Integrand size = 25, antiderivative size = 105 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {1}{12 a^6 c^3 (1-a x)^3}-\frac {1}{2 a^6 c^3 (1-a x)^2}+\frac {23}{16 a^6 c^3 (1-a x)}+\frac {1}{16 a^6 c^3 (1+a x)}+\frac {13 \log (1-a x)}{16 a^6 c^3}+\frac {3 \log (1+a x)}{16 a^6 c^3} \] Output:
1/12/a^6/c^3/(-a*x+1)^3-1/2/a^6/c^3/(-a*x+1)^2+23/16/a^6/c^3/(-a*x+1)+1/16 /a^6/c^3/(a*x+1)+13/16*ln(-a*x+1)/a^6/c^3+3/16*ln(a*x+1)/a^6/c^3
Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.83 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {-52+74 a x+36 a^2 x^2-66 a^3 x^3+39 (-1+a x)^3 (1+a x) \log (1-a x)+9 (-1+a x)^3 (1+a x) \log (1+a x)}{48 a^6 c^3 (-1+a x)^3 (1+a x)} \] Input:
Integrate[(E^(2*ArcTanh[a*x])*x^5)/(c - a^2*c*x^2)^3,x]
Output:
(-52 + 74*a*x + 36*a^2*x^2 - 66*a^3*x^3 + 39*(-1 + a*x)^3*(1 + a*x)*Log[1 - a*x] + 9*(-1 + a*x)^3*(1 + a*x)*Log[1 + a*x])/(48*a^6*c^3*(-1 + a*x)^3*( 1 + a*x))
Time = 0.36 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87, 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.
\(\displaystyle \int \frac {x^5 e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int \frac {x^5}{(1-a x)^4 (a x+1)^2}dx}{c^3}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {3}{16 a^5 (a x+1)}-\frac {1}{16 a^5 (a x+1)^2}+\frac {13}{16 a^5 (a x-1)}+\frac {23}{16 a^5 (a x-1)^2}+\frac {1}{a^5 (a x-1)^3}+\frac {1}{4 a^5 (a x-1)^4}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {23}{16 a^6 (1-a x)}+\frac {1}{16 a^6 (a x+1)}-\frac {1}{2 a^6 (1-a x)^2}+\frac {1}{12 a^6 (1-a x)^3}+\frac {13 \log (1-a x)}{16 a^6}+\frac {3 \log (a x+1)}{16 a^6}}{c^3}\) |
Input:
Int[(E^(2*ArcTanh[a*x])*x^5)/(c - a^2*c*x^2)^3,x]
Output:
(1/(12*a^6*(1 - a*x)^3) - 1/(2*a^6*(1 - a*x)^2) + 23/(16*a^6*(1 - a*x)) + 1/(16*a^6*(1 + a*x)) + (13*Log[1 - a*x])/(16*a^6) + (3*Log[1 + a*x])/(16*a ^6))/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 = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {\frac {1}{16 a^{6} \left (a x +1\right )}+\frac {3 \ln \left (a x +1\right )}{16 a^{6}}-\frac {1}{2 a^{6} \left (a x -1\right )^{2}}-\frac {1}{12 a^{6} \left (a x -1\right )^{3}}+\frac {13 \ln \left (a x -1\right )}{16 a^{6}}-\frac {23}{16 a^{6} \left (a x -1\right )}}{c^{3}}\) | \(76\) |
risch | \(\frac {-\frac {11 x^{3}}{8 a^{3}}+\frac {3 x^{2}}{4 a^{4}}+\frac {37 x}{24 a^{5}}-\frac {13}{12 a^{6}}}{c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}+\frac {13 \ln \left (-a x +1\right )}{16 a^{6} c^{3}}+\frac {3 \ln \left (a x +1\right )}{16 a^{6} c^{3}}\) | \(81\) |
norman | \(\frac {\frac {5 x^{3}}{3 a^{3} c}-\frac {11 x^{5}}{8 a c}-\frac {5 x}{8 c \,a^{5}}-\frac {2 x^{6}}{3 c}-\frac {5}{12 a^{6} c}+\frac {3 x^{2}}{4 a^{4} c}}{\left (a^{2} x^{2}-1\right )^{3} c^{2}}+\frac {13 \ln \left (a x -1\right )}{16 a^{6} c^{3}}+\frac {3 \ln \left (a x +1\right )}{16 a^{6} c^{3}}\) | \(104\) |
parallelrisch | \(\frac {-52+39 \ln \left (a x -1\right ) x^{4} a^{4}+9 \ln \left (a x +1\right ) x^{4} a^{4}-78 a^{3} \ln \left (a x -1\right ) x^{3}-18 \ln \left (a x +1\right ) x^{3} a^{3}-66 a^{3} x^{3}+36 a^{2} x^{2}+78 a \ln \left (a x -1\right ) x +18 \ln \left (a x +1\right ) x a +74 a x -39 \ln \left (a x -1\right )-9 \ln \left (a x +1\right )}{48 a^{6} c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}\) | \(141\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^5/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/c^3*(1/16/a^6/(a*x+1)+3/16*ln(a*x+1)/a^6-1/2/a^6/(a*x-1)^2-1/12/a^6/(a*x -1)^3+13/16/a^6*ln(a*x-1)-23/16/a^6/(a*x-1))
Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.17 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {66 \, a^{3} x^{3} - 36 \, a^{2} x^{2} - 74 \, a x - 9 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) - 39 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 52}{48 \, {\left (a^{10} c^{3} x^{4} - 2 \, a^{9} c^{3} x^{3} + 2 \, a^{7} c^{3} x - a^{6} c^{3}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^5/(-a^2*c*x^2+c)^3,x, algorithm="fricas ")
Output:
-1/48*(66*a^3*x^3 - 36*a^2*x^2 - 74*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) + 52) /(a^10*c^3*x^4 - 2*a^9*c^3*x^3 + 2*a^7*c^3*x - a^6*c^3)
Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {- 33 a^{3} x^{3} + 18 a^{2} x^{2} + 37 a x - 26}{24 a^{10} c^{3} x^{4} - 48 a^{9} c^{3} x^{3} + 48 a^{7} c^{3} x - 24 a^{6} c^{3}} + \frac {\frac {13 \log {\left (x - \frac {1}{a} \right )}}{16} + \frac {3 \log {\left (x + \frac {1}{a} \right )}}{16}}{a^{6} c^{3}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*x**5/(-a**2*c*x**2+c)**3,x)
Output:
(-33*a**3*x**3 + 18*a**2*x**2 + 37*a*x - 26)/(24*a**10*c**3*x**4 - 48*a**9 *c**3*x**3 + 48*a**7*c**3*x - 24*a**6*c**3) + (13*log(x - 1/a)/16 + 3*log( x + 1/a)/16)/(a**6*c**3)
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {33 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 37 \, a x + 26}{24 \, {\left (a^{10} c^{3} x^{4} - 2 \, a^{9} c^{3} x^{3} + 2 \, a^{7} c^{3} x - a^{6} c^{3}\right )}} + \frac {3 \, \log \left (a x + 1\right )}{16 \, a^{6} c^{3}} + \frac {13 \, \log \left (a x - 1\right )}{16 \, a^{6} c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^5/(-a^2*c*x^2+c)^3,x, algorithm="maxima ")
Output:
-1/24*(33*a^3*x^3 - 18*a^2*x^2 - 37*a*x + 26)/(a^10*c^3*x^4 - 2*a^9*c^3*x^ 3 + 2*a^7*c^3*x - a^6*c^3) + 3/16*log(a*x + 1)/(a^6*c^3) + 13/16*log(a*x - 1)/(a^6*c^3)
Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{6} c^{3}} + \frac {13 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{6} c^{3}} - \frac {33 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 37 \, a x + 26}{24 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} a^{6} c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x^5/(-a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
3/16*log(abs(a*x + 1))/(a^6*c^3) + 13/16*log(abs(a*x - 1))/(a^6*c^3) - 1/2 4*(33*a^3*x^3 - 18*a^2*x^2 - 37*a*x + 26)/((a*x + 1)*(a*x - 1)^3*a^6*c^3)
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {13\,\ln \left (a\,x-1\right )}{16\,a^6\,c^3}-\frac {\frac {37\,x}{24\,a^5}-\frac {13}{12\,a^6}-\frac {11\,x^3}{8\,a^3}+\frac {3\,x^2}{4\,a^4}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3}+\frac {3\,\ln \left (a\,x+1\right )}{16\,a^6\,c^3} \] Input:
int(-(x^5*(a*x + 1)^2)/((c - a^2*c*x^2)^3*(a^2*x^2 - 1)),x)
Output:
(13*log(a*x - 1))/(16*a^6*c^3) - ((37*x)/(24*a^5) - 13/(12*a^6) - (11*x^3) /(8*a^3) + (3*x^2)/(4*a^4))/(c^3 + 2*a^3*c^3*x^3 - a^4*c^3*x^4 - 2*a*c^3*x ) + (3*log(a*x + 1))/(16*a^6*c^3)
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.38 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^5}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {39 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-78 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+78 \,\mathrm {log}\left (a x -1\right ) a x -39 \,\mathrm {log}\left (a x -1\right )+9 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}-18 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+18 \,\mathrm {log}\left (a x +1\right ) a x -9 \,\mathrm {log}\left (a x +1\right )-33 a^{4} x^{4}+36 a^{2} x^{2}+8 a x -19}{48 a^{6} c^{3} \left (a^{4} x^{4}-2 a^{3} x^{3}+2 a x -1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*x^5/(-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) - 33*a**4*x**4 + 36*a**2*x**2 + 8 *a*x - 19)/(48*a**6*c**3*(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 1))