Integrand size = 24, antiderivative size = 89 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {1}{2 x^2}-\frac {a}{x}+\frac {a^2}{8 (1-a x)^2}+\frac {a^2}{1-a x}+\frac {a^2}{8 (1+a x)}+3 a^2 \log (x)-\frac {39}{16} a^2 \log (1-a x)-\frac {9}{16} a^2 \log (1+a x) \] Output:
-1/2/x^2-a/x+1/8*a^2/(-a*x+1)^2+a^2/(-a*x+1)+a^2/(8*a*x+8)+3*a^2*ln(x)-39/ 16*a^2*ln(-a*x+1)-9/16*a^2*ln(a*x+1)
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {1}{16} \left (-\frac {8}{x^2}-\frac {16 a}{x}+\frac {16 a^2}{1-a x}+\frac {2 a^2}{(-1+a x)^2}+\frac {2 a^2}{1+a x}+48 a^2 \log (x)-39 a^2 \log (1-a x)-9 a^2 \log (1+a x)\right ) \] Input:
Integrate[E^ArcTanh[a*x]/(x^3*(1 - a^2*x^2)^(5/2)),x]
Output:
(-8/x^2 - (16*a)/x + (16*a^2)/(1 - a*x) + (2*a^2)/(-1 + a*x)^2 + (2*a^2)/( 1 + a*x) + 48*a^2*Log[x] - 39*a^2*Log[1 - a*x] - 9*a^2*Log[1 + a*x])/16
Time = 0.37 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \int \frac {1}{x^3 (1-a x)^3 (a x+1)^2}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {39 a^3}{16 (a x-1)}-\frac {9 a^3}{16 (a x+1)}+\frac {a^3}{(a x-1)^2}-\frac {a^3}{8 (a x+1)^2}-\frac {a^3}{4 (a x-1)^3}+\frac {3 a^2}{x}+\frac {a}{x^2}+\frac {1}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2}{1-a x}+\frac {a^2}{8 (a x+1)}+\frac {a^2}{8 (1-a x)^2}+3 a^2 \log (x)-\frac {39}{16} a^2 \log (1-a x)-\frac {9}{16} a^2 \log (a x+1)-\frac {a}{x}-\frac {1}{2 x^2}\) |
Input:
Int[E^ArcTanh[a*x]/(x^3*(1 - a^2*x^2)^(5/2)),x]
Output:
-1/2*1/x^2 - a/x + a^2/(8*(1 - a*x)^2) + a^2/(1 - a*x) + a^2/(8*(1 + a*x)) + 3*a^2*Log[x] - (39*a^2*Log[1 - a*x])/16 - (9*a^2*Log[1 + a*x])/16
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])
Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {a^{2}}{8 a x +8}-\frac {9 a^{2} \ln \left (a x +1\right )}{16}-\frac {a^{2}}{a x -1}+\frac {a^{2}}{8 \left (a x -1\right )^{2}}-\frac {39 a^{2} \ln \left (a x -1\right )}{16}-\frac {1}{2 x^{2}}-\frac {a}{x}+3 a^{2} \ln \left (x \right )\) | \(78\) |
norman | \(\frac {-\frac {1}{2}-\frac {9}{4} x^{6} a^{6}+3 a^{4} x^{4}-a x +\frac {25}{8} a^{3} x^{3}-\frac {15}{8} a^{5} x^{5}}{x^{2} \left (a^{2} x^{2}-1\right )^{2}}+3 a^{2} \ln \left (x \right )-\frac {39 a^{2} \ln \left (a x -1\right )}{16}-\frac {9 a^{2} \ln \left (a x +1\right )}{16}\) | \(84\) |
risch | \(\frac {-\frac {15}{8} a^{4} x^{4}+\frac {3}{8} a^{3} x^{3}+\frac {11}{4} a^{2} x^{2}-\frac {1}{2} a x -\frac {1}{2}}{x^{2} \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}-\frac {39 a^{2} \ln \left (-a x +1\right )}{16}+3 a^{2} \ln \left (-x \right )-\frac {9 a^{2} \ln \left (a x +1\right )}{16}\) | \(86\) |
meijerg | \(-\frac {a^{2} \left (\frac {2}{a^{2} x^{2}}-5-12 \ln \left (x \right )-6 \ln \left (-a^{2}\right )-\frac {a^{2} x^{2} \left (-5 a^{2} x^{2}+6\right )}{\left (-a^{2} x^{2}+1\right )^{2}}+6 \ln \left (-a^{2} x^{2}+1\right )\right )}{4}-\frac {a^{3} \left (-\frac {15 a^{4} x^{4}-25 a^{2} x^{2}+8}{2 x \sqrt {-a^{2}}\, \left (-a^{2} x^{2}+1\right )^{2}}+\frac {15 a \,\operatorname {arctanh}\left (a x \right )}{2 \sqrt {-a^{2}}}\right )}{4 \sqrt {-a^{2}}}\) | \(141\) |
parallelrisch | \(\frac {48 a^{5} \ln \left (x \right ) x^{5}-39 \ln \left (a x -1\right ) x^{5} a^{5}-9 \ln \left (a x +1\right ) x^{5} a^{5}-44 a^{5} x^{5}-48 \ln \left (x \right ) x^{4} a^{4}+39 \ln \left (a x -1\right ) x^{4} a^{4}+9 \ln \left (a x +1\right ) x^{4} a^{4}-8+14 a^{4} x^{4}-48 \ln \left (x \right ) x^{3} a^{3}+39 a^{3} \ln \left (a x -1\right ) x^{3}+9 \ln \left (a x +1\right ) x^{3} a^{3}+50 a^{3} x^{3}+48 a^{2} \ln \left (x \right ) x^{2}-39 a^{2} \ln \left (a x -1\right ) x^{2}-9 \ln \left (a x +1\right ) x^{2} a^{2}-8 a x}{16 x^{2} \left (a^{2} x^{2}-1\right ) \left (a x -1\right )}\) | \(206\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^3/x^3,x,method=_RETURNVERBOSE)
Output:
1/8*a^2/(a*x+1)-9/16*a^2*ln(a*x+1)-a^2/(a*x-1)+1/8*a^2/(a*x-1)^2-39/16*a^2 *ln(a*x-1)-1/2/x^2-a/x+3*a^2*ln(x)
Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (78) = 156\).
Time = 0.08 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.93 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {30 \, a^{4} x^{4} - 6 \, a^{3} x^{3} - 44 \, a^{2} x^{2} + 8 \, a x + 9 \, {\left (a^{5} x^{5} - a^{4} x^{4} - a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 39 \, {\left (a^{5} x^{5} - a^{4} x^{4} - a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 48 \, {\left (a^{5} x^{5} - a^{4} x^{4} - a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (x\right ) + 8}{16 \, {\left (a^{3} x^{5} - a^{2} x^{4} - a x^{3} + x^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^3,x, algorithm="fricas")
Output:
-1/16*(30*a^4*x^4 - 6*a^3*x^3 - 44*a^2*x^2 + 8*a*x + 9*(a^5*x^5 - a^4*x^4 - a^3*x^3 + a^2*x^2)*log(a*x + 1) + 39*(a^5*x^5 - a^4*x^4 - a^3*x^3 + a^2* x^2)*log(a*x - 1) - 48*(a^5*x^5 - a^4*x^4 - a^3*x^3 + a^2*x^2)*log(x) + 8) /(a^3*x^5 - a^2*x^4 - a*x^3 + x^2)
Time = 0.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx=3 a^{2} \log {\left (x \right )} - \frac {39 a^{2} \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {9 a^{2} \log {\left (x + \frac {1}{a} \right )}}{16} - \frac {15 a^{4} x^{4} - 3 a^{3} x^{3} - 22 a^{2} x^{2} + 4 a x + 4}{8 a^{3} x^{5} - 8 a^{2} x^{4} - 8 a x^{3} + 8 x^{2}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**3/x**3,x)
Output:
3*a**2*log(x) - 39*a**2*log(x - 1/a)/16 - 9*a**2*log(x + 1/a)/16 - (15*a** 4*x**4 - 3*a**3*x**3 - 22*a**2*x**2 + 4*a*x + 4)/(8*a**3*x**5 - 8*a**2*x** 4 - 8*a*x**3 + 8*x**2)
Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {9}{16} \, a^{2} \log \left (a x + 1\right ) - \frac {39}{16} \, a^{2} \log \left (a x - 1\right ) + 3 \, a^{2} \log \left (x\right ) - \frac {15 \, a^{4} x^{4} - 3 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 4}{8 \, {\left (a^{3} x^{5} - a^{2} x^{4} - a x^{3} + x^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^3,x, algorithm="maxima")
Output:
-9/16*a^2*log(a*x + 1) - 39/16*a^2*log(a*x - 1) + 3*a^2*log(x) - 1/8*(15*a ^4*x^4 - 3*a^3*x^3 - 22*a^2*x^2 + 4*a*x + 4)/(a^3*x^5 - a^2*x^4 - a*x^3 + x^2)
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {9}{16} \, a^{2} \log \left ({\left | a x + 1 \right |}\right ) - \frac {39}{16} \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) + 3 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {15 \, a^{4} x^{4} - 3 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 4}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} x^{2}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^3,x, algorithm="giac")
Output:
-9/16*a^2*log(abs(a*x + 1)) - 39/16*a^2*log(abs(a*x - 1)) + 3*a^2*log(abs( x)) - 1/8*(15*a^4*x^4 - 3*a^3*x^3 - 22*a^2*x^2 + 4*a*x + 4)/((a*x + 1)*(a* x - 1)^2*x^2)
Time = 14.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx=3\,a^2\,\ln \left (x\right )-\frac {39\,a^2\,\ln \left (a\,x-1\right )}{16}-\frac {9\,a^2\,\ln \left (a\,x+1\right )}{16}+\frac {\frac {15\,a^4\,x^4}{8}-\frac {3\,a^3\,x^3}{8}-\frac {11\,a^2\,x^2}{4}+\frac {a\,x}{2}+\frac {1}{2}}{-a^3\,x^5+a^2\,x^4+a\,x^3-x^2} \] Input:
int(-(a*x + 1)/(x^3*(a^2*x^2 - 1)^3),x)
Output:
3*a^2*log(x) - (39*a^2*log(a*x - 1))/16 - (9*a^2*log(a*x + 1))/16 + ((a*x) /2 - (11*a^2*x^2)/4 - (3*a^3*x^3)/8 + (15*a^4*x^4)/8 + 1/2)/(a*x^3 - x^2 + a^2*x^4 - a^3*x^5)
Time = 0.15 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.36 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {-39 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}+39 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+39 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-39 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-9 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}+9 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+9 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-9 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+48 \,\mathrm {log}\left (x \right ) a^{5} x^{5}-48 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-48 \,\mathrm {log}\left (x \right ) a^{3} x^{3}+48 \,\mathrm {log}\left (x \right ) a^{2} x^{2}-30 a^{5} x^{5}+36 a^{3} x^{3}+14 a^{2} x^{2}-8 a x -8}{16 x^{2} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^3/x^3,x)
Output:
( - 39*log(a*x - 1)*a**5*x**5 + 39*log(a*x - 1)*a**4*x**4 + 39*log(a*x - 1 )*a**3*x**3 - 39*log(a*x - 1)*a**2*x**2 - 9*log(a*x + 1)*a**5*x**5 + 9*log (a*x + 1)*a**4*x**4 + 9*log(a*x + 1)*a**3*x**3 - 9*log(a*x + 1)*a**2*x**2 + 48*log(x)*a**5*x**5 - 48*log(x)*a**4*x**4 - 48*log(x)*a**3*x**3 + 48*log (x)*a**2*x**2 - 30*a**5*x**5 + 36*a**3*x**3 + 14*a**2*x**2 - 8*a*x - 8)/(1 6*x**2*(a**3*x**3 - a**2*x**2 - a*x + 1))