Integrand size = 25, antiderivative size = 134 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx=-\frac {1}{2 c^3 x^2}-\frac {2 a}{c^3 x}+\frac {a^2}{12 c^3 (1-a x)^3}+\frac {a^2}{2 c^3 (1-a x)^2}+\frac {39 a^2}{16 c^3 (1-a x)}+\frac {a^2}{16 c^3 (1+a x)}+\frac {5 a^2 \log (x)}{c^3}-\frac {75 a^2 \log (1-a x)}{16 c^3}-\frac {5 a^2 \log (1+a x)}{16 c^3} \] Output:
-1/2/c^3/x^2-2*a/c^3/x+1/12*a^2/c^3/(-a*x+1)^3+1/2*a^2/c^3/(-a*x+1)^2+39/1 6*a^2/c^3/(-a*x+1)+1/16*a^2/c^3/(a*x+1)+5*a^2*ln(x)/c^3-75/16*a^2*ln(-a*x+ 1)/c^3-5/16*a^2*ln(a*x+1)/c^3
Time = 0.12 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.73 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx=\frac {-\frac {24}{x^2}-\frac {96 a}{x}+\frac {117 a^2}{1-a x}-\frac {4 a^2}{(-1+a x)^3}+\frac {24 a^2}{(-1+a x)^2}+\frac {3 a^2}{1+a x}+240 a^2 \log (x)-225 a^2 \log (1-a x)-15 a^2 \log (1+a x)}{48 c^3} \] Input:
Integrate[E^(2*ArcTanh[a*x])/(x^3*(c - a^2*c*x^2)^3),x]
Output:
(-24/x^2 - (96*a)/x + (117*a^2)/(1 - a*x) - (4*a^2)/(-1 + a*x)^3 + (24*a^2 )/(-1 + a*x)^2 + (3*a^2)/(1 + a*x) + 240*a^2*Log[x] - 225*a^2*Log[1 - a*x] - 15*a^2*Log[1 + a*x])/(48*c^3)
Time = 0.62 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int \frac {1}{x^3 (1-a x)^4 (a x+1)^2}dx}{c^3}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {75 a^3}{16 (a x-1)}-\frac {5 a^3}{16 (a x+1)}+\frac {39 a^3}{16 (a x-1)^2}-\frac {a^3}{16 (a x+1)^2}-\frac {a^3}{(a x-1)^3}+\frac {a^3}{4 (a x-1)^4}+\frac {5 a^2}{x}+\frac {2 a}{x^2}+\frac {1}{x^3}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {39 a^2}{16 (1-a x)}+\frac {a^2}{16 (a x+1)}+\frac {a^2}{2 (1-a x)^2}+\frac {a^2}{12 (1-a x)^3}+5 a^2 \log (x)-\frac {75}{16} a^2 \log (1-a x)-\frac {5}{16} a^2 \log (a x+1)-\frac {2 a}{x}-\frac {1}{2 x^2}}{c^3}\) |
Input:
Int[E^(2*ArcTanh[a*x])/(x^3*(c - a^2*c*x^2)^3),x]
Output:
(-1/2*1/x^2 - (2*a)/x + a^2/(12*(1 - a*x)^3) + a^2/(2*(1 - a*x)^2) + (39*a ^2)/(16*(1 - a*x)) + a^2/(16*(1 + a*x)) + 5*a^2*Log[x] - (75*a^2*Log[1 - a *x])/16 - (5*a^2*Log[1 + a*x])/16)/c^3
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 = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.70
method | result | size |
default | \(\frac {\frac {a^{2}}{16 a x +16}-\frac {5 a^{2} \ln \left (a x +1\right )}{16}-\frac {a^{2}}{12 \left (a x -1\right )^{3}}+\frac {a^{2}}{2 \left (a x -1\right )^{2}}-\frac {39 a^{2}}{16 \left (a x -1\right )}-\frac {75 a^{2} \ln \left (a x -1\right )}{16}-\frac {1}{2 x^{2}}-\frac {2 a}{x}+5 a^{2} \ln \left (x \right )}{c^{3}}\) | \(94\) |
risch | \(\frac {-\frac {35}{8} a^{5} x^{5}+\frac {25}{4} a^{4} x^{4}+\frac {85}{24} a^{3} x^{3}-\frac {85}{12} a^{2} x^{2}+a x +\frac {1}{2}}{c^{3} x^{2} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}-\frac {5 a^{2} \ln \left (a x +1\right )}{16 c^{3}}+\frac {5 a^{2} \ln \left (-x \right )}{c^{3}}-\frac {75 a^{2} \ln \left (-a x +1\right )}{16 c^{3}}\) | \(105\) |
norman | \(\frac {\frac {1}{2 c}+\frac {2 a x}{c}-\frac {77 a^{3} x^{3}}{8 c}+\frac {35 a^{5} x^{5}}{3 c}-\frac {35 a^{7} x^{7}}{8 c}-\frac {15 a^{4} x^{4}}{2 c}+\frac {45 a^{6} x^{6}}{4 c}-\frac {55 a^{8} x^{8}}{12 c}}{\left (a^{2} x^{2}-1\right )^{3} x^{2} c^{2}}+\frac {5 a^{2} \ln \left (x \right )}{c^{3}}-\frac {75 a^{2} \ln \left (a x -1\right )}{16 c^{3}}-\frac {5 a^{2} \ln \left (a x +1\right )}{16 c^{3}}\) | \(137\) |
parallelrisch | \(\frac {24-510 a^{3} x^{3}+48 a x +300 a^{4} x^{4}+470 a^{5} x^{5}-340 x^{6} a^{6}+480 \ln \left (x \right ) x^{3} a^{3}-240 a^{2} \ln \left (x \right ) x^{2}+15 \ln \left (a x +1\right ) x^{2} a^{2}-225 \ln \left (a x -1\right ) x^{6} a^{6}-15 \ln \left (a x +1\right ) x^{6} a^{6}-30 \ln \left (a x +1\right ) x^{3} a^{3}+240 a^{6} \ln \left (x \right ) x^{6}-450 a^{3} \ln \left (a x -1\right ) x^{3}+450 \ln \left (a x -1\right ) x^{5} a^{5}+30 \ln \left (a x +1\right ) x^{5} a^{5}+225 a^{2} \ln \left (a x -1\right ) x^{2}-480 a^{5} \ln \left (x \right ) x^{5}}{48 c^{3} x^{2} \left (a^{2} x^{2}-1\right ) \left (a x -1\right )^{2}}\) | \(217\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/c^3*(1/16*a^2/(a*x+1)-5/16*a^2*ln(a*x+1)-1/12*a^2/(a*x-1)^3+1/2*a^2/(a*x -1)^2-39/16*a^2/(a*x-1)-75/16*a^2*ln(a*x-1)-1/2/x^2-2*a/x+5*a^2*ln(x))
Time = 0.08 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.47 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx=-\frac {210 \, a^{5} x^{5} - 300 \, a^{4} x^{4} - 170 \, a^{3} x^{3} + 340 \, a^{2} x^{2} - 48 \, a x + 15 \, {\left (a^{6} x^{6} - 2 \, a^{5} x^{5} + 2 \, a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 225 \, {\left (a^{6} x^{6} - 2 \, a^{5} x^{5} + 2 \, a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 240 \, {\left (a^{6} x^{6} - 2 \, a^{5} x^{5} + 2 \, a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (x\right ) - 24}{48 \, {\left (a^{4} c^{3} x^{6} - 2 \, a^{3} c^{3} x^{5} + 2 \, a c^{3} x^{3} - c^{3} x^{2}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^3,x, algorithm="fricas ")
Output:
-1/48*(210*a^5*x^5 - 300*a^4*x^4 - 170*a^3*x^3 + 340*a^2*x^2 - 48*a*x + 15 *(a^6*x^6 - 2*a^5*x^5 + 2*a^3*x^3 - a^2*x^2)*log(a*x + 1) + 225*(a^6*x^6 - 2*a^5*x^5 + 2*a^3*x^3 - a^2*x^2)*log(a*x - 1) - 240*(a^6*x^6 - 2*a^5*x^5 + 2*a^3*x^3 - a^2*x^2)*log(x) - 24)/(a^4*c^3*x^6 - 2*a^3*c^3*x^5 + 2*a*c^3 *x^3 - c^3*x^2)
Time = 0.56 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx=\frac {- 105 a^{5} x^{5} + 150 a^{4} x^{4} + 85 a^{3} x^{3} - 170 a^{2} x^{2} + 24 a x + 12}{24 a^{4} c^{3} x^{6} - 48 a^{3} c^{3} x^{5} + 48 a c^{3} x^{3} - 24 c^{3} x^{2}} + \frac {5 a^{2} \log {\left (x \right )} - \frac {75 a^{2} \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {5 a^{2} \log {\left (x + \frac {1}{a} \right )}}{16}}{c^{3}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)/x**3/(-a**2*c*x**2+c)**3,x)
Output:
(-105*a**5*x**5 + 150*a**4*x**4 + 85*a**3*x**3 - 170*a**2*x**2 + 24*a*x + 12)/(24*a**4*c**3*x**6 - 48*a**3*c**3*x**5 + 48*a*c**3*x**3 - 24*c**3*x**2 ) + (5*a**2*log(x) - 75*a**2*log(x - 1/a)/16 - 5*a**2*log(x + 1/a)/16)/c** 3
Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx=-\frac {105 \, a^{5} x^{5} - 150 \, a^{4} x^{4} - 85 \, a^{3} x^{3} + 170 \, a^{2} x^{2} - 24 \, a x - 12}{24 \, {\left (a^{4} c^{3} x^{6} - 2 \, a^{3} c^{3} x^{5} + 2 \, a c^{3} x^{3} - c^{3} x^{2}\right )}} - \frac {5 \, a^{2} \log \left (a x + 1\right )}{16 \, c^{3}} - \frac {75 \, a^{2} \log \left (a x - 1\right )}{16 \, c^{3}} + \frac {5 \, a^{2} \log \left (x\right )}{c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^3,x, algorithm="maxima ")
Output:
-1/24*(105*a^5*x^5 - 150*a^4*x^4 - 85*a^3*x^3 + 170*a^2*x^2 - 24*a*x - 12) /(a^4*c^3*x^6 - 2*a^3*c^3*x^5 + 2*a*c^3*x^3 - c^3*x^2) - 5/16*a^2*log(a*x + 1)/c^3 - 75/16*a^2*log(a*x - 1)/c^3 + 5*a^2*log(x)/c^3
Time = 0.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx=-\frac {5 \, a^{2} \log \left ({\left | a x + 1 \right |}\right )}{16 \, c^{3}} - \frac {75 \, a^{2} \log \left ({\left | a x - 1 \right |}\right )}{16 \, c^{3}} + \frac {5 \, a^{2} \log \left ({\left | x \right |}\right )}{c^{3}} - \frac {105 \, a^{5} x^{5} - 150 \, a^{4} x^{4} - 85 \, a^{3} x^{3} + 170 \, a^{2} x^{2} - 24 \, a x - 12}{24 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} c^{3} x^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
-5/16*a^2*log(abs(a*x + 1))/c^3 - 75/16*a^2*log(abs(a*x - 1))/c^3 + 5*a^2* log(abs(x))/c^3 - 1/24*(105*a^5*x^5 - 150*a^4*x^4 - 85*a^3*x^3 + 170*a^2*x ^2 - 24*a*x - 12)/((a*x + 1)*(a*x - 1)^3*c^3*x^2)
Time = 23.62 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx=\frac {5\,a^2\,\ln \left (x\right )}{c^3}-\frac {-\frac {35\,a^5\,x^5}{8}+\frac {25\,a^4\,x^4}{4}+\frac {85\,a^3\,x^3}{24}-\frac {85\,a^2\,x^2}{12}+a\,x+\frac {1}{2}}{-a^4\,c^3\,x^6+2\,a^3\,c^3\,x^5-2\,a\,c^3\,x^3+c^3\,x^2}-\frac {75\,a^2\,\ln \left (a\,x-1\right )}{16\,c^3}-\frac {5\,a^2\,\ln \left (a\,x+1\right )}{16\,c^3} \] Input:
int(-(a*x + 1)^2/(x^3*(c - a^2*c*x^2)^3*(a^2*x^2 - 1)),x)
Output:
(5*a^2*log(x))/c^3 - (a*x - (85*a^2*x^2)/12 + (85*a^3*x^3)/24 + (25*a^4*x^ 4)/4 - (35*a^5*x^5)/8 + 1/2)/(c^3*x^2 - 2*a*c^3*x^3 + 2*a^3*c^3*x^5 - a^4* c^3*x^6) - (75*a^2*log(a*x - 1))/(16*c^3) - (5*a^2*log(a*x + 1))/(16*c^3)
Time = 0.15 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.65 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^3} \, dx=\frac {-225 \,\mathrm {log}\left (a x -1\right ) a^{6} x^{6}+450 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}-450 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+225 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-15 \,\mathrm {log}\left (a x +1\right ) a^{6} x^{6}+30 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}-30 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+15 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+240 \,\mathrm {log}\left (x \right ) a^{6} x^{6}-480 \,\mathrm {log}\left (x \right ) a^{5} x^{5}+480 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-240 \,\mathrm {log}\left (x \right ) a^{2} x^{2}-105 a^{6} x^{6}+300 a^{4} x^{4}-40 a^{3} x^{3}-235 a^{2} x^{2}+48 a x +24}{48 c^{3} x^{2} \left (a^{4} x^{4}-2 a^{3} x^{3}+2 a x -1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^3,x)
Output:
( - 225*log(a*x - 1)*a**6*x**6 + 450*log(a*x - 1)*a**5*x**5 - 450*log(a*x - 1)*a**3*x**3 + 225*log(a*x - 1)*a**2*x**2 - 15*log(a*x + 1)*a**6*x**6 + 30*log(a*x + 1)*a**5*x**5 - 30*log(a*x + 1)*a**3*x**3 + 15*log(a*x + 1)*a* *2*x**2 + 240*log(x)*a**6*x**6 - 480*log(x)*a**5*x**5 + 480*log(x)*a**3*x* *3 - 240*log(x)*a**2*x**2 - 105*a**6*x**6 + 300*a**4*x**4 - 40*a**3*x**3 - 235*a**2*x**2 + 48*a*x + 24)/(48*c**3*x**2*(a**4*x**4 - 2*a**3*x**3 + 2*a *x - 1))