Integrand size = 24, antiderivative size = 72 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {1}{8 a^5 (1-a x)^2}-\frac {3}{4 a^5 (1-a x)}-\frac {1}{8 a^5 (1+a x)}-\frac {11 \log (1-a x)}{16 a^5}-\frac {5 \log (1+a x)}{16 a^5} \] Output:
1/8/a^5/(-a*x+1)^2-3/4/a^5/(-a*x+1)-1/8/a^5/(a*x+1)-11/16*ln(-a*x+1)/a^5-5 /16*ln(a*x+1)/a^5
Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {\frac {2 \left (-6+3 a x+5 a^2 x^2\right )}{(-1+a x)^2 (1+a x)}-11 \log (1-a x)-5 \log (1+a x)}{16 a^5} \] Input:
Integrate[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(5/2),x]
Output:
((2*(-6 + 3*a*x + 5*a^2*x^2))/((-1 + a*x)^2*(1 + a*x)) - 11*Log[1 - a*x] - 5*Log[1 + a*x])/(16*a^5)
Time = 0.55 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.
\(\displaystyle \int \frac {x^4 e^{\text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \int \frac {x^4}{(1-a x)^3 (a x+1)^2}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {5}{16 a^4 (a x+1)}+\frac {1}{8 a^4 (a x+1)^2}-\frac {11}{16 a^4 (a x-1)}-\frac {3}{4 a^4 (a x-1)^2}-\frac {1}{4 a^4 (a x-1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{4 a^5 (1-a x)}-\frac {1}{8 a^5 (a x+1)}+\frac {1}{8 a^5 (1-a x)^2}-\frac {11 \log (1-a x)}{16 a^5}-\frac {5 \log (a x+1)}{16 a^5}\) |
Input:
Int[(E^ArcTanh[a*x]*x^4)/(1 - a^2*x^2)^(5/2),x]
Output:
1/(8*a^5*(1 - a*x)^2) - 3/(4*a^5*(1 - a*x)) - 1/(8*a^5*(1 + a*x)) - (11*Lo g[1 - a*x])/(16*a^5) - (5*Log[1 + a*x])/(16*a^5)
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 = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {1}{8 a^{5} \left (a x +1\right )}-\frac {5 \ln \left (a x +1\right )}{16 a^{5}}+\frac {1}{8 a^{5} \left (a x -1\right )^{2}}+\frac {3}{4 a^{5} \left (a x -1\right )}-\frac {11 \ln \left (a x -1\right )}{16 a^{5}}\) | \(60\) |
norman | \(\frac {-\frac {3 x}{8 a^{4}}+\frac {5 x^{3}}{8 a^{2}}-\frac {1}{4 a^{5}}+\frac {x^{4}}{2 a}}{\left (a^{2} x^{2}-1\right )^{2}}-\frac {11 \ln \left (a x -1\right )}{16 a^{5}}-\frac {5 \ln \left (a x +1\right )}{16 a^{5}}\) | \(64\) |
risch | \(\frac {\frac {5 x^{2}}{8 a^{3}}+\frac {3 x}{8 a^{4}}-\frac {3}{4 a^{5}}}{\left (a x -1\right ) \left (a^{2} x^{2}-1\right )}-\frac {5 \ln \left (-a x -1\right )}{16 a^{5}}-\frac {11 \ln \left (a x -1\right )}{16 a^{5}}\) | \(64\) |
meijerg | \(\frac {-\frac {x \left (-a^{2}\right )^{\frac {5}{2}} \left (-25 a^{2} x^{2}+15\right )}{10 a^{4} \left (-a^{2} x^{2}+1\right )^{2}}+\frac {3 \left (-a^{2}\right )^{\frac {5}{2}} \operatorname {arctanh}\left (a x \right )}{2 a^{5}}}{4 a^{4} \sqrt {-a^{2}}}-\frac {\frac {x^{2} a^{2} \left (-9 a^{2} x^{2}+6\right )}{3 \left (-a^{2} x^{2}+1\right )^{2}}+2 \ln \left (-a^{2} x^{2}+1\right )}{4 a^{5}}\) | \(115\) |
parallelrisch | \(-\frac {12+11 a^{3} \ln \left (a x -1\right ) x^{3}+5 \ln \left (a x +1\right ) x^{3} a^{3}-11 a^{2} \ln \left (a x -1\right ) x^{2}-5 \ln \left (a x +1\right ) x^{2} a^{2}-10 a^{2} x^{2}-11 a \ln \left (a x -1\right ) x -5 \ln \left (a x +1\right ) x a -6 a x +11 \ln \left (a x -1\right )+5 \ln \left (a x +1\right )}{16 a^{5} \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}\) | \(130\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^3*x^4,x,method=_RETURNVERBOSE)
Output:
-1/8/a^5/(a*x+1)-5/16*ln(a*x+1)/a^5+1/8/a^5/(a*x-1)^2+3/4/a^5/(a*x-1)-11/1 6/a^5*ln(a*x-1)
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {10 \, a^{2} x^{2} + 6 \, a x - 5 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) - 11 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 12}{16 \, {\left (a^{8} x^{3} - a^{7} x^{2} - a^{6} x + a^{5}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3*x^4,x, algorithm="fricas")
Output:
1/16*(10*a^2*x^2 + 6*a*x - 5*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) - 11*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x - 1) - 12)/(a^8*x^3 - a^7*x^2 - a ^6*x + a^5)
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx=- \frac {- 5 a^{2} x^{2} - 3 a x + 6}{8 a^{8} x^{3} - 8 a^{7} x^{2} - 8 a^{6} x + 8 a^{5}} - \frac {\frac {11 \log {\left (x - \frac {1}{a} \right )}}{16} + \frac {5 \log {\left (x + \frac {1}{a} \right )}}{16}}{a^{5}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**3*x**4,x)
Output:
-(-5*a**2*x**2 - 3*a*x + 6)/(8*a**8*x**3 - 8*a**7*x**2 - 8*a**6*x + 8*a**5 ) - (11*log(x - 1/a)/16 + 5*log(x + 1/a)/16)/a**5
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {5 \, a^{2} x^{2} + 3 \, a x - 6}{8 \, {\left (a^{8} x^{3} - a^{7} x^{2} - a^{6} x + a^{5}\right )}} - \frac {5 \, \log \left (a x + 1\right )}{16 \, a^{5}} - \frac {11 \, \log \left (a x - 1\right )}{16 \, a^{5}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3*x^4,x, algorithm="maxima")
Output:
1/8*(5*a^2*x^2 + 3*a*x - 6)/(a^8*x^3 - a^7*x^2 - a^6*x + a^5) - 5/16*log(a *x + 1)/a^5 - 11/16*log(a*x - 1)/a^5
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {5 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{5}} - \frac {11 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{5}} + \frac {5 \, a^{2} x^{2} + 3 \, a x - 6}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} a^{5}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^3*x^4,x, algorithm="giac")
Output:
-5/16*log(abs(a*x + 1))/a^5 - 11/16*log(abs(a*x - 1))/a^5 + 1/8*(5*a^2*x^2 + 3*a*x - 6)/((a*x + 1)*(a*x - 1)^2*a^5)
Time = 23.95 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {11\,\ln \left (a\,x-1\right )}{16\,a^5}-\frac {5\,\ln \left (a\,x+1\right )}{16\,a^5}-\frac {\frac {3\,x}{8\,a^4}-\frac {3}{4\,a^5}+\frac {5\,x^2}{8\,a^3}}{-a^3\,x^3+a^2\,x^2+a\,x-1} \] Input:
int(-(x^4*(a*x + 1))/(a^2*x^2 - 1)^3,x)
Output:
- (11*log(a*x - 1))/(16*a^5) - (5*log(a*x + 1))/(16*a^5) - ((3*x)/(8*a^4) - 3/(4*a^5) + (5*x^2)/(8*a^3))/(a*x + a^2*x^2 - a^3*x^3 - 1)
Time = 0.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.86 \[ \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {-11 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+11 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+11 \,\mathrm {log}\left (a x -1\right ) a x -11 \,\mathrm {log}\left (a x -1\right )-5 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+5 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+5 \,\mathrm {log}\left (a x +1\right ) a x -5 \,\mathrm {log}\left (a x +1\right )+10 a^{3} x^{3}-4 a x -2}{16 a^{5} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^3*x^4,x)
Output:
( - 11*log(a*x - 1)*a**3*x**3 + 11*log(a*x - 1)*a**2*x**2 + 11*log(a*x - 1 )*a*x - 11*log(a*x - 1) - 5*log(a*x + 1)*a**3*x**3 + 5*log(a*x + 1)*a**2*x **2 + 5*log(a*x + 1)*a*x - 5*log(a*x + 1) + 10*a**3*x**3 - 4*a*x - 2)/(16* a**5*(a**3*x**3 - a**2*x**2 - a*x + 1))