Integrand size = 24, antiderivative size = 71 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {1}{x}+\frac {a}{8 (1-a x)^2}+\frac {3 a}{4 (1-a x)}-\frac {a}{8 (1+a x)}+a \log (x)-\frac {23}{16} a \log (1-a x)+\frac {7}{16} a \log (1+a x) \] Output:
-1/x+1/8*a/(-a*x+1)^2+3*a/(-4*a*x+4)-a/(8*a*x+8)+a*ln(x)-23/16*a*ln(-a*x+1 )+7/16*a*ln(a*x+1)
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {1}{16} \left (-\frac {16}{x}+\frac {12 a}{1-a x}+\frac {2 a}{(-1+a x)^2}-\frac {2 a}{1+a x}+16 a \log (x)-23 a \log (1-a x)+7 a \log (1+a x)\right ) \] Input:
Integrate[E^ArcTanh[a*x]/(x^2*(1 - a^2*x^2)^(5/2)),x]
Output:
(-16/x + (12*a)/(1 - a*x) + (2*a)/(-1 + a*x)^2 - (2*a)/(1 + a*x) + 16*a*Lo g[x] - 23*a*Log[1 - a*x] + 7*a*Log[1 + a*x])/16
Time = 0.35 (sec) , antiderivative size = 71, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \int \frac {1}{x^2 (1-a x)^3 (a x+1)^2}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {23 a^2}{16 (a x-1)}+\frac {7 a^2}{16 (a x+1)}+\frac {3 a^2}{4 (a x-1)^2}+\frac {a^2}{8 (a x+1)^2}-\frac {a^2}{4 (a x-1)^3}+\frac {a}{x}+\frac {1}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 a}{4 (1-a x)}-\frac {a}{8 (a x+1)}+\frac {a}{8 (1-a x)^2}+a \log (x)-\frac {23}{16} a \log (1-a x)+\frac {7}{16} a \log (a x+1)-\frac {1}{x}\) |
Input:
Int[E^ArcTanh[a*x]/(x^2*(1 - a^2*x^2)^(5/2)),x]
Output:
-x^(-1) + a/(8*(1 - a*x)^2) + (3*a)/(4*(1 - a*x)) - a/(8*(1 + a*x)) + a*Lo g[x] - (23*a*Log[1 - a*x])/16 + (7*a*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 = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {a}{8 \left (a x +1\right )}+\frac {7 a \ln \left (a x +1\right )}{16}+\frac {a}{8 \left (a x -1\right )^{2}}-\frac {3 a}{4 \left (a x -1\right )}-\frac {23 a \ln \left (a x -1\right )}{16}-\frac {1}{x}+\ln \left (x \right ) a\) | \(59\) |
risch | \(\frac {-\frac {15}{8} a^{3} x^{3}+\frac {11}{8} a^{2} x^{2}+\frac {7}{4} a x -1}{x \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}+a \ln \left (-x \right )+\frac {7 a \ln \left (a x +1\right )}{16}-\frac {23 a \ln \left (-a x +1\right )}{16}\) | \(71\) |
norman | \(\frac {-1-\frac {3}{4} a^{5} x^{5}+a^{3} x^{3}+\frac {25}{8} a^{2} x^{2}-\frac {15}{8} a^{4} x^{4}}{\left (a^{2} x^{2}-1\right )^{2} x}+\ln \left (x \right ) a -\frac {23 a \ln \left (a x -1\right )}{16}+\frac {7 a \ln \left (a x +1\right )}{16}\) | \(72\) |
meijerg | \(-\frac {a^{2} \left (-\frac {15 a^{4} x^{4}-25 a^{2} x^{2}+8}{2 x \sqrt {-a^{2}}\, \left (-a^{2} x^{2}+1\right )^{2}}+\frac {15 a \,\operatorname {arctanh}\left (a x \right )}{2 \sqrt {-a^{2}}}\right )}{4 \sqrt {-a^{2}}}+\frac {a \left (3+4 \ln \left (x \right )+2 \ln \left (-a^{2}\right )+\frac {a^{2} x^{2} \left (-3 a^{2} x^{2}+4\right )}{\left (-a^{2} x^{2}+1\right )^{2}}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{4}\) | \(130\) |
parallelrisch | \(\frac {16 \ln \left (x \right ) x^{4} a^{4}-23 \ln \left (a x -1\right ) x^{4} a^{4}+7 \ln \left (a x +1\right ) x^{4} a^{4}-28 a^{4} x^{4}-16-16 \ln \left (x \right ) x^{3} a^{3}+23 a^{3} \ln \left (a x -1\right ) x^{3}-7 \ln \left (a x +1\right ) x^{3} a^{3}-2 a^{3} x^{3}-16 a^{2} \ln \left (x \right ) x^{2}+23 a^{2} \ln \left (a x -1\right ) x^{2}-7 \ln \left (a x +1\right ) x^{2} a^{2}+50 a^{2} x^{2}+16 a \ln \left (x \right ) x -23 a \ln \left (a x -1\right ) x +7 \ln \left (a x +1\right ) x a}{16 x \left (a^{2} x^{2}-1\right ) \left (a x -1\right )}\) | \(190\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^3/x^2,x,method=_RETURNVERBOSE)
Output:
-1/8*a/(a*x+1)+7/16*a*ln(a*x+1)+1/8*a/(a*x-1)^2-3/4*a/(a*x-1)-23/16*a*ln(a *x-1)-1/x+ln(x)*a
Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (59) = 118\).
Time = 0.08 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.11 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {30 \, a^{3} x^{3} - 22 \, a^{2} x^{2} - 28 \, a x - 7 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (a x + 1\right ) + 23 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (x\right ) + 16}{16 \, {\left (a^{3} x^{4} - a^{2} x^{3} - a x^{2} + x\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^2,x, algorithm="fricas")
Output:
-1/16*(30*a^3*x^3 - 22*a^2*x^2 - 28*a*x - 7*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(a*x + 1) + 23*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(a*x - 1) - 16*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(x) + 16)/(a^3*x^4 - a^2*x^3 - a*x^2 + x)
Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx=a \log {\left (x \right )} - \frac {23 a \log {\left (x - \frac {1}{a} \right )}}{16} + \frac {7 a \log {\left (x + \frac {1}{a} \right )}}{16} - \frac {15 a^{3} x^{3} - 11 a^{2} x^{2} - 14 a x + 8}{8 a^{3} x^{4} - 8 a^{2} x^{3} - 8 a x^{2} + 8 x} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**3/x**2,x)
Output:
a*log(x) - 23*a*log(x - 1/a)/16 + 7*a*log(x + 1/a)/16 - (15*a**3*x**3 - 11 *a**2*x**2 - 14*a*x + 8)/(8*a**3*x**4 - 8*a**2*x**3 - 8*a*x**2 + 8*x)
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {7}{16} \, a \log \left (a x + 1\right ) - \frac {23}{16} \, a \log \left (a x - 1\right ) + a \log \left (x\right ) - \frac {15 \, a^{3} x^{3} - 11 \, a^{2} x^{2} - 14 \, a x + 8}{8 \, {\left (a^{3} x^{4} - a^{2} x^{3} - a x^{2} + x\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^2,x, algorithm="maxima")
Output:
7/16*a*log(a*x + 1) - 23/16*a*log(a*x - 1) + a*log(x) - 1/8*(15*a^3*x^3 - 11*a^2*x^2 - 14*a*x + 8)/(a^3*x^4 - a^2*x^3 - a*x^2 + x)
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {7}{16} \, a \log \left ({\left | a x + 1 \right |}\right ) - \frac {23}{16} \, a \log \left ({\left | a x - 1 \right |}\right ) + a \log \left ({\left | x \right |}\right ) - \frac {15 \, a^{3} x^{3} - 11 \, a^{2} x^{2} - 14 \, a x + 8}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} x} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^2,x, algorithm="giac")
Output:
7/16*a*log(abs(a*x + 1)) - 23/16*a*log(abs(a*x - 1)) + a*log(abs(x)) - 1/8 *(15*a^3*x^3 - 11*a^2*x^2 - 14*a*x + 8)/((a*x + 1)*(a*x - 1)^2*x)
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {15\,a^3\,x^3}{8}+\frac {11\,a^2\,x^2}{8}+\frac {7\,a\,x}{4}-1}{a^3\,x^4-a^2\,x^3-a\,x^2+x}+a\,\ln \left (x\right )-\frac {23\,a\,\ln \left (a\,x-1\right )}{16}+\frac {7\,a\,\ln \left (a\,x+1\right )}{16} \] Input:
int(-(a*x + 1)/(x^2*(a^2*x^2 - 1)^3),x)
Output:
((7*a*x)/4 + (11*a^2*x^2)/8 - (15*a^3*x^3)/8 - 1)/(x - a*x^2 - a^2*x^3 + a ^3*x^4) + a*log(x) - (23*a*log(a*x - 1))/16 + (7*a*log(a*x + 1))/16
Time = 0.15 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.68 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {-23 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+23 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+23 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-23 \,\mathrm {log}\left (a x -1\right ) a x +7 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}-7 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-7 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+7 \,\mathrm {log}\left (a x +1\right ) a x +16 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-16 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-16 \,\mathrm {log}\left (x \right ) a^{2} x^{2}+16 \,\mathrm {log}\left (x \right ) a x -30 a^{4} x^{4}+52 a^{2} x^{2}-2 a x -16}{16 x \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^3/x^2,x)
Output:
( - 23*log(a*x - 1)*a**4*x**4 + 23*log(a*x - 1)*a**3*x**3 + 23*log(a*x - 1 )*a**2*x**2 - 23*log(a*x - 1)*a*x + 7*log(a*x + 1)*a**4*x**4 - 7*log(a*x + 1)*a**3*x**3 - 7*log(a*x + 1)*a**2*x**2 + 7*log(a*x + 1)*a*x + 16*log(x)* a**4*x**4 - 16*log(x)*a**3*x**3 - 16*log(x)*a**2*x**2 + 16*log(x)*a*x - 30 *a**4*x**4 + 52*a**2*x**2 - 2*a*x - 16)/(16*x*(a**3*x**3 - a**2*x**2 - a*x + 1))