Integrand size = 22, antiderivative size = 64 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=\frac {c^5}{4 a^5 x^4}-\frac {c^5}{3 a^4 x^3}-\frac {c^5}{a^3 x^2}+\frac {2 c^5}{a^2 x}+c^5 x-\frac {c^5 \log (x)}{a} \] Output:
1/4*c^5/a^5/x^4-1/3*c^5/a^4/x^3-c^5/a^3/x^2+2*c^5/a^2/x+c^5*x-c^5*ln(x)/a
Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=\frac {c^5 \left (\frac {1}{4 x^4}-\frac {a}{3 x^3}-\frac {a^2}{x^2}+\frac {2 a^3}{x}+a^5 x-a^4 \log (x)\right )}{a^5} \] Input:
Integrate[E^(4*ArcTanh[a*x])*(c - c/(a*x))^5,x]
Output:
(c^5*(1/(4*x^4) - a/(3*x^3) - a^2/x^2 + (2*a^3)/x + a^5*x - a^4*Log[x]))/a ^5
Time = 0.37 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80, 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.
\(\displaystyle \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle -\frac {c^5 \int \frac {e^{4 \text {arctanh}(a x)} (1-a x)^5}{x^5}dx}{a^5}\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle -\frac {c^5 \int \frac {(1-a x)^3 (a x+1)^2}{x^5}dx}{a^5}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {c^5 \int \left (-a^5+\frac {a^4}{x}+\frac {2 a^3}{x^2}-\frac {2 a^2}{x^3}-\frac {a}{x^4}+\frac {1}{x^5}\right )dx}{a^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c^5 \left (a^5 (-x)+a^4 \log (x)-\frac {2 a^3}{x}+\frac {a^2}{x^2}+\frac {a}{3 x^3}-\frac {1}{4 x^4}\right )}{a^5}\) |
Input:
Int[E^(4*ArcTanh[a*x])*(c - c/(a*x))^5,x]
Output:
-((c^5*(-1/4*1/x^4 + a/(3*x^3) + a^2/x^2 - (2*a^3)/x - a^5*x + a^4*Log[x]) )/a^5)
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 = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.75
method | result | size |
default | \(\frac {c^{5} \left (x \,a^{5}-\frac {a^{2}}{x^{2}}+\frac {2 a^{3}}{x}+\frac {1}{4 x^{4}}-a^{4} \ln \left (x \right )-\frac {a}{3 x^{3}}\right )}{a^{5}}\) | \(48\) |
risch | \(c^{5} x +\frac {2 a^{3} c^{5} x^{3}-a^{2} c^{5} x^{2}-\frac {1}{3} a \,c^{5} x +\frac {1}{4} c^{5}}{a^{5} x^{4}}-\frac {c^{5} \ln \left (x \right )}{a}\) | \(59\) |
parallelrisch | \(-\frac {-12 a^{5} c^{5} x^{5}+12 c^{5} \ln \left (x \right ) a^{4} x^{4}-24 a^{3} c^{5} x^{3}+12 a^{2} c^{5} x^{2}+4 a \,c^{5} x -3 c^{5}}{12 a^{5} x^{4}}\) | \(68\) |
norman | \(\frac {-a^{5} c^{5} x^{6}+a^{4} c^{5} x^{5}+a^{6} c^{5} x^{7}-\frac {c^{5}}{4 a}+\frac {c^{5} x}{3}+\frac {5 c^{5} a \,x^{2}}{4}-\frac {7 c^{5} a^{2} x^{3}}{3}}{\left (a^{2} x^{2}-1\right ) a^{4} x^{4}}-\frac {c^{5} \ln \left (x \right )}{a}\) | \(96\) |
meijerg | \(-\frac {2 c^{5} \left (\frac {1}{a^{2} x^{2}}-1-4 \ln \left (x \right )-2 \ln \left (-a^{2}\right )-\frac {3 a^{2} x^{2}}{-3 a^{2} x^{2}+3}+2 \ln \left (-a^{2} x^{2}+1\right )\right )}{a}+\frac {c^{5} \left (-\frac {2 \left (-15 a^{4} x^{4}+10 a^{2} x^{2}+2\right )}{3 x^{3} \left (-a^{2}\right )^{\frac {3}{2}} \left (-2 a^{2} x^{2}+2\right )}+\frac {5 a^{3} \operatorname {arctanh}\left (a x \right )}{\left (-a^{2}\right )^{\frac {3}{2}}}\right )}{2 \sqrt {-a^{2}}}-\frac {c^{5} \left (\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )\right )}{2 a}+\frac {c^{5} \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}-\frac {3 c^{5} \left (1+2 \ln \left (x \right )+\ln \left (-a^{2}\right )+\frac {2 a^{2} x^{2}}{-2 a^{2} x^{2}+2}-\ln \left (-a^{2} x^{2}+1\right )\right )}{a}+\frac {2 a \,c^{5} x^{2}}{-a^{2} x^{2}+1}-\frac {c^{5} \left (-\frac {1}{2 a^{4} x^{4}}-\frac {2}{a^{2} x^{2}}+1+6 \ln \left (x \right )+3 \ln \left (-a^{2}\right )+\frac {4 a^{2} x^{2}}{-4 a^{2} x^{2}+4}-3 \ln \left (-a^{2} x^{2}+1\right )\right )}{2 a}+\frac {3 c^{5} \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \operatorname {arctanh}\left (a x \right )}{a}\right )}{\sqrt {-a^{2}}}+\frac {2 c^{5} \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \,\operatorname {arctanh}\left (a x \right )}{\sqrt {-a^{2}}}\right )}{\sqrt {-a^{2}}}+\frac {2 c^{5} \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \operatorname {arctanh}\left (a x \right )}{a^{3}}\right )}{\sqrt {-a^{2}}}\) | \(547\) |
Input:
int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^5,x,method=_RETURNVERBOSE)
Output:
c^5/a^5*(x*a^5-a^2/x^2+2*a^3/x+1/4/x^4-a^4*ln(x)-1/3*a/x^3)
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=\frac {12 \, a^{5} c^{5} x^{5} - 12 \, a^{4} c^{5} x^{4} \log \left (x\right ) + 24 \, a^{3} c^{5} x^{3} - 12 \, a^{2} c^{5} x^{2} - 4 \, a c^{5} x + 3 \, c^{5}}{12 \, a^{5} x^{4}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^5,x, algorithm="fricas")
Output:
1/12*(12*a^5*c^5*x^5 - 12*a^4*c^5*x^4*log(x) + 24*a^3*c^5*x^3 - 12*a^2*c^5 *x^2 - 4*a*c^5*x + 3*c^5)/(a^5*x^4)
Time = 0.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=\frac {a^{5} c^{5} x - a^{4} c^{5} \log {\left (x \right )} + \frac {24 a^{3} c^{5} x^{3} - 12 a^{2} c^{5} x^{2} - 4 a c^{5} x + 3 c^{5}}{12 x^{4}}}{a^{5}} \] Input:
integrate((a*x+1)**4/(-a**2*x**2+1)**2*(c-c/a/x)**5,x)
Output:
(a**5*c**5*x - a**4*c**5*log(x) + (24*a**3*c**5*x**3 - 12*a**2*c**5*x**2 - 4*a*c**5*x + 3*c**5)/(12*x**4))/a**5
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=c^{5} x - \frac {c^{5} \log \left (x\right )}{a} + \frac {24 \, a^{3} c^{5} x^{3} - 12 \, a^{2} c^{5} x^{2} - 4 \, a c^{5} x + 3 \, c^{5}}{12 \, a^{5} x^{4}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^5,x, algorithm="maxima")
Output:
c^5*x - c^5*log(x)/a + 1/12*(24*a^3*c^5*x^3 - 12*a^2*c^5*x^2 - 4*a*c^5*x + 3*c^5)/(a^5*x^4)
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=c^{5} x - \frac {c^{5} \log \left ({\left | x \right |}\right )}{a} + \frac {24 \, a^{3} c^{5} x^{3} - 12 \, a^{2} c^{5} x^{2} - 4 \, a c^{5} x + 3 \, c^{5}}{12 \, a^{5} x^{4}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^5,x, algorithm="giac")
Output:
c^5*x - c^5*log(abs(x))/a + 1/12*(24*a^3*c^5*x^3 - 12*a^2*c^5*x^2 - 4*a*c^ 5*x + 3*c^5)/(a^5*x^4)
Time = 14.96 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=-\frac {c^5\,\left (4\,a\,x+12\,a^2\,x^2-24\,a^3\,x^3-12\,a^5\,x^5+12\,a^4\,x^4\,\ln \left (x\right )-3\right )}{12\,a^5\,x^4} \] Input:
int(((c - c/(a*x))^5*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)
Output:
-(c^5*(4*a*x + 12*a^2*x^2 - 24*a^3*x^3 - 12*a^5*x^5 + 12*a^4*x^4*log(x) - 3))/(12*a^5*x^4)
Time = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^5 \, dx=\frac {c^{5} \left (-12 \,\mathrm {log}\left (x \right ) a^{4} x^{4}+12 a^{5} x^{5}+24 a^{3} x^{3}-12 a^{2} x^{2}-4 a x +3\right )}{12 a^{5} x^{4}} \] Input:
int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^5,x)
Output:
(c**5*( - 12*log(x)*a**4*x**4 + 12*a**5*x**5 + 24*a**3*x**3 - 12*a**2*x**2 - 4*a*x + 3))/(12*a**5*x**4)