Integrand size = 25, antiderivative size = 99 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {1}{2 c^2 x^2}-\frac {2 a}{c^2 x}+\frac {a^2}{4 c^2 (1-a x)^2}+\frac {7 a^2}{4 c^2 (1-a x)}+\frac {4 a^2 \log (x)}{c^2}-\frac {31 a^2 \log (1-a x)}{8 c^2}-\frac {a^2 \log (1+a x)}{8 c^2} \] Output:
-1/2/c^2/x^2-2*a/c^2/x+1/4*a^2/c^2/(-a*x+1)^2+7/4*a^2/c^2/(-a*x+1)+4*a^2*l n(x)/c^2-31/8*a^2*ln(-a*x+1)/c^2-1/8*a^2*ln(a*x+1)/c^2
Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.73 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {\frac {4}{x^2}+\frac {16 a}{x}-\frac {2 a^2}{(-1+a x)^2}+\frac {14 a^2}{-1+a x}-32 a^2 \log (x)+31 a^2 \log (1-a x)+a^2 \log (1+a x)}{8 c^2} \] Input:
Integrate[E^(2*ArcTanh[a*x])/(x^3*(c - a^2*c*x^2)^2),x]
Output:
-1/8*(4/x^2 + (16*a)/x - (2*a^2)/(-1 + a*x)^2 + (14*a^2)/(-1 + a*x) - 32*a ^2*Log[x] + 31*a^2*Log[1 - a*x] + a^2*Log[1 + a*x])/c^2
Time = 0.35 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int \frac {1}{x^3 (1-a x)^3 (a x+1)}dx}{c^2}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {31 a^3}{8 (a x-1)}-\frac {a^3}{8 (a x+1)}+\frac {7 a^3}{4 (a x-1)^2}-\frac {a^3}{2 (a x-1)^3}+\frac {4 a^2}{x}+\frac {2 a}{x^2}+\frac {1}{x^3}\right )dx}{c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {7 a^2}{4 (1-a x)}+\frac {a^2}{4 (1-a x)^2}+4 a^2 \log (x)-\frac {31}{8} a^2 \log (1-a x)-\frac {1}{8} a^2 \log (a x+1)-\frac {2 a}{x}-\frac {1}{2 x^2}}{c^2}\) |
Input:
Int[E^(2*ArcTanh[a*x])/(x^3*(c - a^2*c*x^2)^2),x]
Output:
(-1/2*1/x^2 - (2*a)/x + a^2/(4*(1 - a*x)^2) + (7*a^2)/(4*(1 - a*x)) + 4*a^ 2*Log[x] - (31*a^2*Log[1 - a*x])/8 - (a^2*Log[1 + a*x])/8)/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 = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71
method | result | size |
default | \(\frac {-\frac {a^{2} \ln \left (a x +1\right )}{8}+\frac {a^{2}}{4 \left (a x -1\right )^{2}}-\frac {7 a^{2}}{4 \left (a x -1\right )}-\frac {31 a^{2} \ln \left (a x -1\right )}{8}-\frac {1}{2 x^{2}}-\frac {2 a}{x}+4 a^{2} \ln \left (x \right )}{c^{2}}\) | \(70\) |
risch | \(\frac {-\frac {15}{4} a^{3} x^{3}+\frac {11}{2} a^{2} x^{2}-a x -\frac {1}{2}}{c^{2} x^{2} \left (a x -1\right )^{2}}-\frac {31 a^{2} \ln \left (-a x +1\right )}{8 c^{2}}+\frac {4 a^{2} \ln \left (-x \right )}{c^{2}}-\frac {a^{2} \ln \left (a x +1\right )}{8 c^{2}}\) | \(79\) |
norman | \(\frac {-\frac {1}{2 c}-\frac {2 a x}{c}+\frac {25 a^{3} x^{3}}{4 c}-\frac {15 a^{5} x^{5}}{4 c}+\frac {4 a^{4} x^{4}}{c}-\frac {3 a^{6} x^{6}}{c}}{\left (a^{2} x^{2}-1\right )^{2} x^{2} c}+\frac {4 a^{2} \ln \left (x \right )}{c^{2}}-\frac {31 a^{2} \ln \left (a x -1\right )}{8 c^{2}}-\frac {a^{2} \ln \left (a x +1\right )}{8 c^{2}}\) | \(115\) |
parallelrisch | \(\frac {-4+32 \ln \left (x \right ) x^{4} a^{4}-31 \ln \left (a x -1\right ) x^{4} a^{4}-\ln \left (a x +1\right ) x^{4} a^{4}-44 a^{4} x^{4}-64 \ln \left (x \right ) x^{3} a^{3}+62 a^{3} \ln \left (a x -1\right ) x^{3}+2 \ln \left (a x +1\right ) x^{3} a^{3}+58 a^{3} x^{3}+32 a^{2} \ln \left (x \right ) x^{2}-31 a^{2} \ln \left (a x -1\right ) x^{2}-\ln \left (a x +1\right ) x^{2} a^{2}-8 a x}{8 c^{2} x^{2} \left (a x -1\right )^{2}}\) | \(152\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)
Output:
1/c^2*(-1/8*a^2*ln(a*x+1)+1/4*a^2/(a*x-1)^2-7/4*a^2/(a*x-1)-31/8*a^2*ln(a* x-1)-1/2/x^2-2*a/x+4*a^2*ln(x))
Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.42 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {30 \, a^{3} x^{3} - 44 \, a^{2} x^{2} + 8 \, a x + {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 31 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 32 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (x\right ) + 4}{8 \, {\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{2}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^2,x, algorithm="fricas ")
Output:
-1/8*(30*a^3*x^3 - 44*a^2*x^2 + 8*a*x + (a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*lo g(a*x + 1) + 31*(a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*log(a*x - 1) - 32*(a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*log(x) + 4)/(a^2*c^2*x^4 - 2*a*c^2*x^3 + c^2*x^2)
Time = 0.46 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.93 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx=- \frac {15 a^{3} x^{3} - 22 a^{2} x^{2} + 4 a x + 2}{4 a^{2} c^{2} x^{4} - 8 a c^{2} x^{3} + 4 c^{2} x^{2}} - \frac {- 4 a^{2} \log {\left (x \right )} + \frac {31 a^{2} \log {\left (x - \frac {1}{a} \right )}}{8} + \frac {a^{2} \log {\left (x + \frac {1}{a} \right )}}{8}}{c^{2}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)/x**3/(-a**2*c*x**2+c)**2,x)
Output:
-(15*a**3*x**3 - 22*a**2*x**2 + 4*a*x + 2)/(4*a**2*c**2*x**4 - 8*a*c**2*x* *3 + 4*c**2*x**2) - (-4*a**2*log(x) + 31*a**2*log(x - 1/a)/8 + a**2*log(x + 1/a)/8)/c**2
Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.93 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {a^{2} \log \left (a x + 1\right )}{8 \, c^{2}} - \frac {31 \, a^{2} \log \left (a x - 1\right )}{8 \, c^{2}} + \frac {4 \, a^{2} \log \left (x\right )}{c^{2}} - \frac {15 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 2}{4 \, {\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{2}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^2,x, algorithm="maxima ")
Output:
-1/8*a^2*log(a*x + 1)/c^2 - 31/8*a^2*log(a*x - 1)/c^2 + 4*a^2*log(x)/c^2 - 1/4*(15*a^3*x^3 - 22*a^2*x^2 + 4*a*x + 2)/(a^2*c^2*x^4 - 2*a*c^2*x^3 + c^ 2*x^2)
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.80 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {a^{2} \log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac {31 \, a^{2} \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac {4 \, a^{2} \log \left ({\left | x \right |}\right )}{c^{2}} - \frac {15 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 2}{4 \, {\left (a x - 1\right )}^{2} c^{2} x^{2}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^2,x, algorithm="giac")
Output:
-1/8*a^2*log(abs(a*x + 1))/c^2 - 31/8*a^2*log(abs(a*x - 1))/c^2 + 4*a^2*lo g(abs(x))/c^2 - 1/4*(15*a^3*x^3 - 22*a^2*x^2 + 4*a*x + 2)/((a*x - 1)^2*c^2 *x^2)
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx=\frac {4\,a^2\,\ln \left (x\right )}{c^2}-\frac {\frac {15\,a^3\,x^3}{4}-\frac {11\,a^2\,x^2}{2}+a\,x+\frac {1}{2}}{a^2\,c^2\,x^4-2\,a\,c^2\,x^3+c^2\,x^2}-\frac {31\,a^2\,\ln \left (a\,x-1\right )}{8\,c^2}-\frac {a^2\,\ln \left (a\,x+1\right )}{8\,c^2} \] Input:
int(-(a*x + 1)^2/(x^3*(c - a^2*c*x^2)^2*(a^2*x^2 - 1)),x)
Output:
(4*a^2*log(x))/c^2 - (a*x - (11*a^2*x^2)/2 + (15*a^3*x^3)/4 + 1/2)/(c^2*x^ 2 - 2*a*c^2*x^3 + a^2*c^2*x^4) - (31*a^2*log(a*x - 1))/(8*c^2) - (a^2*log( a*x + 1))/(8*c^2)
Time = 0.16 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.61 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^2} \, dx=\frac {-31 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+62 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-31 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+2 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+32 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-64 \,\mathrm {log}\left (x \right ) a^{3} x^{3}+32 \,\mathrm {log}\left (x \right ) a^{2} x^{2}-15 a^{4} x^{4}+29 a^{2} x^{2}-8 a x -4}{8 c^{2} x^{2} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)/x^3/(-a^2*c*x^2+c)^2,x)
Output:
( - 31*log(a*x - 1)*a**4*x**4 + 62*log(a*x - 1)*a**3*x**3 - 31*log(a*x - 1 )*a**2*x**2 - log(a*x + 1)*a**4*x**4 + 2*log(a*x + 1)*a**3*x**3 - log(a*x + 1)*a**2*x**2 + 32*log(x)*a**4*x**4 - 64*log(x)*a**3*x**3 + 32*log(x)*a** 2*x**2 - 15*a**4*x**4 + 29*a**2*x**2 - 8*a*x - 4)/(8*c**2*x**2*(a**2*x**2 - 2*a*x + 1))