Integrand size = 24, antiderivative size = 102 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {1}{3 x^3}-\frac {a}{2 x^2}-\frac {3 a^2}{x}+\frac {a^3}{8 (1-a x)^2}+\frac {5 a^3}{4 (1-a x)}-\frac {a^3}{8 (1+a x)}+3 a^3 \log (x)-\frac {59}{16} a^3 \log (1-a x)+\frac {11}{16} a^3 \log (1+a x) \] Output:
-1/3/x^3-1/2*a/x^2-3*a^2/x+1/8*a^3/(-a*x+1)^2+5*a^3/(-4*a*x+4)-a^3/(8*a*x+ 8)+3*a^3*ln(x)-59/16*a^3*ln(-a*x+1)+11/16*a^3*ln(a*x+1)
Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {1}{48} \left (-\frac {16}{x^3}-\frac {24 a}{x^2}-\frac {144 a^2}{x}+\frac {60 a^3}{1-a x}+\frac {6 a^3}{(-1+a x)^2}-\frac {6 a^3}{1+a x}+144 a^3 \log (x)-177 a^3 \log (1-a x)+33 a^3 \log (1+a x)\right ) \] Input:
Integrate[E^ArcTanh[a*x]/(x^4*(1 - a^2*x^2)^(5/2)),x]
Output:
(-16/x^3 - (24*a)/x^2 - (144*a^2)/x + (60*a^3)/(1 - a*x) + (6*a^3)/(-1 + a *x)^2 - (6*a^3)/(1 + a*x) + 144*a^3*Log[x] - 177*a^3*Log[1 - a*x] + 33*a^3 *Log[1 + a*x])/48
Time = 0.39 (sec) , antiderivative size = 102, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \int \frac {1}{x^4 (1-a x)^3 (a x+1)^2}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {59 a^4}{16 (a x-1)}+\frac {11 a^4}{16 (a x+1)}+\frac {5 a^4}{4 (a x-1)^2}+\frac {a^4}{8 (a x+1)^2}-\frac {a^4}{4 (a x-1)^3}+\frac {3 a^3}{x}+\frac {3 a^2}{x^2}+\frac {a}{x^3}+\frac {1}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 a^3}{4 (1-a x)}-\frac {a^3}{8 (a x+1)}+\frac {a^3}{8 (1-a x)^2}+3 a^3 \log (x)-\frac {59}{16} a^3 \log (1-a x)+\frac {11}{16} a^3 \log (a x+1)-\frac {3 a^2}{x}-\frac {a}{2 x^2}-\frac {1}{3 x^3}\) |
Input:
Int[E^ArcTanh[a*x]/(x^4*(1 - a^2*x^2)^(5/2)),x]
Output:
-1/3*1/x^3 - a/(2*x^2) - (3*a^2)/x + a^3/(8*(1 - a*x)^2) + (5*a^3)/(4*(1 - a*x)) - a^3/(8*(1 + a*x)) + 3*a^3*Log[x] - (59*a^3*Log[1 - a*x])/16 + (11 *a^3*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 = 86, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {a^{3}}{8 \left (a x +1\right )}+\frac {11 a^{3} \ln \left (a x +1\right )}{16}+\frac {a^{3}}{8 \left (a x -1\right )^{2}}-\frac {5 a^{3}}{4 \left (a x -1\right )}-\frac {59 a^{3} \ln \left (a x -1\right )}{16}-\frac {1}{3 x^{3}}-\frac {a}{2 x^{2}}-\frac {3 a^{2}}{x}+3 a^{3} \ln \left (x \right )\) | \(86\) |
norman | \(\frac {-\frac {1}{3}-\frac {9}{4} a^{7} x^{7}+3 a^{5} x^{5}-\frac {1}{2} a x -\frac {7}{3} a^{2} x^{2}+\frac {175}{24} a^{4} x^{4}-\frac {35}{8} x^{6} a^{6}}{\left (a^{2} x^{2}-1\right )^{2} x^{3}}+3 a^{3} \ln \left (x \right )-\frac {59 a^{3} \ln \left (a x -1\right )}{16}+\frac {11 a^{3} \ln \left (a x +1\right )}{16}\) | \(92\) |
risch | \(\frac {-\frac {35}{8} a^{5} x^{5}+\frac {23}{8} a^{4} x^{4}+\frac {53}{12} a^{3} x^{3}-\frac {13}{6} a^{2} x^{2}-\frac {1}{6} a x -\frac {1}{3}}{x^{3} \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}+3 a^{3} \ln \left (-x \right )+\frac {11 a^{3} \ln \left (a x +1\right )}{16}-\frac {59 a^{3} \ln \left (-a x +1\right )}{16}\) | \(94\) |
meijerg | \(\frac {a^{4} \left (-\frac {105 x^{6} a^{6}-175 a^{4} x^{4}+56 a^{2} x^{2}+8}{6 x^{3} \left (-a^{2}\right )^{\frac {3}{2}} \left (-a^{2} x^{2}+1\right )^{2}}+\frac {35 a^{3} \operatorname {arctanh}\left (a x \right )}{2 \left (-a^{2}\right )^{\frac {3}{2}}}\right )}{4 \sqrt {-a^{2}}}-\frac {a^{3} \left (\frac {2}{a^{2} x^{2}}-5-12 \ln \left (x \right )-6 \ln \left (-a^{2}\right )-\frac {a^{2} x^{2} \left (-5 a^{2} x^{2}+6\right )}{\left (-a^{2} x^{2}+1\right )^{2}}+6 \ln \left (-a^{2} x^{2}+1\right )\right )}{4}\) | \(151\) |
parallelrisch | \(\frac {144 a^{6} \ln \left (x \right ) x^{6}-177 \ln \left (a x -1\right ) x^{6} a^{6}+33 \ln \left (a x +1\right ) x^{6} a^{6}-212 x^{6} a^{6}-144 a^{5} \ln \left (x \right ) x^{5}+177 \ln \left (a x -1\right ) x^{5} a^{5}-33 \ln \left (a x +1\right ) x^{5} a^{5}-16+2 a^{5} x^{5}-144 \ln \left (x \right ) x^{4} a^{4}+177 \ln \left (a x -1\right ) x^{4} a^{4}-33 \ln \left (a x +1\right ) x^{4} a^{4}+350 a^{4} x^{4}+144 \ln \left (x \right ) x^{3} a^{3}-177 a^{3} \ln \left (a x -1\right ) x^{3}+33 \ln \left (a x +1\right ) x^{3} a^{3}-104 a^{2} x^{2}-8 a x}{48 x^{3} \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}\) | \(214\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^3/x^4,x,method=_RETURNVERBOSE)
Output:
-1/8*a^3/(a*x+1)+11/16*a^3*ln(a*x+1)+1/8*a^3/(a*x-1)^2-5/4*a^3/(a*x-1)-59/ 16*a^3*ln(a*x-1)-1/3/x^3-1/2*a/x^2-3*a^2/x+3*a^3*ln(x)
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (86) = 172\).
Time = 0.08 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.76 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {210 \, a^{5} x^{5} - 138 \, a^{4} x^{4} - 212 \, a^{3} x^{3} + 104 \, a^{2} x^{2} + 8 \, a x - 33 \, {\left (a^{6} x^{6} - a^{5} x^{5} - a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (a x + 1\right ) + 177 \, {\left (a^{6} x^{6} - a^{5} x^{5} - a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (a x - 1\right ) - 144 \, {\left (a^{6} x^{6} - a^{5} x^{5} - a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (x\right ) + 16}{48 \, {\left (a^{3} x^{6} - a^{2} x^{5} - a x^{4} + x^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^4,x, algorithm="fricas")
Output:
-1/48*(210*a^5*x^5 - 138*a^4*x^4 - 212*a^3*x^3 + 104*a^2*x^2 + 8*a*x - 33* (a^6*x^6 - a^5*x^5 - a^4*x^4 + a^3*x^3)*log(a*x + 1) + 177*(a^6*x^6 - a^5* x^5 - a^4*x^4 + a^3*x^3)*log(a*x - 1) - 144*(a^6*x^6 - a^5*x^5 - a^4*x^4 + a^3*x^3)*log(x) + 16)/(a^3*x^6 - a^2*x^5 - a*x^4 + x^3)
Time = 0.46 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx=3 a^{3} \log {\left (x \right )} - \frac {59 a^{3} \log {\left (x - \frac {1}{a} \right )}}{16} + \frac {11 a^{3} \log {\left (x + \frac {1}{a} \right )}}{16} - \frac {105 a^{5} x^{5} - 69 a^{4} x^{4} - 106 a^{3} x^{3} + 52 a^{2} x^{2} + 4 a x + 8}{24 a^{3} x^{6} - 24 a^{2} x^{5} - 24 a x^{4} + 24 x^{3}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**3/x**4,x)
Output:
3*a**3*log(x) - 59*a**3*log(x - 1/a)/16 + 11*a**3*log(x + 1/a)/16 - (105*a **5*x**5 - 69*a**4*x**4 - 106*a**3*x**3 + 52*a**2*x**2 + 4*a*x + 8)/(24*a* *3*x**6 - 24*a**2*x**5 - 24*a*x**4 + 24*x**3)
Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {11}{16} \, a^{3} \log \left (a x + 1\right ) - \frac {59}{16} \, a^{3} \log \left (a x - 1\right ) + 3 \, a^{3} \log \left (x\right ) - \frac {105 \, a^{5} x^{5} - 69 \, a^{4} x^{4} - 106 \, a^{3} x^{3} + 52 \, a^{2} x^{2} + 4 \, a x + 8}{24 \, {\left (a^{3} x^{6} - a^{2} x^{5} - a x^{4} + x^{3}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^4,x, algorithm="maxima")
Output:
11/16*a^3*log(a*x + 1) - 59/16*a^3*log(a*x - 1) + 3*a^3*log(x) - 1/24*(105 *a^5*x^5 - 69*a^4*x^4 - 106*a^3*x^3 + 52*a^2*x^2 + 4*a*x + 8)/(a^3*x^6 - a ^2*x^5 - a*x^4 + x^3)
Time = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {11}{16} \, a^{3} \log \left ({\left | a x + 1 \right |}\right ) - \frac {59}{16} \, a^{3} \log \left ({\left | a x - 1 \right |}\right ) + 3 \, a^{3} \log \left ({\left | x \right |}\right ) - \frac {105 \, a^{5} x^{5} - 69 \, a^{4} x^{4} - 106 \, a^{3} x^{3} + 52 \, a^{2} x^{2} + 4 \, a x + 8}{24 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} x^{3}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3/x^4,x, algorithm="giac")
Output:
11/16*a^3*log(abs(a*x + 1)) - 59/16*a^3*log(abs(a*x - 1)) + 3*a^3*log(abs( x)) - 1/24*(105*a^5*x^5 - 69*a^4*x^4 - 106*a^3*x^3 + 52*a^2*x^2 + 4*a*x + 8)/((a*x + 1)*(a*x - 1)^2*x^3)
Time = 0.06 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx=3\,a^3\,\ln \left (x\right )-\frac {59\,a^3\,\ln \left (a\,x-1\right )}{16}+\frac {11\,a^3\,\ln \left (a\,x+1\right )}{16}+\frac {\frac {35\,a^5\,x^5}{8}-\frac {23\,a^4\,x^4}{8}-\frac {53\,a^3\,x^3}{12}+\frac {13\,a^2\,x^2}{6}+\frac {a\,x}{6}+\frac {1}{3}}{-a^3\,x^6+a^2\,x^5+a\,x^4-x^3} \] Input:
int(-(a*x + 1)/(x^4*(a^2*x^2 - 1)^3),x)
Output:
3*a^3*log(x) - (59*a^3*log(a*x - 1))/16 + (11*a^3*log(a*x + 1))/16 + ((a*x )/6 + (13*a^2*x^2)/6 - (53*a^3*x^3)/12 - (23*a^4*x^4)/8 + (35*a^5*x^5)/8 + 1/3)/(a*x^4 - x^3 + a^2*x^5 - a^3*x^6)
Time = 0.15 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.14 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {-177 \,\mathrm {log}\left (a x -1\right ) a^{6} x^{6}+177 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}+177 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-177 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+33 \,\mathrm {log}\left (a x +1\right ) a^{6} x^{6}-33 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}-33 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+33 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+144 \,\mathrm {log}\left (x \right ) a^{6} x^{6}-144 \,\mathrm {log}\left (x \right ) a^{5} x^{5}-144 \,\mathrm {log}\left (x \right ) a^{4} x^{4}+144 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-210 a^{6} x^{6}+348 a^{4} x^{4}+2 a^{3} x^{3}-104 a^{2} x^{2}-8 a x -16}{48 x^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^3/x^4,x)
Output:
( - 177*log(a*x - 1)*a**6*x**6 + 177*log(a*x - 1)*a**5*x**5 + 177*log(a*x - 1)*a**4*x**4 - 177*log(a*x - 1)*a**3*x**3 + 33*log(a*x + 1)*a**6*x**6 - 33*log(a*x + 1)*a**5*x**5 - 33*log(a*x + 1)*a**4*x**4 + 33*log(a*x + 1)*a* *3*x**3 + 144*log(x)*a**6*x**6 - 144*log(x)*a**5*x**5 - 144*log(x)*a**4*x* *4 + 144*log(x)*a**3*x**3 - 210*a**6*x**6 + 348*a**4*x**4 + 2*a**3*x**3 - 104*a**2*x**2 - 8*a*x - 16)/(48*x**3*(a**3*x**3 - a**2*x**2 - a*x + 1))