Integrand size = 24, antiderivative size = 59 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {1}{8 (1-a x)^2}+\frac {1}{2 (1-a x)}+\frac {1}{8 (1+a x)}+\log (x)-\frac {11}{16} \log (1-a x)-\frac {5}{16} \log (1+a x) \] Output:
1/8/(-a*x+1)^2+1/(-2*a*x+2)+1/(8*a*x+8)+ln(x)-11/16*ln(-a*x+1)-5/16*ln(a*x +1)
Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {1}{16} \left (\frac {8}{1-a x}+\frac {2}{(-1+a x)^2}+\frac {2}{1+a x}+16 \log (x)-11 \log (1-a x)-5 \log (1+a x)\right ) \] Input:
Integrate[E^ArcTanh[a*x]/(x*(1 - a^2*x^2)^(5/2)),x]
Output:
(8/(1 - a*x) + 2/(-1 + a*x)^2 + 2/(1 + a*x) + 16*Log[x] - 11*Log[1 - a*x] - 5*Log[1 + a*x])/16
Time = 0.53 (sec) , antiderivative size = 59, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \int \frac {1}{x (1-a x)^3 (a x+1)^2}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {11 a}{16 (a x-1)}-\frac {5 a}{16 (a x+1)}+\frac {a}{2 (a x-1)^2}-\frac {a}{8 (a x+1)^2}-\frac {a}{4 (a x-1)^3}+\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2 (1-a x)}+\frac {1}{8 (a x+1)}+\frac {1}{8 (1-a x)^2}-\frac {11}{16} \log (1-a x)-\frac {5}{16} \log (a x+1)+\log (x)\) |
Input:
Int[E^ArcTanh[a*x]/(x*(1 - a^2*x^2)^(5/2)),x]
Output:
1/(8*(1 - a*x)^2) + 1/(2*(1 - a*x)) + 1/(8*(1 + a*x)) + Log[x] - (11*Log[1 - a*x])/16 - (5*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 = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {1}{8 a x +8}-\frac {5 \ln \left (a x +1\right )}{16}+\frac {1}{8 \left (a x -1\right )^{2}}-\frac {1}{2 \left (a x -1\right )}-\frac {11 \ln \left (a x -1\right )}{16}+\ln \left (x \right )\) | \(47\) |
risch | \(\frac {-\frac {3}{8} a^{2} x^{2}-\frac {1}{8} a x +\frac {3}{4}}{\left (a x -1\right ) \left (a^{2} x^{2}-1\right )}-\frac {5 \ln \left (a x +1\right )}{16}-\frac {11 \ln \left (-a x +1\right )}{16}+\ln \left (-x \right )\) | \(56\) |
norman | \(\frac {-\frac {3}{4} a^{4} x^{4}+a^{2} x^{2}+\frac {5}{8} a x -\frac {3}{8} a^{3} x^{3}}{\left (a^{2} x^{2}-1\right )^{2}}-\frac {11 \ln \left (a x -1\right )}{16}-\frac {5 \ln \left (a x +1\right )}{16}+\ln \left (x \right )\) | \(60\) |
meijerg | \(\frac {3}{4}+\ln \left (x \right )+\frac {\ln \left (-a^{2}\right )}{2}+\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+4\right )}{4 \left (-a^{2} x^{2}+1\right )^{2}}-\frac {\ln \left (-a^{2} x^{2}+1\right )}{2}+\frac {a \left (\frac {x \sqrt {-a^{2}}\, \left (-3 a^{2} x^{2}+5\right )}{2 \left (-a^{2} x^{2}+1\right )^{2}}+\frac {3 \sqrt {-a^{2}}\, \operatorname {arctanh}\left (a x \right )}{2 a}\right )}{4 \sqrt {-a^{2}}}\) | \(115\) |
parallelrisch | \(\frac {16 \ln \left (x \right ) x^{3} a^{3}-11 a^{3} \ln \left (a x -1\right ) x^{3}-5 \ln \left (a x +1\right ) x^{3} a^{3}-12 a^{3} x^{3}-16 a^{2} \ln \left (x \right ) x^{2}+11 a^{2} \ln \left (a x -1\right ) x^{2}+5 \ln \left (a x +1\right ) x^{2} a^{2}+6 a^{2} x^{2}-16 a \ln \left (x \right ) x +11 a \ln \left (a x -1\right ) x +5 \ln \left (a x +1\right ) x a +10 a x +16 \ln \left (x \right )-11 \ln \left (a x -1\right )-5 \ln \left (a x +1\right )}{16 \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}\) | \(164\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^3/x,x,method=_RETURNVERBOSE)
Output:
1/8/(a*x+1)-5/16*ln(a*x+1)+1/8/(a*x-1)^2-1/2/(a*x-1)-11/16*ln(a*x-1)+ln(x)
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (47) = 94\).
Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.07 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, a^{2} x^{2} + 2 \, a x + 5 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 11 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (x\right ) - 12}{16 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x,x, algorithm="fricas")
Output:
-1/16*(6*a^2*x^2 + 2*a*x + 5*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) + 11*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x - 1) - 16*(a^3*x^3 - a^2*x^2 - a* x + 1)*log(x) - 12)/(a^3*x^3 - a^2*x^2 - a*x + 1)
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.02 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx=- \frac {3 a^{2} x^{2} + a x - 6}{8 a^{3} x^{3} - 8 a^{2} x^{2} - 8 a x + 8} + \log {\left (x \right )} - \frac {11 \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {5 \log {\left (x + \frac {1}{a} \right )}}{16} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**3/x,x)
Output:
-(3*a**2*x**2 + a*x - 6)/(8*a**3*x**3 - 8*a**2*x**2 - 8*a*x + 8) + log(x) - 11*log(x - 1/a)/16 - 5*log(x + 1/a)/16
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {3 \, a^{2} x^{2} + a x - 6}{8 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )}} - \frac {5}{16} \, \log \left (a x + 1\right ) - \frac {11}{16} \, \log \left (a x - 1\right ) + \log \left (x\right ) \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x,x, algorithm="maxima")
Output:
-1/8*(3*a^2*x^2 + a*x - 6)/(a^3*x^3 - a^2*x^2 - a*x + 1) - 5/16*log(a*x + 1) - 11/16*log(a*x - 1) + log(x)
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {3 \, a^{2} x^{2} + a x - 6}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2}} - \frac {5}{16} \, \log \left ({\left | a x + 1 \right |}\right ) - \frac {11}{16} \, \log \left ({\left | a x - 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x,x, algorithm="giac")
Output:
-1/8*(3*a^2*x^2 + a*x - 6)/((a*x + 1)*(a*x - 1)^2) - 5/16*log(abs(a*x + 1) ) - 11/16*log(abs(a*x - 1)) + log(abs(x))
Time = 23.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx=\ln \left (x\right )-\frac {11\,\ln \left (1-a\,x\right )}{16}-\frac {5\,\ln \left (a\,x+1\right )}{16}+\frac {\frac {3\,a^2\,x^2}{8}+\frac {a\,x}{8}-\frac {3}{4}}{-a^3\,x^3+a^2\,x^2+a\,x-1} \] Input:
int(-(a*x + 1)/(x*(a^2*x^2 - 1)^3),x)
Output:
log(x) - (11*log(1 - a*x))/16 - (5*log(a*x + 1))/16 + ((a*x)/8 + (3*a^2*x^ 2)/8 - 3/4)/(a*x + a^2*x^2 - a^3*x^3 - 1)
Time = 0.15 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.73 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {-11 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+11 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+11 \,\mathrm {log}\left (a x -1\right ) a x -11 \,\mathrm {log}\left (a x -1\right )-5 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+5 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+5 \,\mathrm {log}\left (a x +1\right ) a x -5 \,\mathrm {log}\left (a x +1\right )+16 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-16 \,\mathrm {log}\left (x \right ) a^{2} x^{2}-16 \,\mathrm {log}\left (x \right ) a x +16 \,\mathrm {log}\left (x \right )-6 a^{3} x^{3}+4 a x +6}{16 a^{3} x^{3}-16 a^{2} x^{2}-16 a x +16} \] Input:
int((a*x+1)/(-a^2*x^2+1)^3/x,x)
Output:
( - 11*log(a*x - 1)*a**3*x**3 + 11*log(a*x - 1)*a**2*x**2 + 11*log(a*x - 1 )*a*x - 11*log(a*x - 1) - 5*log(a*x + 1)*a**3*x**3 + 5*log(a*x + 1)*a**2*x **2 + 5*log(a*x + 1)*a*x - 5*log(a*x + 1) + 16*log(x)*a**3*x**3 - 16*log(x )*a**2*x**2 - 16*log(x)*a*x + 16*log(x) - 6*a**3*x**3 + 4*a*x + 6)/(16*(a* *3*x**3 - a**2*x**2 - a*x + 1))