Integrand size = 22, antiderivative size = 76 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=-\frac {c^3}{5 a^6 x^5}+\frac {c^3}{2 a^5 x^4}+\frac {c^3}{3 a^4 x^3}-\frac {2 c^3}{a^3 x^2}+\frac {c^3}{a^2 x}+c^3 x-\frac {2 c^3 \log (x)}{a} \] Output:
-1/5*c^3/a^6/x^5+1/2*c^3/a^5/x^4+1/3*c^3/a^4/x^3-2*c^3/a^3/x^2+c^3/a^2/x+c ^3*x-2*c^3*ln(x)/a
Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=-\frac {c^3}{5 a^6 x^5}+\frac {c^3}{2 a^5 x^4}+\frac {c^3}{3 a^4 x^3}-\frac {2 c^3}{a^3 x^2}+\frac {c^3}{a^2 x}+c^3 x-\frac {2 c^3 \log (x)}{a} \] Input:
Integrate[(c - c/(a^2*x^2))^3/E^(2*ArcCoth[a*x]),x]
Output:
-1/5*c^3/(a^6*x^5) + c^3/(2*a^5*x^4) + c^3/(3*a^4*x^3) - (2*c^3)/(a^3*x^2) + c^3/(a^2*x) + c^3*x - (2*c^3*Log[x])/a
Time = 0.79 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6717, 27, 6707, 6700, 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.
\(\displaystyle \int \left (c-\frac {c}{a^2 x^2}\right )^3 e^{-2 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {c^3 e^{-2 \text {arctanh}(a x)} \left (a^2-\frac {1}{x^2}\right )^3}{a^6}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {c^3 \int e^{-2 \text {arctanh}(a x)} \left (a^2-\frac {1}{x^2}\right )^3dx}{a^6}\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle \frac {c^3 \int \frac {e^{-2 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^3}{x^6}dx}{a^6}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {c^3 \int \frac {(1-a x)^4 (a x+1)^2}{x^6}dx}{a^6}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {c^3 \int \left (a^6-\frac {2 a^5}{x}-\frac {a^4}{x^2}+\frac {4 a^3}{x^3}-\frac {a^2}{x^4}-\frac {2 a}{x^5}+\frac {1}{x^6}\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^3 \left (a^6 x-2 a^5 \log (x)+\frac {a^4}{x}-\frac {2 a^3}{x^2}+\frac {a^2}{3 x^3}+\frac {a}{2 x^4}-\frac {1}{5 x^5}\right )}{a^6}\) |
Input:
Int[(c - c/(a^2*x^2))^3/E^(2*ArcCoth[a*x]),x]
Output:
(c^3*(-1/5*1/x^5 + a/(2*x^4) + a^2/(3*x^3) - (2*a^3)/x^2 + a^4/x + a^6*x - 2*a^5*Log[x]))/a^6
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_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 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 = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {c^{3} \left (x \,a^{6}-\frac {2 a^{3}}{x^{2}}+\frac {a^{4}}{x}+\frac {a}{2 x^{4}}-2 a^{5} \ln \left (x \right )+\frac {a^{2}}{3 x^{3}}-\frac {1}{5 x^{5}}\right )}{a^{6}}\) | \(55\) |
risch | \(c^{3} x +\frac {a^{4} c^{3} x^{4}-2 a^{3} c^{3} x^{3}+\frac {1}{3} a^{2} c^{3} x^{2}+\frac {1}{2} a \,c^{3} x -\frac {1}{5} c^{3}}{a^{6} x^{5}}-\frac {2 c^{3} \ln \left (x \right )}{a}\) | \(69\) |
norman | \(\frac {a^{3} c^{3} x^{4}+a^{5} c^{3} x^{6}-\frac {c^{3}}{5 a}+\frac {c^{3} x}{2}+\frac {a \,c^{3} x^{2}}{3}-2 a^{2} c^{3} x^{3}}{a^{5} x^{5}}-\frac {2 c^{3} \ln \left (x \right )}{a}\) | \(74\) |
parallelrisch | \(-\frac {-30 a^{6} c^{3} x^{6}+60 c^{3} \ln \left (x \right ) a^{5} x^{5}-30 a^{4} c^{3} x^{4}+60 a^{3} c^{3} x^{3}-10 a^{2} c^{3} x^{2}-15 a \,c^{3} x +6 c^{3}}{30 a^{6} x^{5}}\) | \(79\) |
meijerg | \(\frac {c^{3} \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {c^{3} \ln \left (a x +1\right )}{a}-\frac {3 c^{3} \left (\ln \left (x \right )+\ln \left (a \right )-\ln \left (a x +1\right )\right )}{a}+\frac {3 c^{3} \left (-\frac {1}{a x}-\ln \left (x \right )-\ln \left (a \right )+\ln \left (a x +1\right )\right )}{a}+\frac {3 c^{3} \left (-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}+\ln \left (x \right )+\ln \left (a \right )-\ln \left (a x +1\right )\right )}{a}-\frac {3 c^{3} \left (-\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}-\ln \left (x \right )-\ln \left (a \right )+\ln \left (a x +1\right )\right )}{a}-\frac {c^{3} \left (-\frac {1}{4 a^{4} x^{4}}+\frac {1}{3 x^{3} a^{3}}-\frac {1}{2 a^{2} x^{2}}+\frac {1}{a x}+\ln \left (x \right )+\ln \left (a \right )-\ln \left (a x +1\right )\right )}{a}+\frac {c^{3} \left (-\frac {1}{5 x^{5} a^{5}}+\frac {1}{4 a^{4} x^{4}}-\frac {1}{3 x^{3} a^{3}}+\frac {1}{2 a^{2} x^{2}}-\frac {1}{a x}-\ln \left (x \right )-\ln \left (a \right )+\ln \left (a x +1\right )\right )}{a}\) | \(284\) |
Input:
int((c-c/a^2/x^2)^3*(a*x-1)/(a*x+1),x,method=_RETURNVERBOSE)
Output:
c^3/a^6*(x*a^6-2*a^3/x^2+a^4/x+1/2*a/x^4-2*a^5*ln(x)+1/3*a^2/x^3-1/5/x^5)
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {30 \, a^{6} c^{3} x^{6} - 60 \, a^{5} c^{3} x^{5} \log \left (x\right ) + 30 \, a^{4} c^{3} x^{4} - 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \] Input:
integrate((c-c/a^2/x^2)^3*(a*x-1)/(a*x+1),x, algorithm="fricas")
Output:
1/30*(30*a^6*c^3*x^6 - 60*a^5*c^3*x^5*log(x) + 30*a^4*c^3*x^4 - 60*a^3*c^3 *x^3 + 10*a^2*c^3*x^2 + 15*a*c^3*x - 6*c^3)/(a^6*x^5)
Time = 0.17 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {a^{6} c^{3} x - 2 a^{5} c^{3} \log {\left (x \right )} + \frac {30 a^{4} c^{3} x^{4} - 60 a^{3} c^{3} x^{3} + 10 a^{2} c^{3} x^{2} + 15 a c^{3} x - 6 c^{3}}{30 x^{5}}}{a^{6}} \] Input:
integrate((c-c/a**2/x**2)**3*(a*x-1)/(a*x+1),x)
Output:
(a**6*c**3*x - 2*a**5*c**3*log(x) + (30*a**4*c**3*x**4 - 60*a**3*c**3*x**3 + 10*a**2*c**3*x**2 + 15*a*c**3*x - 6*c**3)/(30*x**5))/a**6
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=c^{3} x - \frac {2 \, c^{3} \log \left (x\right )}{a} + \frac {30 \, a^{4} c^{3} x^{4} - 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \] Input:
integrate((c-c/a^2/x^2)^3*(a*x-1)/(a*x+1),x, algorithm="maxima")
Output:
c^3*x - 2*c^3*log(x)/a + 1/30*(30*a^4*c^3*x^4 - 60*a^3*c^3*x^3 + 10*a^2*c^ 3*x^2 + 15*a*c^3*x - 6*c^3)/(a^6*x^5)
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=c^{3} x - \frac {2 \, c^{3} \log \left ({\left | x \right |}\right )}{a} + \frac {30 \, a^{4} c^{3} x^{4} - 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \] Input:
integrate((c-c/a^2/x^2)^3*(a*x-1)/(a*x+1),x, algorithm="giac")
Output:
c^3*x - 2*c^3*log(abs(x))/a + 1/30*(30*a^4*c^3*x^4 - 60*a^3*c^3*x^3 + 10*a ^2*c^3*x^2 + 15*a*c^3*x - 6*c^3)/(a^6*x^5)
Time = 13.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^3\,\left (\frac {a\,x}{2}+\frac {a^2\,x^2}{3}-2\,a^3\,x^3+a^4\,x^4+a^6\,x^6-2\,a^5\,x^5\,\ln \left (x\right )-\frac {1}{5}\right )}{a^6\,x^5} \] Input:
int(((c - c/(a^2*x^2))^3*(a*x - 1))/(a*x + 1),x)
Output:
(c^3*((a*x)/2 + (a^2*x^2)/3 - 2*a^3*x^3 + a^4*x^4 + a^6*x^6 - 2*a^5*x^5*lo g(x) - 1/5))/(a^6*x^5)
Time = 0.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^{3} \left (-60 \,\mathrm {log}\left (x \right ) a^{5} x^{5}+30 a^{6} x^{6}+30 a^{4} x^{4}-60 a^{3} x^{3}+10 a^{2} x^{2}+15 a x -6\right )}{30 a^{6} x^{5}} \] Input:
int((c-c/a^2/x^2)^3*(a*x-1)/(a*x+1),x)
Output:
(c**3*( - 60*log(x)*a**5*x**5 + 30*a**6*x**6 + 30*a**4*x**4 - 60*a**3*x**3 + 10*a**2*x**2 + 15*a*x - 6))/(30*a**6*x**5)