Integrand size = 25, antiderivative size = 110 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {1}{3 c^2 x^3}-\frac {a}{c^2 x^2}-\frac {4 a^2}{c^2 x}+\frac {a^3}{4 c^2 (1-a x)^2}+\frac {9 a^3}{4 c^2 (1-a x)}+\frac {6 a^3 \log (x)}{c^2}-\frac {49 a^3 \log (1-a x)}{8 c^2}+\frac {a^3 \log (1+a x)}{8 c^2} \] Output:
-1/3/c^2/x^3-a/c^2/x^2-4*a^2/c^2/x+1/4*a^3/c^2/(-a*x+1)^2+9/4*a^3/c^2/(-a* x+1)+6*a^3*ln(x)/c^2-49/8*a^3*ln(-a*x+1)/c^2+1/8*a^3*ln(a*x+1)/c^2
Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx=\frac {-\frac {1}{3 x^3}-\frac {a}{x^2}-\frac {4 a^2}{x}+\frac {9 a^3}{4-4 a x}+\frac {a^3}{4 (-1+a x)^2}+6 a^3 \log (x)-\frac {49}{8} a^3 \log (1-a x)+\frac {1}{8} a^3 \log (1+a x)}{c^2} \] Input:
Integrate[E^(2*ArcTanh[a*x])/(x^4*(c - a^2*c*x^2)^2),x]
Output:
(-1/3*1/x^3 - a/x^2 - (4*a^2)/x + (9*a^3)/(4 - 4*a*x) + a^3/(4*(-1 + a*x)^ 2) + 6*a^3*Log[x] - (49*a^3*Log[1 - a*x])/8 + (a^3*Log[1 + a*x])/8)/c^2
Time = 0.58 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.82, 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.
\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int \frac {1}{x^4 (1-a x)^3 (a x+1)}dx}{c^2}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (-\frac {49 a^4}{8 (a x-1)}+\frac {a^4}{8 (a x+1)}+\frac {9 a^4}{4 (a x-1)^2}-\frac {a^4}{2 (a x-1)^3}+\frac {6 a^3}{x}+\frac {4 a^2}{x^2}+\frac {2 a}{x^3}+\frac {1}{x^4}\right )dx}{c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {9 a^3}{4 (1-a x)}+\frac {a^3}{4 (1-a x)^2}+6 a^3 \log (x)-\frac {49}{8} a^3 \log (1-a x)+\frac {1}{8} a^3 \log (a x+1)-\frac {4 a^2}{x}-\frac {a}{x^2}-\frac {1}{3 x^3}}{c^2}\) |
Input:
Int[E^(2*ArcTanh[a*x])/(x^4*(c - a^2*c*x^2)^2),x]
Output:
(-1/3*1/x^3 - a/x^2 - (4*a^2)/x + a^3/(4*(1 - a*x)^2) + (9*a^3)/(4*(1 - a* x)) + 6*a^3*Log[x] - (49*a^3*Log[1 - a*x])/8 + (a^3*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 = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.71
method | result | size |
default | \(\frac {\frac {a^{3} \ln \left (a x +1\right )}{8}+\frac {a^{3}}{4 \left (a x -1\right )^{2}}-\frac {9 a^{3}}{4 \left (a x -1\right )}-\frac {49 a^{3} \ln \left (a x -1\right )}{8}-\frac {1}{3 x^{3}}-\frac {a}{x^{2}}-\frac {4 a^{2}}{x}+6 a^{3} \ln \left (x \right )}{c^{2}}\) | \(78\) |
risch | \(\frac {-\frac {25}{4} a^{4} x^{4}+\frac {19}{2} a^{3} x^{3}-\frac {7}{3} a^{2} x^{2}-\frac {1}{3} a x -\frac {1}{3}}{c^{2} x^{3} \left (a x -1\right )^{2}}+\frac {a^{3} \ln \left (a x +1\right )}{8 c^{2}}-\frac {49 a^{3} \ln \left (-a x +1\right )}{8 c^{2}}+\frac {6 a^{3} \ln \left (-x \right )}{c^{2}}\) | \(87\) |
norman | \(\frac {-\frac {1}{3 c}-\frac {a x}{c}-\frac {10 a^{2} x^{2}}{3 c}+\frac {125 a^{4} x^{4}}{12 c}-\frac {25 a^{6} x^{6}}{4 c}+\frac {6 a^{5} x^{5}}{c}-\frac {9 a^{7} x^{7}}{2 c}}{\left (a^{2} x^{2}-1\right )^{2} x^{3} c}+\frac {6 a^{3} \ln \left (x \right )}{c^{2}}-\frac {49 a^{3} \ln \left (a x -1\right )}{8 c^{2}}+\frac {a^{3} \ln \left (a x +1\right )}{8 c^{2}}\) | \(126\) |
parallelrisch | \(\frac {144 a^{5} \ln \left (x \right ) x^{5}-8-147 \ln \left (a x -1\right ) x^{5} a^{5}+3 \ln \left (a x +1\right ) x^{5} a^{5}-228 a^{5} x^{5}-288 \ln \left (x \right ) x^{4} a^{4}+294 \ln \left (a x -1\right ) x^{4} a^{4}-6 \ln \left (a x +1\right ) x^{4} a^{4}+306 a^{4} x^{4}+144 \ln \left (x \right ) x^{3} a^{3}-147 a^{3} \ln \left (a x -1\right ) x^{3}+3 \ln \left (a x +1\right ) x^{3} a^{3}-56 a^{2} x^{2}-8 a x}{24 c^{2} x^{3} \left (a x -1\right )^{2}}\) | \(160\) |
Input:
int((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)
Output:
1/c^2*(1/8*a^3*ln(a*x+1)+1/4*a^3/(a*x-1)^2-9/4*a^3/(a*x-1)-49/8*a^3*ln(a*x -1)-1/3/x^3-a/x^2-4*a^2/x+6*a^3*ln(x))
Time = 0.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.36 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx=-\frac {150 \, a^{4} x^{4} - 228 \, a^{3} x^{3} + 56 \, a^{2} x^{2} + 8 \, a x - 3 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (a x + 1\right ) + 147 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (a x - 1\right ) - 144 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (x\right ) + 8}{24 \, {\left (a^{2} c^{2} x^{5} - 2 \, a c^{2} x^{4} + c^{2} x^{3}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^2,x, algorithm="fricas ")
Output:
-1/24*(150*a^4*x^4 - 228*a^3*x^3 + 56*a^2*x^2 + 8*a*x - 3*(a^5*x^5 - 2*a^4 *x^4 + a^3*x^3)*log(a*x + 1) + 147*(a^5*x^5 - 2*a^4*x^4 + a^3*x^3)*log(a*x - 1) - 144*(a^5*x^5 - 2*a^4*x^4 + a^3*x^3)*log(x) + 8)/(a^2*c^2*x^5 - 2*a *c^2*x^4 + c^2*x^3)
Time = 0.40 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx=- \frac {75 a^{4} x^{4} - 114 a^{3} x^{3} + 28 a^{2} x^{2} + 4 a x + 4}{12 a^{2} c^{2} x^{5} - 24 a c^{2} x^{4} + 12 c^{2} x^{3}} - \frac {- 6 a^{3} \log {\left (x \right )} + \frac {49 a^{3} \log {\left (x - \frac {1}{a} \right )}}{8} - \frac {a^{3} \log {\left (x + \frac {1}{a} \right )}}{8}}{c^{2}} \] Input:
integrate((a*x+1)**2/(-a**2*x**2+1)/x**4/(-a**2*c*x**2+c)**2,x)
Output:
-(75*a**4*x**4 - 114*a**3*x**3 + 28*a**2*x**2 + 4*a*x + 4)/(12*a**2*c**2*x **5 - 24*a*c**2*x**4 + 12*c**2*x**3) - (-6*a**3*log(x) + 49*a**3*log(x - 1 /a)/8 - a**3*log(x + 1/a)/8)/c**2
Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx=\frac {a^{3} \log \left (a x + 1\right )}{8 \, c^{2}} - \frac {49 \, a^{3} \log \left (a x - 1\right )}{8 \, c^{2}} + \frac {6 \, a^{3} \log \left (x\right )}{c^{2}} - \frac {75 \, a^{4} x^{4} - 114 \, a^{3} x^{3} + 28 \, a^{2} x^{2} + 4 \, a x + 4}{12 \, {\left (a^{2} c^{2} x^{5} - 2 \, a c^{2} x^{4} + c^{2} x^{3}\right )}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^2,x, algorithm="maxima ")
Output:
1/8*a^3*log(a*x + 1)/c^2 - 49/8*a^3*log(a*x - 1)/c^2 + 6*a^3*log(x)/c^2 - 1/12*(75*a^4*x^4 - 114*a^3*x^3 + 28*a^2*x^2 + 4*a*x + 4)/(a^2*c^2*x^5 - 2* a*c^2*x^4 + c^2*x^3)
Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx=\frac {a^{3} \log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac {49 \, a^{3} \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac {6 \, a^{3} \log \left ({\left | x \right |}\right )}{c^{2}} - \frac {75 \, a^{4} x^{4} - 114 \, a^{3} x^{3} + 28 \, a^{2} x^{2} + 4 \, a x + 4}{12 \, {\left (a x - 1\right )}^{2} c^{2} x^{3}} \] Input:
integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^2,x, algorithm="giac")
Output:
1/8*a^3*log(abs(a*x + 1))/c^2 - 49/8*a^3*log(abs(a*x - 1))/c^2 + 6*a^3*log (abs(x))/c^2 - 1/12*(75*a^4*x^4 - 114*a^3*x^3 + 28*a^2*x^2 + 4*a*x + 4)/(( a*x - 1)^2*c^2*x^3)
Time = 23.97 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.91 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx=\frac {6\,a^3\,\ln \left (x\right )}{c^2}-\frac {\frac {25\,a^4\,x^4}{4}-\frac {19\,a^3\,x^3}{2}+\frac {7\,a^2\,x^2}{3}+\frac {a\,x}{3}+\frac {1}{3}}{a^2\,c^2\,x^5-2\,a\,c^2\,x^4+c^2\,x^3}-\frac {49\,a^3\,\ln \left (a\,x-1\right )}{8\,c^2}+\frac {a^3\,\ln \left (a\,x+1\right )}{8\,c^2} \] Input:
int(-(a*x + 1)^2/(x^4*(c - a^2*c*x^2)^2*(a^2*x^2 - 1)),x)
Output:
(6*a^3*log(x))/c^2 - ((a*x)/3 + (7*a^2*x^2)/3 - (19*a^3*x^3)/2 + (25*a^4*x ^4)/4 + 1/3)/(c^2*x^3 - 2*a*c^2*x^4 + a^2*c^2*x^5) - (49*a^3*log(a*x - 1)) /(8*c^2) + (a^3*log(a*x + 1))/(8*c^2)
Time = 0.15 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.52 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx=\frac {-147 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}+294 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-147 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+3 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}-6 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+3 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+144 \,\mathrm {log}\left (x \right ) a^{5} x^{5}-288 \,\mathrm {log}\left (x \right ) a^{4} x^{4}+144 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-75 a^{5} x^{5}+153 a^{3} x^{3}-56 a^{2} x^{2}-8 a x -8}{24 c^{2} x^{3} \left (a^{2} x^{2}-2 a x +1\right )} \] Input:
int((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^2,x)
Output:
( - 147*log(a*x - 1)*a**5*x**5 + 294*log(a*x - 1)*a**4*x**4 - 147*log(a*x - 1)*a**3*x**3 + 3*log(a*x + 1)*a**5*x**5 - 6*log(a*x + 1)*a**4*x**4 + 3*l og(a*x + 1)*a**3*x**3 + 144*log(x)*a**5*x**5 - 288*log(x)*a**4*x**4 + 144* log(x)*a**3*x**3 - 75*a**5*x**5 + 153*a**3*x**3 - 56*a**2*x**2 - 8*a*x - 8 )/(24*c**2*x**3*(a**2*x**2 - 2*a*x + 1))