Integrand size = 22, antiderivative size = 76 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=-\frac {x}{c^2}+\frac {1}{4 a c^2 (1-a x)^2}-\frac {7}{4 a c^2 (1-a x)}-\frac {17 \log (1-a x)}{8 a c^2}+\frac {\log (1+a x)}{8 a c^2} \] Output:
-x/c^2+1/4/a/c^2/(-a*x+1)^2-7/4/a/c^2/(-a*x+1)-17/8*ln(-a*x+1)/a/c^2+1/8*l n(a*x+1)/a/c^2
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {-12+6 a x+16 a^2 x^2-8 a^3 x^3-17 (-1+a x)^2 \log (1-a x)+(-1+a x)^2 \log (1+a x)}{8 a c^2 (-1+a x)^2} \] Input:
Integrate[E^(2*ArcTanh[a*x])/(c - c/(a^2*x^2))^2,x]
Output:
(-12 + 6*a*x + 16*a^2*x^2 - 8*a^3*x^3 - 17*(-1 + a*x)^2*Log[1 - a*x] + (-1 + a*x)^2*Log[1 + a*x])/(8*a*c^2*(-1 + a*x)^2)
Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6707, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle \frac {a^4 \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^2}dx}{c^2}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {a^4 \int \frac {x^4}{(1-a x)^3 (a x+1)}dx}{c^2}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^4 \int \left (\frac {1}{8 a^4 (a x+1)}-\frac {1}{a^4}-\frac {17}{8 a^4 (a x-1)}-\frac {7}{4 a^4 (a x-1)^2}-\frac {1}{2 a^4 (a x-1)^3}\right )dx}{c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^4 \left (-\frac {7}{4 a^5 (1-a x)}+\frac {1}{4 a^5 (1-a x)^2}-\frac {17 \log (1-a x)}{8 a^5}+\frac {\log (a x+1)}{8 a^5}-\frac {x}{a^4}\right )}{c^2}\) |
Input:
Int[E^(2*ArcTanh[a*x])/(c - c/(a^2*x^2))^2,x]
Output:
(a^4*(-(x/a^4) + 1/(4*a^5*(1 - a*x)^2) - 7/(4*a^5*(1 - a*x)) - (17*Log[1 - a*x])/(8*a^5) + Log[1 + a*x]/(8*a^5)))/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])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {a^{4} \left (-\frac {x}{a^{4}}+\frac {\ln \left (a x +1\right )}{8 a^{5}}+\frac {7}{4 a^{5} \left (a x -1\right )}-\frac {17 \ln \left (a x -1\right )}{8 a^{5}}+\frac {1}{4 a^{5} \left (a x -1\right )^{2}}\right )}{c^{2}}\) | \(61\) |
risch | \(-\frac {x}{c^{2}}+\frac {\frac {7 c^{2} x}{4}-\frac {3 c^{2}}{2 a}}{c^{4} \left (a x -1\right )^{2}}-\frac {17 \ln \left (a x -1\right )}{8 a \,c^{2}}+\frac {\ln \left (-a x -1\right )}{8 a \,c^{2}}\) | \(63\) |
norman | \(\frac {-\frac {9 x}{4 c}-\frac {a^{4} x^{5}}{c}+\frac {15 a^{2} x^{3}}{4 c}-\frac {a \,x^{2}}{c}+\frac {3 a^{3} x^{4}}{2 c}}{\left (a^{2} x^{2}-1\right )^{2} c}-\frac {17 \ln \left (a x -1\right )}{8 a \,c^{2}}+\frac {\ln \left (a x +1\right )}{8 a \,c^{2}}\) | \(94\) |
parallelrisch | \(\frac {-8 a^{3} x^{3}-17 a^{2} \ln \left (a x -1\right ) x^{2}+\ln \left (a x +1\right ) x^{2} a^{2}+28 a^{2} x^{2}+34 a \ln \left (a x -1\right ) x -2 \ln \left (a x +1\right ) x a -18 a x -17 \ln \left (a x -1\right )+\ln \left (a x +1\right )}{8 c^{2} \left (a x -1\right )^{2} a}\) | \(98\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)/(c-c/a^2/x^2)^2,x,method=_RETURNVERBOSE)
Output:
a^4/c^2*(-x/a^4+1/8*ln(a*x+1)/a^5+7/4/a^5/(a*x-1)-17/8/a^5*ln(a*x-1)+1/4/a ^5/(a*x-1)^2)
Time = 0.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.22 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=-\frac {8 \, a^{3} x^{3} - 16 \, a^{2} x^{2} - 6 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 17 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 12}{8 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/(c-c/a^2/x^2)^2,x, algorithm="fricas")
Output:
-1/8*(8*a^3*x^3 - 16*a^2*x^2 - 6*a*x - (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + 17*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) + 12)/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2)
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=- a^{4} \left (\frac {- 7 a x + 6}{4 a^{7} c^{2} x^{2} - 8 a^{6} c^{2} x + 4 a^{5} c^{2}} + \frac {x}{a^{4} c^{2}} + \frac {\frac {17 \log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a^{5} c^{2}}\right ) \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)/(c-c/a**2/x**2)**2,x)
Output:
-a**4*((-7*a*x + 6)/(4*a**7*c**2*x**2 - 8*a**6*c**2*x + 4*a**5*c**2) + x/( a**4*c**2) + (17*log(x - 1/a)/8 - log(x + 1/a)/8)/(a**5*c**2))
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {7 \, a x - 6}{4 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} - \frac {x}{c^{2}} + \frac {\log \left (a x + 1\right )}{8 \, a c^{2}} - \frac {17 \, \log \left (a x - 1\right )}{8 \, a c^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/(c-c/a^2/x^2)^2,x, algorithm="maxima")
Output:
1/4*(7*a*x - 6)/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2) - x/c^2 + 1/8*log(a*x + 1)/(a*c^2) - 17/8*log(a*x - 1)/(a*c^2)
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=-\frac {x}{c^{2}} + \frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{2}} - \frac {17 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{2}} + \frac {7 \, a x - 6}{4 \, {\left (a x - 1\right )}^{2} a c^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/(c-c/a^2/x^2)^2,x, algorithm="giac")
Output:
-x/c^2 + 1/8*log(abs(a*x + 1))/(a*c^2) - 17/8*log(abs(a*x - 1))/(a*c^2) + 1/4*(7*a*x - 6)/((a*x - 1)^2*a*c^2)
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {\frac {7\,x}{4}-\frac {3}{2\,a}}{a^2\,c^2\,x^2-2\,a\,c^2\,x+c^2}-\frac {x}{c^2}-\frac {17\,\ln \left (a\,x-1\right )}{8\,a\,c^2}+\frac {\ln \left (a\,x+1\right )}{8\,a\,c^2} \] Input:
int(-(a*x + 1)^2/((c - c/(a^2*x^2))^2*(a^2*x^2 - 1)),x)
Output:
((7*x)/4 - 3/(2*a))/(c^2 + a^2*c^2*x^2 - 2*a*c^2*x) - x/c^2 - (17*log(a*x - 1))/(8*a*c^2) + log(a*x + 1)/(8*a*c^2)
Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.34 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {-17 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+34 \,\mathrm {log}\left (a x -1\right ) a x -17 \,\mathrm {log}\left (a x -1\right )+\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-2 \,\mathrm {log}\left (a x +1\right ) a x +\mathrm {log}\left (a x +1\right )-8 a^{3} x^{3}+19 a^{2} x^{2}-9}{8 a \,c^{2} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)/(c-c/a^2/x^2)^2,x)
Output:
( - 17*log(a*x - 1)*a**2*x**2 + 34*log(a*x - 1)*a*x - 17*log(a*x - 1) + lo g(a*x + 1)*a**2*x**2 - 2*log(a*x + 1)*a*x + log(a*x + 1) - 8*a**3*x**3 + 1 9*a**2*x**2 - 9)/(8*a*c**2*(a**2*x**2 - 2*a*x + 1))