Integrand size = 25, antiderivative size = 93 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx=\frac {1}{12 c^3 (1-a x)^3}+\frac {1}{4 c^3 (1-a x)^2}+\frac {11}{16 c^3 (1-a x)}+\frac {1}{16 c^3 (1+a x)}+\frac {\log (x)}{c^3}-\frac {13 \log (1-a x)}{16 c^3}-\frac {3 \log (1+a x)}{16 c^3} \] Output:
1/12/c^3/(-a*x+1)^3+1/4/c^3/(-a*x+1)^2+11/16/c^3/(-a*x+1)+1/16/c^3/(a*x+1) +ln(x)/c^3-13/16*ln(-a*x+1)/c^3-3/16*ln(a*x+1)/c^3
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.71 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx=\frac {\frac {33}{1-a x}-\frac {4}{(-1+a x)^3}+\frac {12}{(-1+a x)^2}+\frac {3}{1+a x}+48 \log (x)-39 \log (1-a x)-9 \log (1+a x)}{48 c^3} \] Input:
Integrate[E^(2*ArcTanh[a*x])/(x*(c - a^2*c*x^2)^3),x]
Output:
(33/(1 - a*x) - 4/(-1 + a*x)^3 + 12/(-1 + a*x)^2 + 3/(1 + a*x) + 48*Log[x] - 39*Log[1 - a*x] - 9*Log[1 + a*x])/(48*c^3)
Time = 0.53 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.81, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int \frac {1}{x (1-a x)^4 (a x+1)^2}dx}{c^3}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {13 a}{16 (a x-1)}-\frac {3 a}{16 (a x+1)}+\frac {11 a}{16 (a x-1)^2}-\frac {a}{16 (a x+1)^2}-\frac {a}{2 (a x-1)^3}+\frac {a}{4 (a x-1)^4}+\frac {1}{x}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {11}{16 (1-a x)}+\frac {1}{16 (a x+1)}+\frac {1}{4 (1-a x)^2}+\frac {1}{12 (1-a x)^3}-\frac {13}{16} \log (1-a x)-\frac {3}{16} \log (a x+1)+\log (x)}{c^3}\) |
Input:
Int[E^(2*ArcTanh[a*x])/(x*(c - a^2*c*x^2)^3),x]
Output:
(1/(12*(1 - a*x)^3) + 1/(4*(1 - a*x)^2) + 11/(16*(1 - a*x)) + 1/(16*(1 + a *x)) + Log[x] - (13*Log[1 - a*x])/16 - (3*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 = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65
method | result | size |
default | \(\frac {\frac {1}{16 a x +16}-\frac {3 \ln \left (a x +1\right )}{16}-\frac {1}{12 \left (a x -1\right )^{3}}+\frac {1}{4 \left (a x -1\right )^{2}}-\frac {11}{16 \left (a x -1\right )}-\frac {13 \ln \left (a x -1\right )}{16}+\ln \left (x \right )}{c^{3}}\) | \(60\) |
risch | \(\frac {-\frac {5}{8} a^{3} x^{3}+\frac {3}{4} a^{2} x^{2}+\frac {19}{24} a x -\frac {13}{12}}{c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}+\frac {\ln \left (-x \right )}{c^{3}}-\frac {13 \ln \left (-a x +1\right )}{16 c^{3}}-\frac {3 \ln \left (a x +1\right )}{16 c^{3}}\) | \(77\) |
norman | \(\frac {-\frac {11 a x}{8 c}+\frac {5 a^{3} x^{3}}{3 c}-\frac {5 a^{5} x^{5}}{8 c}-\frac {2 a^{2} x^{2}}{c}-\frac {13 a^{6} x^{6}}{12 c}+\frac {11 a^{4} x^{4}}{4 c}}{\left (a^{2} x^{2}-1\right )^{3} c^{2}}+\frac {\ln \left (x \right )}{c^{3}}-\frac {13 \ln \left (a x -1\right )}{16 c^{3}}-\frac {3 \ln \left (a x +1\right )}{16 c^{3}}\) | \(108\) |
parallelrisch | \(\frac {48 \ln \left (x \right ) x^{4} a^{4}-39 \ln \left (a x -1\right ) x^{4} a^{4}-9 \ln \left (a x +1\right ) x^{4} a^{4}-52 a^{4} x^{4}-96 \ln \left (x \right ) x^{3} a^{3}+78 a^{3} \ln \left (a x -1\right ) x^{3}+18 \ln \left (a x +1\right ) x^{3} a^{3}+74 a^{3} x^{3}+36 a^{2} x^{2}+96 a \ln \left (x \right ) x -78 a \ln \left (a x -1\right ) x -18 \ln \left (a x +1\right ) x a -66 a x -48 \ln \left (x \right )+39 \ln \left (a x -1\right )+9 \ln \left (a x +1\right )}{48 c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}\) | \(175\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/c^3*(1/16/(a*x+1)-3/16*ln(a*x+1)-1/12/(a*x-1)^3+1/4/(a*x-1)^2-11/16/(a*x -1)-13/16*ln(a*x-1)+ln(x))
Time = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.54 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx=-\frac {30 \, a^{3} x^{3} - 36 \, a^{2} x^{2} - 38 \, a x + 9 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 39 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) - 48 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (x\right ) + 52}{48 \, {\left (a^{4} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{3} + 2 \, a c^{3} x - c^{3}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^3,x, algorithm="fricas")
Output:
-1/48*(30*a^3*x^3 - 36*a^2*x^2 - 38*a*x + 9*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x + 1) + 39*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x - 1) - 48* (a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(x) + 52)/(a^4*c^3*x^4 - 2*a^3*c^3*x^ 3 + 2*a*c^3*x - c^3)
Time = 0.39 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx=\frac {- 15 a^{3} x^{3} + 18 a^{2} x^{2} + 19 a x - 26}{24 a^{4} c^{3} x^{4} - 48 a^{3} c^{3} x^{3} + 48 a c^{3} x - 24 c^{3}} + \frac {\log {\left (x \right )} - \frac {13 \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {3 \log {\left (x + \frac {1}{a} \right )}}{16}}{c^{3}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)/x/(-a**2*c*x**2+c)**3,x)
Output:
(-15*a**3*x**3 + 18*a**2*x**2 + 19*a*x - 26)/(24*a**4*c**3*x**4 - 48*a**3* c**3*x**3 + 48*a*c**3*x - 24*c**3) + (log(x) - 13*log(x - 1/a)/16 - 3*log( x + 1/a)/16)/c**3
Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.96 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx=-\frac {15 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 19 \, a x + 26}{24 \, {\left (a^{4} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{3} + 2 \, a c^{3} x - c^{3}\right )}} - \frac {3 \, \log \left (a x + 1\right )}{16 \, c^{3}} - \frac {13 \, \log \left (a x - 1\right )}{16 \, c^{3}} + \frac {\log \left (x\right )}{c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^3,x, algorithm="maxima")
Output:
-1/24*(15*a^3*x^3 - 18*a^2*x^2 - 19*a*x + 26)/(a^4*c^3*x^4 - 2*a^3*c^3*x^3 + 2*a*c^3*x - c^3) - 3/16*log(a*x + 1)/c^3 - 13/16*log(a*x - 1)/c^3 + log (x)/c^3
Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx=-\frac {3 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, c^{3}} - \frac {13 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, c^{3}} + \frac {\log \left ({\left | x \right |}\right )}{c^{3}} - \frac {15 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 19 \, a x + 26}{24 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
-3/16*log(abs(a*x + 1))/c^3 - 13/16*log(abs(a*x - 1))/c^3 + log(abs(x))/c^ 3 - 1/24*(15*a^3*x^3 - 18*a^2*x^2 - 19*a*x + 26)/((a*x + 1)*(a*x - 1)^3*c^ 3)
Time = 23.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx=\frac {\ln \left (x\right )}{c^3}-\frac {13\,\ln \left (a\,x-1\right )}{16\,c^3}-\frac {3\,\ln \left (a\,x+1\right )}{16\,c^3}-\frac {-\frac {5\,a^3\,x^3}{8}+\frac {3\,a^2\,x^2}{4}+\frac {19\,a\,x}{24}-\frac {13}{12}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3} \] Input:
int(-(a*x + 1)^2/(x*(c - a^2*c*x^2)^3*(a^2*x^2 - 1)),x)
Output:
log(x)/c^3 - (13*log(a*x - 1))/(16*c^3) - (3*log(a*x + 1))/(16*c^3) - ((19 *a*x)/24 + (3*a^2*x^2)/4 - (5*a^3*x^3)/8 - 13/12)/(c^3 + 2*a^3*c^3*x^3 - a ^4*c^3*x^4 - 2*a*c^3*x)
Time = 0.15 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.85 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx=\frac {-39 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+78 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-78 \,\mathrm {log}\left (a x -1\right ) a x +39 \,\mathrm {log}\left (a x -1\right )-9 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+18 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-18 \,\mathrm {log}\left (a x +1\right ) a x +9 \,\mathrm {log}\left (a x +1\right )+48 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-96 \,\mathrm {log}\left (x \right ) a^{3} x^{3}+96 \,\mathrm {log}\left (x \right ) a x -48 \,\mathrm {log}\left (x \right )-15 a^{4} x^{4}+36 a^{2} x^{2}+8 a x -37}{48 c^{3} \left (a^{4} x^{4}-2 a^{3} x^{3}+2 a x -1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)/x/(-a^2*c*x^2+c)^3,x)
Output:
( - 39*log(a*x - 1)*a**4*x**4 + 78*log(a*x - 1)*a**3*x**3 - 78*log(a*x - 1 )*a*x + 39*log(a*x - 1) - 9*log(a*x + 1)*a**4*x**4 + 18*log(a*x + 1)*a**3* x**3 - 18*log(a*x + 1)*a*x + 9*log(a*x + 1) + 48*log(x)*a**4*x**4 - 96*log (x)*a**3*x**3 + 96*log(x)*a*x - 48*log(x) - 15*a**4*x**4 + 36*a**2*x**2 + 8*a*x - 37)/(48*c**3*(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 1))