Integrand size = 23, antiderivative size = 68 \[ \int \frac {e^{2 \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {1}{12 a^2 c^3 (1-a x)^3}-\frac {1}{16 a^2 c^3 (1-a x)}+\frac {1}{16 a^2 c^3 (1+a x)}-\frac {\text {arctanh}(a x)}{8 a^2 c^3} \] Output:
1/12/a^2/c^3/(-a*x+1)^3-1/16/a^2/c^3/(-a*x+1)+1/16/a^2/c^3/(a*x+1)-1/8*arc tanh(a*x)/a^2/c^3
Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\frac {1}{12 a^2 (1-a x)^3}-\frac {1}{16 a^2 (1-a x)}+\frac {1}{16 a^2 (1+a x)}-\frac {\text {arctanh}(a x)}{8 a^2}}{c^3} \] Input:
Integrate[(E^(2*ArcTanh[a*x])*x)/(c - a^2*c*x^2)^3,x]
Output:
(1/(12*a^2*(1 - a*x)^3) - 1/(16*a^2*(1 - a*x)) + 1/(16*a^2*(1 + a*x)) - Ar cTanh[a*x]/(8*a^2))/c^3
Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6700, 86, 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.
\(\displaystyle \int \frac {x e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int \frac {x}{(1-a x)^4 (a x+1)^2}dx}{c^3}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {1}{8 \left (a^2 x^2-1\right ) a}-\frac {1}{16 (a x-1)^2 a}-\frac {1}{16 (a x+1)^2 a}+\frac {1}{4 (a x-1)^4 a}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\text {arctanh}(a x)}{8 a^2}-\frac {1}{16 a^2 (1-a x)}+\frac {1}{16 a^2 (a x+1)}+\frac {1}{12 a^2 (1-a x)^3}}{c^3}\) |
Input:
Int[(E^(2*ArcTanh[a*x])*x)/(c - a^2*c*x^2)^3,x]
Output:
(1/(12*a^2*(1 - a*x)^3) - 1/(16*a^2*(1 - a*x)) + 1/(16*a^2*(1 + a*x)) - Ar cTanh[a*x]/(8*a^2))/c^3
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 64, normalized size of antiderivative = 0.94
method | result | size |
default | \(\frac {\frac {1}{16 a^{2} \left (a x +1\right )}-\frac {\ln \left (a x +1\right )}{16 a^{2}}-\frac {1}{12 a^{2} \left (a x -1\right )^{3}}+\frac {1}{16 \left (a x -1\right ) a^{2}}+\frac {\ln \left (a x -1\right )}{16 a^{2}}}{c^{3}}\) | \(64\) |
risch | \(\frac {\frac {a \,x^{3}}{8}-\frac {x^{2}}{4}+\frac {x}{24 a}-\frac {1}{12 a^{2}}}{c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}+\frac {\ln \left (-a x +1\right )}{16 a^{2} c^{3}}-\frac {\ln \left (a x +1\right )}{16 a^{2} c^{3}}\) | \(76\) |
norman | \(\frac {-\frac {x}{8 a c}-\frac {a \,x^{3}}{3 c}+\frac {a^{3} x^{5}}{8 c}-\frac {x^{2}}{2 c}+\frac {a^{2} x^{4}}{4 c}-\frac {a^{4} x^{6}}{12 c}}{\left (a^{2} x^{2}-1\right )^{3} c^{2}}+\frac {\ln \left (a x -1\right )}{16 a^{2} c^{3}}-\frac {\ln \left (a x +1\right )}{16 a^{2} c^{3}}\) | \(105\) |
parallelrisch | \(\frac {3 \ln \left (a x -1\right ) x^{4} a^{4}-3 \ln \left (a x +1\right ) x^{4} a^{4}-4 a^{4} x^{4}-6 a^{3} \ln \left (a x -1\right ) x^{3}+6 \ln \left (a x +1\right ) x^{3} a^{3}+14 a^{3} x^{3}-12 a^{2} x^{2}+6 a \ln \left (a x -1\right ) x -6 \ln \left (a x +1\right ) x a -6 a x -3 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )}{48 a^{2} c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}\) | \(148\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)*x/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/c^3*(1/16/a^2/(a*x+1)-1/16*ln(a*x+1)/a^2-1/12/a^2/(a*x-1)^3+1/16/(a*x-1) /a^2+1/16/a^2*ln(a*x-1))
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (58) = 116\).
Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.81 \[ \int \frac {e^{2 \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {6 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 2 \, a x - 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) - 4}{48 \, {\left (a^{6} c^{3} x^{4} - 2 \, a^{5} c^{3} x^{3} + 2 \, a^{3} c^{3} x - a^{2} c^{3}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x/(-a^2*c*x^2+c)^3,x, algorithm="fricas")
Output:
1/48*(6*a^3*x^3 - 12*a^2*x^2 + 2*a*x - 3*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1) *log(a*x + 1) + 3*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x - 1) - 4)/(a^6 *c^3*x^4 - 2*a^5*c^3*x^3 + 2*a^3*c^3*x - a^2*c^3)
Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.28 \[ \int \frac {e^{2 \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 a^{3} x^{3} - 6 a^{2} x^{2} + a x - 2}{24 a^{6} c^{3} x^{4} - 48 a^{5} c^{3} x^{3} + 48 a^{3} c^{3} x - 24 a^{2} c^{3}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{16} - \frac {\log {\left (x + \frac {1}{a} \right )}}{16}}{a^{2} c^{3}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)*x/(-a**2*c*x**2+c)**3,x)
Output:
(3*a**3*x**3 - 6*a**2*x**2 + a*x - 2)/(24*a**6*c**3*x**4 - 48*a**5*c**3*x* *3 + 48*a**3*c**3*x - 24*a**2*c**3) + (log(x - 1/a)/16 - log(x + 1/a)/16)/ (a**2*c**3)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.37 \[ \int \frac {e^{2 \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x - 2}{24 \, {\left (a^{6} c^{3} x^{4} - 2 \, a^{5} c^{3} x^{3} + 2 \, a^{3} c^{3} x - a^{2} c^{3}\right )}} - \frac {\log \left (a x + 1\right )}{16 \, a^{2} c^{3}} + \frac {\log \left (a x - 1\right )}{16 \, a^{2} c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x/(-a^2*c*x^2+c)^3,x, algorithm="maxima")
Output:
1/24*(3*a^3*x^3 - 6*a^2*x^2 + a*x - 2)/(a^6*c^3*x^4 - 2*a^5*c^3*x^3 + 2*a^ 3*c^3*x - a^2*c^3) - 1/16*log(a*x + 1)/(a^2*c^3) + 1/16*log(a*x - 1)/(a^2* c^3)
Time = 0.13 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \frac {e^{2 \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {\log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{2} c^{3}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{2} c^{3}} + \frac {3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x - 2}{24 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} a^{2} c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)*x/(-a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
-1/16*log(abs(a*x + 1))/(a^2*c^3) + 1/16*log(abs(a*x - 1))/(a^2*c^3) + 1/2 4*(3*a^3*x^3 - 6*a^2*x^2 + a*x - 2)/((a*x + 1)*(a*x - 1)^3*a^2*c^3)
Time = 24.89 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {\frac {x}{24\,a}+\frac {a\,x^3}{8}-\frac {1}{12\,a^2}-\frac {x^2}{4}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3}-\frac {\mathrm {atanh}\left (a\,x\right )}{8\,a^2\,c^3} \] Input:
int(-(x*(a*x + 1)^2)/((c - a^2*c*x^2)^3*(a^2*x^2 - 1)),x)
Output:
- (x/(24*a) + (a*x^3)/8 - 1/(12*a^2) - x^2/4)/(c^3 + 2*a^3*c^3*x^3 - a^4*c ^3*x^4 - 2*a*c^3*x) - atanh(a*x)/(8*a^2*c^3)
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.13 \[ \int \frac {e^{2 \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-6 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+6 \,\mathrm {log}\left (a x -1\right ) a x -3 \,\mathrm {log}\left (a x -1\right )-3 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+6 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-6 \,\mathrm {log}\left (a x +1\right ) a x +3 \,\mathrm {log}\left (a x +1\right )+3 a^{4} x^{4}-12 a^{2} x^{2}+8 a x -7}{48 a^{2} c^{3} \left (a^{4} x^{4}-2 a^{3} x^{3}+2 a x -1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)*x/(-a^2*c*x^2+c)^3,x)
Output:
(3*log(a*x - 1)*a**4*x**4 - 6*log(a*x - 1)*a**3*x**3 + 6*log(a*x - 1)*a*x - 3*log(a*x - 1) - 3*log(a*x + 1)*a**4*x**4 + 6*log(a*x + 1)*a**3*x**3 - 6 *log(a*x + 1)*a*x + 3*log(a*x + 1) + 3*a**4*x**4 - 12*a**2*x**2 + 8*a*x - 7)/(48*a**2*c**3*(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 1))