2*ArcCoth[a*x])*(c - c/(a*x))^4),x]
Output:
-((a^4*(-(x/a^4) + 1/(4*a^5*(1 - a*x)^2) - 7/(4*a^5*(1 - a*x)) - (17*Log[1 - a*x])/(8*a^5) + Log[1 + a*x]/(8*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_))^(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 = 0.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {a^{4} \left (\frac {x}{a^{4}}-\frac {\ln \left (a x +1\right )}{8 a^{5}}-\frac {1}{4 a^{5} \left (a x -1\right )^{2}}-\frac {7}{4 a^{5} \left (a x -1\right )}+\frac {17 \ln \left (a x -1\right )}{8 a^{5}}\right )}{c^{4}}\) | \(60\) |
risch | \(\frac {x}{c^{4}}+\frac {-\frac {7 c^{4} x}{4}+\frac {3 c^{4}}{2 a}}{c^{8} \left (a x -1\right )^{2}}+\frac {17 \ln \left (-a x +1\right )}{8 a \,c^{4}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{4}}\) | \(62\) |
norman | \(\frac {\frac {a^{3} x^{4}}{c}-\frac {9 x}{4 c}+\frac {23 a \,x^{2}}{4 c}-\frac {9 a^{2} x^{3}}{2 c}}{c^{3} \left (a x -1\right )^{3}}+\frac {17 \ln \left (a x -1\right )}{8 a \,c^{4}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{4}}\) | \(78\) |
parallelrisch | \(\frac {8 a^{3} x^{3}+17 a^{2} \ln \left (a x -1\right ) x^{2}-\ln \left (a x +1\right ) x^{2} a^{2}-28 a^{2} x^{2}-34 a \ln \left (a x -1\right ) x +2 \ln \left (a x +1\right ) x a +18 a x +17 \ln \left (a x -1\right )-\ln \left (a x +1\right )}{8 c^{4} \left (a x -1\right )^{2} a}\) | \(101\) |
Input:
int((a*x-1)/(a*x+1)/(c-c/a/x)^4,x,method=_RETURNVERBOSE)
Output:
a^4/c^4*(x/a^4-1/8*ln(a*x+1)/a^5-1/4/a^5/(a*x-1)^2-7/4/a^5/(a*x-1)+17/8/a^ 5*ln(a*x-1))
Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {8 \, a^{3} x^{3} - 16 \, a^{2} x^{2} - 6 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 17 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 12}{8 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \] Input:
integrate((a*x-1)/(a*x+1)/(c-c/a/x)^4,x, algorithm="fricas")
Output:
1/8*(8*a^3*x^3 - 16*a^2*x^2 - 6*a*x - (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + 17*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) + 12)/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4)
Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=a^{4} \left (\frac {- 7 a x + 6}{4 a^{7} c^{4} x^{2} - 8 a^{6} c^{4} x + 4 a^{5} c^{4}} + \frac {x}{a^{4} c^{4}} + \frac {\frac {17 \log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a^{5} c^{4}}\right ) \] Input:
integrate((a*x-1)/(a*x+1)/(c-c/a/x)**4,x)
Output:
a**4*((-7*a*x + 6)/(4*a**7*c**4*x**2 - 8*a**6*c**4*x + 4*a**5*c**4) + x/(a **4*c**4) + (17*log(x - 1/a)/8 - log(x + 1/a)/8)/(a**5*c**4))
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=-\frac {7 \, a x - 6}{4 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} + \frac {x}{c^{4}} - \frac {\log \left (a x + 1\right )}{8 \, a c^{4}} + \frac {17 \, \log \left (a x - 1\right )}{8 \, a c^{4}} \] Input:
integrate((a*x-1)/(a*x+1)/(c-c/a/x)^4,x, algorithm="maxima")
Output:
-1/4*(7*a*x - 6)/(a^3*c^4*x^2 - 2*a^2*c^4*x + a*c^4) + x/c^4 - 1/8*log(a*x + 1)/(a*c^4) + 17/8*log(a*x - 1)/(a*c^4)
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^{4}} - \frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{4}} + \frac {17 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{4}} - \frac {7 \, a x - 6}{4 \, {\left (a x - 1\right )}^{2} a c^{4}} \] Input:
integrate((a*x-1)/(a*x+1)/(c-c/a/x)^4,x, algorithm="giac")
Output:
x/c^4 - 1/8*log(abs(a*x + 1))/(a*c^4) + 17/8*log(abs(a*x - 1))/(a*c^4) - 1 /4*(7*a*x - 6)/((a*x - 1)^2*a*c^4)
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {x}{c^4}-\frac {\frac {7\,x}{4}-\frac {3}{2\,a}}{a^2\,c^4\,x^2-2\,a\,c^4\,x+c^4}+\frac {17\,\ln \left (a\,x-1\right )}{8\,a\,c^4}-\frac {\ln \left (a\,x+1\right )}{8\,a\,c^4} \] Input:
int((a*x - 1)/((c - c/(a*x))^4*(a*x + 1)),x)
Output:
x/c^4 - ((7*x)/4 - 3/(2*a))/(c^4 + a^2*c^4*x^2 - 2*a*c^4*x) + (17*log(a*x - 1))/(8*a*c^4) - log(a*x + 1)/(8*a*c^4)
Time = 0.17 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^4} \, dx=\frac {17 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-34 \,\mathrm {log}\left (a x -1\right ) a x +17 \,\mathrm {log}\left (a x -1\right )-\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+2 \,\mathrm {log}\left (a x +1\right ) a x -\mathrm {log}\left (a x +1\right )+8 a^{3} x^{3}-19 a^{2} x^{2}+9}{8 a \,c^{4} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int((a*x-1)/(a*x+1)/(c-c/a/x)^4,x)
Output:
(17*log(a*x - 1)*a**2*x**2 - 34*log(a*x - 1)*a*x + 17*log(a*x - 1) - log(a *x + 1)*a**2*x**2 + 2*log(a*x + 1)*a*x - log(a*x + 1) + 8*a**3*x**3 - 19*a **2*x**2 + 9)/(8*a*c**4*(a**2*x**2 - 2*a*x + 1))