Integrand size = 22, antiderivative size = 86 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {1}{12 a c^3 (1-a x)^3}+\frac {1}{8 a c^3 (1-a x)^2}+\frac {3}{16 a c^3 (1-a x)}-\frac {1}{16 a c^3 (1+a x)}+\frac {\text {arctanh}(a x)}{4 a c^3} \] Output:
1/12/a/c^3/(-a*x+1)^3+1/8/a/c^3/(-a*x+1)^2+3/16/a/c^3/(-a*x+1)-1/16/a/c^3/ (a*x+1)+1/4*arctanh(a*x)/a/c^3
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {4+a x-6 a^2 x^2+3 a^3 x^3-3 (-1+a x)^3 (1+a x) \text {arctanh}(a x)}{12 a c^3 (-1+a x)^3 (1+a x)} \] Input:
Integrate[E^(2*ArcTanh[a*x])/(c - a^2*c*x^2)^3,x]
Output:
-1/12*(4 + a*x - 6*a^2*x^2 + 3*a^3*x^3 - 3*(-1 + a*x)^3*(1 + a*x)*ArcTanh[ a*x])/(a*c^3*(-1 + a*x)^3*(1 + a*x))
Time = 0.46 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6690, 54, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {\int \frac {1}{(1-a x)^4 (a x+1)^2}dx}{c^3}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {3}{16 (a x-1)^2}+\frac {1}{16 (a x+1)^2}-\frac {1}{4 (a x-1)^3}+\frac {1}{4 (a x-1)^4}-\frac {1}{4 \left (a^2 x^2-1\right )}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\text {arctanh}(a x)}{4 a}+\frac {3}{16 a (1-a x)}-\frac {1}{16 a (a x+1)}+\frac {1}{8 a (1-a x)^2}+\frac {1}{12 a (1-a x)^3}}{c^3}\) |
Input:
Int[E^(2*ArcTanh[a*x])/(c - a^2*c*x^2)^3,x]
Output:
(1/(12*a*(1 - a*x)^3) + 1/(8*a*(1 - a*x)^2) + 3/(16*a*(1 - a*x)) - 1/(16*a *(1 + a*x)) + ArcTanh[a*x]/(4*a))/c^3
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p Int[(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a , c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {-\frac {1}{16 a \left (a x +1\right )}+\frac {\ln \left (a x +1\right )}{8 a}-\frac {1}{12 a \left (a x -1\right )^{3}}+\frac {1}{8 a \left (a x -1\right )^{2}}-\frac {3}{16 \left (a x -1\right ) a}-\frac {\ln \left (a x -1\right )}{8 a}}{c^{3}}\) | \(76\) |
risch | \(\frac {-\frac {a^{2} x^{3}}{4}+\frac {a \,x^{2}}{2}-\frac {x}{12}-\frac {1}{3 a}}{c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}+\frac {\ln \left (-a x -1\right )}{8 a \,c^{3}}-\frac {\ln \left (a x -1\right )}{8 a \,c^{3}}\) | \(76\) |
norman | \(\frac {\frac {a^{3} x^{4}}{c}-\frac {3 x}{4 c}-\frac {a^{4} x^{5}}{4 c}+\frac {2 a^{2} x^{3}}{3 c}-\frac {a \,x^{2}}{c}-\frac {a^{5} x^{6}}{3 c}}{\left (a^{2} x^{2}-1\right )^{3} c^{2}}-\frac {\ln \left (a x -1\right )}{8 a \,c^{3}}+\frac {\ln \left (a x +1\right )}{8 a \,c^{3}}\) | \(104\) |
parallelrisch | \(\frac {-3 \ln \left (a x -1\right ) x^{4} a^{4}+3 \ln \left (a x +1\right ) x^{4} a^{4}-8 a^{4} x^{4}+6 a^{3} \ln \left (a x -1\right ) x^{3}-6 \ln \left (a x +1\right ) x^{3} a^{3}+10 a^{3} x^{3}+12 a^{2} x^{2}-6 a \ln \left (a x -1\right ) x +6 \ln \left (a x +1\right ) x a -18 a x +3 \ln \left (a x -1\right )-3 \ln \left (a x +1\right )}{24 c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right ) a}\) | \(148\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/c^3*(-1/16/a/(a*x+1)+1/8*ln(a*x+1)/a-1/12/a/(a*x-1)^3+1/8/a/(a*x-1)^2-3/ 16/(a*x-1)/a-1/8/a*ln(a*x-1))
Time = 0.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.41 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {6 \, a^{3} x^{3} - 12 \, a^{2} x^{2} + 2 \, a x - 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 8}{24 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="fricas")
Output:
-1/24*(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) + 8)/(a^ 5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3)
Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {- 3 a^{3} x^{3} + 6 a^{2} x^{2} - a x - 4}{12 a^{5} c^{3} x^{4} - 24 a^{4} c^{3} x^{3} + 24 a^{2} c^{3} x - 12 a c^{3}} + \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{8} + \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a c^{3}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)/(-a**2*c*x**2+c)**3,x)
Output:
(-3*a**3*x**3 + 6*a**2*x**2 - a*x - 4)/(12*a**5*c**3*x**4 - 24*a**4*c**3*x **3 + 24*a**2*c**3*x - 12*a*c**3) + (-log(x - 1/a)/8 + log(x + 1/a)/8)/(a* c**3)
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x + 4}{12 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} + \frac {\log \left (a x + 1\right )}{8 \, a c^{3}} - \frac {\log \left (a x - 1\right )}{8 \, a c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="maxima")
Output:
-1/12*(3*a^3*x^3 - 6*a^2*x^2 + a*x + 4)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a ^2*c^3*x - a*c^3) + 1/8*log(a*x + 1)/(a*c^3) - 1/8*log(a*x - 1)/(a*c^3)
Time = 0.14 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{3}} - \frac {\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{3}} - \frac {3 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + a x + 4}{12 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} a c^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/(-a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
1/8*log(abs(a*x + 1))/(a*c^3) - 1/8*log(abs(a*x - 1))/(a*c^3) - 1/12*(3*a^ 3*x^3 - 6*a^2*x^2 + a*x + 4)/((a*x + 1)*(a*x - 1)^3*a*c^3)
Time = 23.58 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\frac {x}{12}-\frac {a\,x^2}{2}+\frac {1}{3\,a}+\frac {a^2\,x^3}{4}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3}+\frac {\mathrm {atanh}\left (a\,x\right )}{4\,a\,c^3} \] Input:
int(-(a*x + 1)^2/((c - a^2*c*x^2)^3*(a^2*x^2 - 1)),x)
Output:
(x/12 - (a*x^2)/2 + 1/(3*a) + (a^2*x^3)/4)/(c^3 + 2*a^3*c^3*x^3 - a^4*c^3* x^4 - 2*a*c^3*x) + atanh(a*x)/(4*a*c^3)
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.69 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {-3 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+6 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-6 \,\mathrm {log}\left (a x -1\right ) a x +3 \,\mathrm {log}\left (a x -1\right )+3 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}-6 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+6 \,\mathrm {log}\left (a x +1\right ) a x -3 \,\mathrm {log}\left (a x +1\right )-3 a^{4} x^{4}+12 a^{2} x^{2}-8 a x -5}{24 a \,c^{3} \left (a^{4} x^{4}-2 a^{3} x^{3}+2 a x -1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)/(-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 - 5)/(24*a*c**3*(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 1))