Integrand size = 25, antiderivative size = 79 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {x}{a^4 c^2}+\frac {1}{4 a^5 c^2 (1-a x)^2}-\frac {7}{4 a^5 c^2 (1-a x)}-\frac {17 \log (1-a x)}{8 a^5 c^2}+\frac {\log (1+a x)}{8 a^5 c^2} \] Output:
-x/a^4/c^2+1/4/a^5/c^2/(-a*x+1)^2-7/4/a^5/c^2/(-a*x+1)-17/8*ln(-a*x+1)/a^5 /c^2+1/8*ln(a*x+1)/a^5/c^2
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.86 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {-\frac {x}{a^4}+\frac {1}{4 a^5 (1-a x)^2}-\frac {7}{4 a^5 (1-a x)}-\frac {17 \log (1-a x)}{8 a^5}+\frac {\log (1+a x)}{8 a^5}}{c^2} \] Input:
Integrate[(E^(2*ArcTanh[a*x])*x^4)/(c - a^2*c*x^2)^2,x]
Output:
(-(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
Time = 0.35 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.86, 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.
\(\displaystyle \int \frac {x^4 e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int \frac {x^4}{(1-a x)^3 (a x+1)}dx}{c^2}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {1}{8 a^4 (a x+1)}-\frac {1}{a^4}-\frac {17}{8 a^4 (a x-1)}-\frac {7}{4 a^4 (a x-1)^2}-\frac {1}{2 a^4 (a x-1)^3}\right )dx}{c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {7}{4 a^5 (1-a x)}+\frac {1}{4 a^5 (1-a x)^2}-\frac {17 \log (1-a x)}{8 a^5}+\frac {\log (a x+1)}{8 a^5}-\frac {x}{a^4}}{c^2}\) |
Input:
Int[(E^(2*ArcTanh[a*x])*x^4)/(c - a^2*c*x^2)^2,x]
Output:
(-(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])
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73
method | result | size |
default | \(\frac {-\frac {x}{a^{4}}+\frac {\ln \left (a x +1\right )}{8 a^{5}}+\frac {7}{4 a^{5} \left (a x -1\right )}-\frac {17 \ln \left (a x -1\right )}{8 a^{5}}+\frac {1}{4 \left (a x -1\right )^{2} a^{5}}}{c^{2}}\) | \(58\) |
risch | \(-\frac {x}{a^{4} c^{2}}+\frac {\frac {7 c^{2} x}{4}-\frac {3 c^{2}}{2 a}}{a^{4} c^{4} \left (a x -1\right )^{2}}-\frac {17 \ln \left (a x -1\right )}{8 c^{2} a^{5}}+\frac {\ln \left (-a x -1\right )}{8 a^{5} c^{2}}\) | \(69\) |
parallelrisch | \(\frac {-28-8 a^{3} x^{3}-17 a^{2} \ln \left (a x -1\right ) x^{2}+\ln \left (a x +1\right ) x^{2} a^{2}+34 a \ln \left (a x -1\right ) x -2 \ln \left (a x +1\right ) x a +38 a x -17 \ln \left (a x -1\right )+\ln \left (a x +1\right )}{8 a^{5} c^{2} \left (a x -1\right )^{2}}\) | \(91\) |
norman | \(\frac {-\frac {x^{5}}{c}-\frac {9 x}{4 a^{4} c}+\frac {15 x^{3}}{4 a^{2} c}-\frac {3}{2 c \,a^{5}}+\frac {2 x^{2}}{a^{3} c}}{\left (a^{2} x^{2}-1\right )^{2} c}-\frac {17 \ln \left (a x -1\right )}{8 c^{2} a^{5}}+\frac {\ln \left (a x +1\right )}{8 a^{5} c^{2}}\) | \(93\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)
Output:
1/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* x-1)^2/a^5)
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {8 \, a^{3} x^{3} - 16 \, a^{2} x^{2} - 6 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 17 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 12}{8 \, {\left (a^{7} c^{2} x^{2} - 2 \, a^{6} c^{2} x + a^{5} c^{2}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^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^7*c^2*x^2 - 2*a^6*c^2*x + a^5*c^2)
Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=- \frac {- 7 a x + 6}{4 a^{7} c^{2} x^{2} - 8 a^{6} c^{2} x + 4 a^{5} c^{2}} - \frac {x}{a^{4} c^{2}} - \frac {\frac {17 \log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a^{5} c^{2}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*x**4/(-a**2*c*x**2+c)**2,x)
Output:
-(-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 = 75, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {7 \, a x - 6}{4 \, {\left (a^{7} c^{2} x^{2} - 2 \, a^{6} c^{2} x + a^{5} c^{2}\right )}} - \frac {x}{a^{4} c^{2}} + \frac {\log \left (a x + 1\right )}{8 \, a^{5} c^{2}} - \frac {17 \, \log \left (a x - 1\right )}{8 \, a^{5} c^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^2,x, algorithm="maxima ")
Output:
1/4*(7*a*x - 6)/(a^7*c^2*x^2 - 2*a^6*c^2*x + a^5*c^2) - x/(a^4*c^2) + 1/8* log(a*x + 1)/(a^5*c^2) - 17/8*log(a*x - 1)/(a^5*c^2)
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.77 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {x}{a^{4} c^{2}} + \frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a^{5} c^{2}} - \frac {17 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, a^{5} c^{2}} + \frac {7 \, a x - 6}{4 \, {\left (a x - 1\right )}^{2} a^{5} c^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^2,x, algorithm="giac")
Output:
-x/(a^4*c^2) + 1/8*log(abs(a*x + 1))/(a^5*c^2) - 17/8*log(abs(a*x - 1))/(a ^5*c^2) + 1/4*(7*a*x - 6)/((a*x - 1)^2*a^5*c^2)
Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {\frac {7\,x}{4}-\frac {3}{2\,a}}{a^6\,c^2\,x^2-2\,a^5\,c^2\,x+a^4\,c^2}-\frac {x}{a^4\,c^2}-\frac {17\,\ln \left (a\,x-1\right )}{8\,a^5\,c^2}+\frac {\ln \left (a\,x+1\right )}{8\,a^5\,c^2} \] Input:
int(-(x^4*(a*x + 1)^2)/((c - a^2*c*x^2)^2*(a^2*x^2 - 1)),x)
Output:
((7*x)/4 - 3/(2*a))/(a^4*c^2 - 2*a^5*c^2*x + a^6*c^2*x^2) - x/(a^4*c^2) - (17*log(a*x - 1))/(8*a^5*c^2) + log(a*x + 1)/(8*a^5*c^2)
Time = 0.16 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.29 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^4}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {-17 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+34 \,\mathrm {log}\left (a x -1\right ) a x -17 \,\mathrm {log}\left (a x -1\right )+\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-2 \,\mathrm {log}\left (a x +1\right ) a x +\mathrm {log}\left (a x +1\right )-8 a^{3} x^{3}+19 a^{2} x^{2}-9}{8 a^{5} c^{2} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^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**5*c**2*(a**2*x**2 - 2*a*x + 1))