Integrand size = 25, antiderivative size = 78 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {1}{c^2 x}+\frac {a}{4 c^2 (1-a x)^2}+\frac {5 a}{4 c^2 (1-a x)}+\frac {2 a \log (x)}{c^2}-\frac {17 a \log (1-a x)}{8 c^2}+\frac {a \log (1+a x)}{8 c^2} \] Output:
-1/c^2/x+1/4*a/c^2/(-a*x+1)^2+5/4*a/c^2/(-a*x+1)+2*a*ln(x)/c^2-17/8*a*ln(- a*x+1)/c^2+1/8*a*ln(a*x+1)/c^2
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.73 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx=\frac {-\frac {8}{x}+\frac {10 a}{1-a x}+\frac {2 a}{(-1+a x)^2}+16 a \log (x)-17 a \log (1-a x)+a \log (1+a x)}{8 c^2} \] Input:
Integrate[E^(2*ArcTanh[a*x])/(x^2*(c - a^2*c*x^2)^2),x]
Output:
(-8/x + (10*a)/(1 - a*x) + (2*a)/(-1 + a*x)^2 + 16*a*Log[x] - 17*a*Log[1 - a*x] + a*Log[1 + a*x])/(8*c^2)
Time = 0.34 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int \frac {1}{x^2 (1-a x)^3 (a x+1)}dx}{c^2}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {17 a^2}{8 (a x-1)}+\frac {a^2}{8 (a x+1)}+\frac {5 a^2}{4 (a x-1)^2}-\frac {a^2}{2 (a x-1)^3}+\frac {2 a}{x}+\frac {1}{x^2}\right )dx}{c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {5 a}{4 (1-a x)}+\frac {a}{4 (1-a x)^2}+2 a \log (x)-\frac {17}{8} a \log (1-a x)+\frac {1}{8} a \log (a x+1)-\frac {1}{x}}{c^2}\) |
Input:
Int[E^(2*ArcTanh[a*x])/(x^2*(c - a^2*c*x^2)^2),x]
Output:
(-x^(-1) + a/(4*(1 - a*x)^2) + (5*a)/(4*(1 - a*x)) + 2*a*Log[x] - (17*a*Lo g[1 - a*x])/8 + (a*Log[1 + a*x])/8)/c^2
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 = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.69
method | result | size |
default | \(\frac {\frac {a \ln \left (a x +1\right )}{8}+\frac {a}{4 \left (a x -1\right )^{2}}-\frac {5 a}{4 \left (a x -1\right )}-\frac {17 a \ln \left (a x -1\right )}{8}-\frac {1}{x}+2 \ln \left (x \right ) a}{c^{2}}\) | \(54\) |
risch | \(\frac {-\frac {9}{4} a^{2} x^{2}+\frac {7}{2} a x -1}{c^{2} x \left (a x -1\right )^{2}}+\frac {a \ln \left (a x +1\right )}{8 c^{2}}-\frac {17 a \ln \left (-a x +1\right )}{8 c^{2}}+\frac {2 a \ln \left (-x \right )}{c^{2}}\) | \(65\) |
norman | \(\frac {-\frac {1}{c}+\frac {15 a^{2} x^{2}}{4 c}-\frac {9 a^{4} x^{4}}{4 c}+\frac {2 a^{3} x^{3}}{c}-\frac {3 a^{5} x^{5}}{2 c}}{\left (a^{2} x^{2}-1\right )^{2} x c}+\frac {2 a \ln \left (x \right )}{c^{2}}-\frac {17 a \ln \left (a x -1\right )}{8 c^{2}}+\frac {a \ln \left (a x +1\right )}{8 c^{2}}\) | \(102\) |
parallelrisch | \(\frac {-8+16 \ln \left (x \right ) x^{3} a^{3}-17 a^{3} \ln \left (a x -1\right ) x^{3}+\ln \left (a x +1\right ) x^{3} a^{3}-28 a^{3} x^{3}-32 a^{2} \ln \left (x \right ) x^{2}+34 a^{2} \ln \left (a x -1\right ) x^{2}-2 \ln \left (a x +1\right ) x^{2} a^{2}+38 a^{2} x^{2}+16 a \ln \left (x \right ) x -17 a \ln \left (a x -1\right ) x +\ln \left (a x +1\right ) x a}{8 c^{2} x \left (a x -1\right )^{2}}\) | \(134\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)/x^2/(-a^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)
Output:
1/c^2*(1/8*a*ln(a*x+1)+1/4*a/(a*x-1)^2-5/4*a/(a*x-1)-17/8*a*ln(a*x-1)-1/x+ 2*ln(x)*a)
Time = 0.07 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.54 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {18 \, a^{2} x^{2} - 28 \, a x - {\left (a^{3} x^{3} - 2 \, a^{2} x^{2} + a x\right )} \log \left (a x + 1\right ) + 17 \, {\left (a^{3} x^{3} - 2 \, a^{2} x^{2} + a x\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{3} x^{3} - 2 \, a^{2} x^{2} + a x\right )} \log \left (x\right ) + 8}{8 \, {\left (a^{2} c^{2} x^{3} - 2 \, a c^{2} x^{2} + c^{2} x\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^2/(-a^2*c*x^2+c)^2,x, algorithm="fricas ")
Output:
-1/8*(18*a^2*x^2 - 28*a*x - (a^3*x^3 - 2*a^2*x^2 + a*x)*log(a*x + 1) + 17* (a^3*x^3 - 2*a^2*x^2 + a*x)*log(a*x - 1) - 16*(a^3*x^3 - 2*a^2*x^2 + a*x)* log(x) + 8)/(a^2*c^2*x^3 - 2*a*c^2*x^2 + c^2*x)
Time = 0.50 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx=- \frac {9 a^{2} x^{2} - 14 a x + 4}{4 a^{2} c^{2} x^{3} - 8 a c^{2} x^{2} + 4 c^{2} x} - \frac {- 2 a \log {\left (x \right )} + \frac {17 a \log {\left (x - \frac {1}{a} \right )}}{8} - \frac {a \log {\left (x + \frac {1}{a} \right )}}{8}}{c^{2}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)/x**2/(-a**2*c*x**2+c)**2,x)
Output:
-(9*a**2*x**2 - 14*a*x + 4)/(4*a**2*c**2*x**3 - 8*a*c**2*x**2 + 4*c**2*x) - (-2*a*log(x) + 17*a*log(x - 1/a)/8 - a*log(x + 1/a)/8)/c**2
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {9 \, a^{2} x^{2} - 14 \, a x + 4}{4 \, {\left (a^{2} c^{2} x^{3} - 2 \, a c^{2} x^{2} + c^{2} x\right )}} + \frac {a \log \left (a x + 1\right )}{8 \, c^{2}} - \frac {17 \, a \log \left (a x - 1\right )}{8 \, c^{2}} + \frac {2 \, a \log \left (x\right )}{c^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^2/(-a^2*c*x^2+c)^2,x, algorithm="maxima ")
Output:
-1/4*(9*a^2*x^2 - 14*a*x + 4)/(a^2*c^2*x^3 - 2*a*c^2*x^2 + c^2*x) + 1/8*a* log(a*x + 1)/c^2 - 17/8*a*log(a*x - 1)/c^2 + 2*a*log(x)/c^2
Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx=\frac {a \log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac {17 \, a \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac {2 \, a \log \left ({\left | x \right |}\right )}{c^{2}} - \frac {9 \, a^{2} x^{2} - 14 \, a x + 4}{4 \, {\left (a x - 1\right )}^{2} c^{2} x} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^2/(-a^2*c*x^2+c)^2,x, algorithm="giac")
Output:
1/8*a*log(abs(a*x + 1))/c^2 - 17/8*a*log(abs(a*x - 1))/c^2 + 2*a*log(abs(x ))/c^2 - 1/4*(9*a^2*x^2 - 14*a*x + 4)/((a*x - 1)^2*c^2*x)
Time = 25.71 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx=\frac {2\,a\,\ln \left (x\right )}{c^2}-\frac {\frac {9\,a^2\,x^2}{4}-\frac {7\,a\,x}{2}+1}{a^2\,c^2\,x^3-2\,a\,c^2\,x^2+c^2\,x}-\frac {17\,a\,\ln \left (a\,x-1\right )}{8\,c^2}+\frac {a\,\ln \left (a\,x+1\right )}{8\,c^2} \] Input:
int(-(a*x + 1)^2/(x^2*(c - a^2*c*x^2)^2*(a^2*x^2 - 1)),x)
Output:
(2*a*log(x))/c^2 - ((9*a^2*x^2)/4 - (7*a*x)/2 + 1)/(c^2*x - 2*a*c^2*x^2 + a^2*c^2*x^3) - (17*a*log(a*x - 1))/(8*c^2) + (a*log(a*x + 1))/(8*c^2)
Time = 0.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.76 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^2} \, dx=\frac {-17 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+34 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-17 \,\mathrm {log}\left (a x -1\right ) a x +\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-2 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+\mathrm {log}\left (a x +1\right ) a x +16 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-32 \,\mathrm {log}\left (x \right ) a^{2} x^{2}+16 \,\mathrm {log}\left (x \right ) a x -9 a^{3} x^{3}+19 a x -8}{8 c^{2} x \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)/x^2/(-a^2*c*x^2+c)^2,x)
Output:
( - 17*log(a*x - 1)*a**3*x**3 + 34*log(a*x - 1)*a**2*x**2 - 17*log(a*x - 1 )*a*x + log(a*x + 1)*a**3*x**3 - 2*log(a*x + 1)*a**2*x**2 + log(a*x + 1)*a *x + 16*log(x)*a**3*x**3 - 32*log(x)*a**2*x**2 + 16*log(x)*a*x - 9*a**3*x* *3 + 19*a*x - 8)/(8*c**2*x*(a**2*x**2 - 2*a*x + 1))