Integrand size = 24, antiderivative size = 63 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx=-\frac {1}{2 x^2}-\frac {a}{x}+\frac {a^2}{2 (1-a x)}+2 a^2 \log (x)-\frac {7}{4} a^2 \log (1-a x)-\frac {1}{4} a^2 \log (1+a x) \] Output:
-1/2/x^2-a/x+a^2/(-2*a*x+2)+2*a^2*ln(x)-7/4*a^2*ln(-a*x+1)-1/4*a^2*ln(a*x+ 1)
Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {1}{4} \left (-\frac {2}{x^2}-\frac {4 a}{x}+\frac {2 a^2}{1-a x}+8 a^2 \log (x)-7 a^2 \log (1-a x)-a^2 \log (1+a x)\right ) \] Input:
Integrate[E^ArcTanh[a*x]/(x^3*(1 - a^2*x^2)^(3/2)),x]
Output:
(-2/x^2 - (4*a)/x + (2*a^2)/(1 - a*x) + 8*a^2*Log[x] - 7*a^2*Log[1 - a*x] - a^2*Log[1 + a*x])/4
Time = 0.54 (sec) , antiderivative size = 63, 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.
\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \int \frac {1}{x^3 (1-a x)^2 (a x+1)}dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (-\frac {7 a^3}{4 (a x-1)}-\frac {a^3}{4 (a x+1)}+\frac {a^3}{2 (a x-1)^2}+\frac {2 a^2}{x}+\frac {a}{x^2}+\frac {1}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2}{2 (1-a x)}+2 a^2 \log (x)-\frac {7}{4} a^2 \log (1-a x)-\frac {1}{4} a^2 \log (a x+1)-\frac {a}{x}-\frac {1}{2 x^2}\) |
Input:
Int[E^ArcTanh[a*x]/(x^3*(1 - a^2*x^2)^(3/2)),x]
Output:
-1/2*1/x^2 - a/x + a^2/(2*(1 - a*x)) + 2*a^2*Log[x] - (7*a^2*Log[1 - a*x]) /4 - (a^2*Log[1 + a*x])/4
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 = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {a^{2} \ln \left (a x +1\right )}{4}-\frac {a^{2}}{2 \left (a x -1\right )}-\frac {7 a^{2} \ln \left (a x -1\right )}{4}-\frac {1}{2 x^{2}}-\frac {a}{x}+2 a^{2} \ln \left (x \right )\) | \(54\) |
risch | \(\frac {-\frac {3}{2} a^{2} x^{2}+\frac {1}{2} a x +\frac {1}{2}}{x^{2} \left (a x -1\right )}-\frac {a^{2} \ln \left (a x +1\right )}{4}-\frac {7 a^{2} \ln \left (-a x +1\right )}{4}+2 a^{2} \ln \left (-x \right )\) | \(59\) |
norman | \(\frac {\frac {1}{2}+a x -a^{4} x^{4}-\frac {3}{2} a^{3} x^{3}}{\left (a^{2} x^{2}-1\right ) x^{2}}+2 a^{2} \ln \left (x \right )-\frac {7 a^{2} \ln \left (a x -1\right )}{4}-\frac {a^{2} \ln \left (a x +1\right )}{4}\) | \(67\) |
parallelrisch | \(\frac {8 \ln \left (x \right ) x^{3} a^{4}-7 \ln \left (a x -1\right ) x^{3} a^{4}-\ln \left (a x +1\right ) x^{3} a^{4}-8 \ln \left (x \right ) x^{2} a^{3}+7 a^{3} \ln \left (a x -1\right ) x^{2}+\ln \left (a x +1\right ) x^{2} a^{3}-6 a^{3} x^{2}+2 a^{2} x +2 a}{4 a \,x^{2} \left (a x -1\right )}\) | \(109\) |
meijerg | \(-\frac {a^{3} \left (-\frac {2 \left (-3 a^{2} x^{2}+2\right )}{x \sqrt {-a^{2}}\, \left (-2 a^{2} x^{2}+2\right )}+\frac {3 a \,\operatorname {arctanh}\left (a x \right )}{\sqrt {-a^{2}}}\right )}{2 \sqrt {-a^{2}}}-\frac {a^{2} \left (\frac {1}{a^{2} x^{2}}-1-4 \ln \left (x \right )-2 \ln \left (-a^{2}\right )-\frac {3 a^{2} x^{2}}{-3 a^{2} x^{2}+3}+2 \ln \left (-a^{2} x^{2}+1\right )\right )}{2}\) | \(122\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^2/x^3,x,method=_RETURNVERBOSE)
Output:
-1/4*a^2*ln(a*x+1)-1/2*a^2/(a*x-1)-7/4*a^2*ln(a*x-1)-1/2/x^2-a/x+2*a^2*ln( x)
Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx=-\frac {6 \, a^{2} x^{2} - 2 \, a x + {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 7 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 8 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (x\right ) - 2}{4 \, {\left (a x^{3} - x^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^2/x^3,x, algorithm="fricas")
Output:
-1/4*(6*a^2*x^2 - 2*a*x + (a^3*x^3 - a^2*x^2)*log(a*x + 1) + 7*(a^3*x^3 - a^2*x^2)*log(a*x - 1) - 8*(a^3*x^3 - a^2*x^2)*log(x) - 2)/(a*x^3 - x^2)
Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx=2 a^{2} \log {\left (x \right )} - \frac {7 a^{2} \log {\left (x - \frac {1}{a} \right )}}{4} - \frac {a^{2} \log {\left (x + \frac {1}{a} \right )}}{4} + \frac {- 3 a^{2} x^{2} + a x + 1}{2 a x^{3} - 2 x^{2}} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**2/x**3,x)
Output:
2*a**2*log(x) - 7*a**2*log(x - 1/a)/4 - a**2*log(x + 1/a)/4 + (-3*a**2*x** 2 + a*x + 1)/(2*a*x**3 - 2*x**2)
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx=-\frac {1}{4} \, a^{2} \log \left (a x + 1\right ) - \frac {7}{4} \, a^{2} \log \left (a x - 1\right ) + 2 \, a^{2} \log \left (x\right ) - \frac {3 \, a^{2} x^{2} - a x - 1}{2 \, {\left (a x^{3} - x^{2}\right )}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^2/x^3,x, algorithm="maxima")
Output:
-1/4*a^2*log(a*x + 1) - 7/4*a^2*log(a*x - 1) + 2*a^2*log(x) - 1/2*(3*a^2*x ^2 - a*x - 1)/(a*x^3 - x^2)
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx=-\frac {1}{4} \, a^{2} \log \left ({\left | a x + 1 \right |}\right ) - \frac {7}{4} \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) + 2 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {3 \, a^{2} x^{2} - a x - 1}{2 \, {\left (a x - 1\right )} x^{2}} \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^2/x^3,x, algorithm="giac")
Output:
-1/4*a^2*log(abs(a*x + 1)) - 7/4*a^2*log(abs(a*x - 1)) + 2*a^2*log(abs(x)) - 1/2*(3*a^2*x^2 - a*x - 1)/((a*x - 1)*x^2)
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx=2\,a^2\,\ln \left (x\right )-\frac {7\,a^2\,\ln \left (a\,x-1\right )}{4}-\frac {a^2\,\ln \left (a\,x+1\right )}{4}+\frac {-\frac {3\,a^2\,x^2}{2}+\frac {a\,x}{2}+\frac {1}{2}}{a\,x^3-x^2} \] Input:
int((a*x + 1)/(x^3*(a^2*x^2 - 1)^2),x)
Output:
2*a^2*log(x) - (7*a^2*log(a*x - 1))/4 - (a^2*log(a*x + 1))/4 + ((a*x)/2 - (3*a^2*x^2)/2 + 1/2)/(a*x^3 - x^2)
Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.60 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {-7 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+7 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+8 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-8 \,\mathrm {log}\left (x \right ) a^{2} x^{2}-6 a^{3} x^{3}+2 a x +2}{4 x^{2} \left (a x -1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^2/x^3,x)
Output:
( - 7*log(a*x - 1)*a**3*x**3 + 7*log(a*x - 1)*a**2*x**2 - log(a*x + 1)*a** 3*x**3 + log(a*x + 1)*a**2*x**2 + 8*log(x)*a**3*x**3 - 8*log(x)*a**2*x**2 - 6*a**3*x**3 + 2*a*x + 2)/(4*x**2*(a*x - 1))