Integrand size = 22, antiderivative size = 63 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+c^3 x+\frac {4 c^3 \log (x)}{a} \] Output:
1/5*c^3/a^6/x^5+c^3/a^5/x^4+5/3*c^3/a^4/x^3-5*c^3/a^2/x+c^3*x+4*c^3*ln(x)/ a
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+c^3 x+\frac {4 c^3 \log (x)}{a} \] Input:
Integrate[E^(4*ArcTanh[a*x])*(c - c/(a^2*x^2))^3,x]
Output:
c^3/(5*a^6*x^5) + c^3/(a^5*x^4) + (5*c^3)/(3*a^4*x^3) - (5*c^3)/(a^2*x) + c^3*x + (4*c^3*Log[x])/a
Time = 0.62 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6707, 6700, 84, 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.
\(\displaystyle \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle -\frac {c^3 \int \frac {e^{4 \text {arctanh}(a x)} \left (1-a^2 x^2\right )^3}{x^6}dx}{a^6}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle -\frac {c^3 \int \frac {(1-a x) (a x+1)^5}{x^6}dx}{a^6}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle -\frac {c^3 \int \left (-a^6-\frac {4 a^5}{x}-\frac {5 a^4}{x^2}+\frac {5 a^2}{x^4}+\frac {4 a}{x^5}+\frac {1}{x^6}\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c^3 \left (a^6 (-x)-4 a^5 \log (x)+\frac {5 a^4}{x}-\frac {5 a^2}{3 x^3}-\frac {a}{x^4}-\frac {1}{5 x^5}\right )}{a^6}\) |
Input:
Int[E^(4*ArcTanh[a*x])*(c - c/(a^2*x^2))^3,x]
Output:
-((c^3*(-1/5*1/x^5 - a/x^4 - (5*a^2)/(3*x^3) + (5*a^4)/x - a^6*x - 4*a^5*L og[x]))/a^6)
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
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]
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.75
method | result | size |
default | \(\frac {c^{3} \left (x \,a^{6}-\frac {5 a^{4}}{x}+\frac {a}{x^{4}}+4 a^{5} \ln \left (x \right )+\frac {5 a^{2}}{3 x^{3}}+\frac {1}{5 x^{5}}\right )}{a^{6}}\) | \(47\) |
risch | \(c^{3} x +\frac {-5 a^{4} c^{3} x^{4}+\frac {5}{3} a^{2} c^{3} x^{2}+a \,c^{3} x +\frac {1}{5} c^{3}}{a^{6} x^{5}}+\frac {4 c^{3} \ln \left (x \right )}{a}\) | \(58\) |
parallelrisch | \(\frac {15 a^{6} c^{3} x^{6}+60 c^{3} \ln \left (x \right ) a^{5} x^{5}-75 a^{4} c^{3} x^{4}+25 a^{2} c^{3} x^{2}+15 a \,c^{3} x +3 c^{3}}{15 a^{6} x^{5}}\) | \(68\) |
norman | \(\frac {a^{2} c^{3} x^{3}+a^{7} c^{3} x^{8}-\frac {c^{3}}{5 a}-c^{3} x -\frac {22 a \,c^{3} x^{2}}{15}+\frac {20 a^{3} c^{3} x^{4}}{3}-6 a^{5} c^{3} x^{6}}{\left (a^{2} x^{2}-1\right ) a^{5} x^{5}}+\frac {4 c^{3} \ln \left (x \right )}{a}\) | \(96\) |
meijerg | \(\frac {c^{3} \left (\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-10 a^{2} x^{2}+15\right )}{5 a^{4} \left (-a^{2} x^{2}+1\right )}-\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{a^{5}}\right )}{2 \sqrt {-a^{2}}}-\frac {3 c^{3} \left (\frac {x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (-a^{2} x^{2}+1\right )}-\frac {\left (-a^{2}\right )^{\frac {3}{2}} \operatorname {arctanh}\left (a x \right )}{a^{3}}\right )}{2 \sqrt {-a^{2}}}-\frac {7 c^{3} \left (\frac {2 x \sqrt {-a^{2}}}{-2 a^{2} x^{2}+2}+\frac {\sqrt {-a^{2}}\, \operatorname {arctanh}\left (a x \right )}{a}\right )}{\sqrt {-a^{2}}}-\frac {7 c^{3} \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \,\operatorname {arctanh}\left (a x \right )}{\sqrt {-a^{2}}}\right )}{\sqrt {-a^{2}}}+\frac {2 c^{3} \left (\frac {a^{2} x^{2}}{-a^{2} x^{2}+1}+\ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {4 a \,c^{3} x^{2}}{-a^{2} x^{2}+1}-\frac {4 c^{3} \left (\frac {1}{a^{2} x^{2}}-1-4 \ln \left (x \right )-2 \ln \left (-a^{2}\right )-\frac {3 a^{2} x^{2}}{-3 a^{2} x^{2}+3}+2 \ln \left (-a^{2} x^{2}+1\right )\right )}{a}-\frac {3 c^{3} \left (-\frac {2 \left (-15 a^{4} x^{4}+10 a^{2} x^{2}+2\right )}{3 x^{3} \left (-a^{2}\right )^{\frac {3}{2}} \left (-2 a^{2} x^{2}+2\right )}+\frac {5 a^{3} \operatorname {arctanh}\left (a x \right )}{\left (-a^{2}\right )^{\frac {3}{2}}}\right )}{2 \sqrt {-a^{2}}}-\frac {2 c^{3} \left (-\frac {1}{2 a^{4} x^{4}}-\frac {2}{a^{2} x^{2}}+1+6 \ln \left (x \right )+3 \ln \left (-a^{2}\right )+\frac {4 a^{2} x^{2}}{-4 a^{2} x^{2}+4}-3 \ln \left (-a^{2} x^{2}+1\right )\right )}{a}+\frac {c^{3} \left (-\frac {2 \left (-105 x^{6} a^{6}+70 a^{4} x^{4}+14 a^{2} x^{2}+6\right )}{5 x^{5} \left (-a^{2}\right )^{\frac {5}{2}} \left (-6 a^{2} x^{2}+6\right )}+\frac {7 a^{5} \operatorname {arctanh}\left (a x \right )}{\left (-a^{2}\right )^{\frac {5}{2}}}\right )}{2 \sqrt {-a^{2}}}\) | \(573\) |
Input:
int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x,method=_RETURNVERBOSE)
Output:
c^3/a^6*(x*a^6-5*a^4/x+a/x^4+4*a^5*ln(x)+5/3*a^2/x^3+1/5/x^5)
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {15 \, a^{6} c^{3} x^{6} + 60 \, a^{5} c^{3} x^{5} \log \left (x\right ) - 75 \, a^{4} c^{3} x^{4} + 25 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x + 3 \, c^{3}}{15 \, a^{6} x^{5}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x, algorithm="fricas")
Output:
1/15*(15*a^6*c^3*x^6 + 60*a^5*c^3*x^5*log(x) - 75*a^4*c^3*x^4 + 25*a^2*c^3 *x^2 + 15*a*c^3*x + 3*c^3)/(a^6*x^5)
Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {a^{6} c^{3} x + 4 a^{5} c^{3} \log {\left (x \right )} + \frac {- 75 a^{4} c^{3} x^{4} + 25 a^{2} c^{3} x^{2} + 15 a c^{3} x + 3 c^{3}}{15 x^{5}}}{a^{6}} \] Input:
integrate((a*x+1)**4/(-a**2*x**2+1)**2*(c-c/a**2/x**2)**3,x)
Output:
(a**6*c**3*x + 4*a**5*c**3*log(x) + (-75*a**4*c**3*x**4 + 25*a**2*c**3*x** 2 + 15*a*c**3*x + 3*c**3)/(15*x**5))/a**6
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=c^{3} x + \frac {4 \, c^{3} \log \left (x\right )}{a} - \frac {75 \, a^{4} c^{3} x^{4} - 25 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 3 \, c^{3}}{15 \, a^{6} x^{5}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x, algorithm="maxima")
Output:
c^3*x + 4*c^3*log(x)/a - 1/15*(75*a^4*c^3*x^4 - 25*a^2*c^3*x^2 - 15*a*c^3* x - 3*c^3)/(a^6*x^5)
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=c^{3} x + \frac {4 \, c^{3} \log \left ({\left | x \right |}\right )}{a} - \frac {75 \, a^{4} c^{3} x^{4} - 25 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 3 \, c^{3}}{15 \, a^{6} x^{5}} \] Input:
integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x, algorithm="giac")
Output:
c^3*x + 4*c^3*log(abs(x))/a - 1/15*(75*a^4*c^3*x^4 - 25*a^2*c^3*x^2 - 15*a *c^3*x - 3*c^3)/(a^6*x^5)
Time = 22.59 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.76 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^3\,\left (a\,x+\frac {5\,a^2\,x^2}{3}-5\,a^4\,x^4+a^6\,x^6+4\,a^5\,x^5\,\ln \left (x\right )+\frac {1}{5}\right )}{a^6\,x^5} \] Input:
int(((c - c/(a^2*x^2))^3*(a*x + 1)^4)/(a^2*x^2 - 1)^2,x)
Output:
(c^3*(a*x + (5*a^2*x^2)/3 - 5*a^4*x^4 + a^6*x^6 + 4*a^5*x^5*log(x) + 1/5)) /(a^6*x^5)
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx=\frac {c^{3} \left (60 \,\mathrm {log}\left (x \right ) a^{5} x^{5}+15 a^{6} x^{6}-75 a^{4} x^{4}+25 a^{2} x^{2}+15 a x +3\right )}{15 a^{6} x^{5}} \] Input:
int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a^2/x^2)^3,x)
Output:
(c**3*(60*log(x)*a**5*x**5 + 15*a**6*x**6 - 75*a**4*x**4 + 25*a**2*x**2 + 15*a*x + 3))/(15*a**6*x**5)