Integrand size = 22, antiderivative size = 119 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {1}{64 a c^4 (1-a x)^2}+\frac {5}{64 a c^4 (1-a x)}-\frac {1}{32 a c^4 (1+a x)^4}-\frac {1}{16 a c^4 (1+a x)^3}-\frac {3}{32 a c^4 (1+a x)^2}-\frac {5}{32 a c^4 (1+a x)}+\frac {15 \text {arctanh}(a x)}{64 a c^4} \] Output:
1/64/a/c^4/(-a*x+1)^2+5/64/a/c^4/(-a*x+1)-1/32/a/c^4/(a*x+1)^4-1/16/a/c^4/ (a*x+1)^3-3/32/a/c^4/(a*x+1)^2-5/32/a/c^4/(a*x+1)+15/64*arctanh(a*x)/a/c^4
Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {-16+17 a x+50 a^2 x^2+10 a^3 x^3-30 a^4 x^4-15 a^5 x^5+15 (-1+a x)^2 (1+a x)^4 \text {arctanh}(a x)}{64 a (-1+a x)^2 (c+a c x)^4} \] Input:
Integrate[1/(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^4),x]
Output:
(-16 + 17*a*x + 50*a^2*x^2 + 10*a^3*x^3 - 30*a^4*x^4 - 15*a^5*x^5 + 15*(-1 + a*x)^2*(1 + a*x)^4*ArcTanh[a*x])/(64*a*(-1 + a*x)^2*(c + a*c*x)^4)
Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86, 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.
\(\displaystyle \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle \frac {\int \frac {1}{(1-a x)^3 (a x+1)^5}dx}{c^4}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {5}{64 (a x-1)^2}+\frac {5}{32 (a x+1)^2}-\frac {1}{32 (a x-1)^3}+\frac {3}{16 (a x+1)^3}+\frac {3}{16 (a x+1)^4}+\frac {1}{8 (a x+1)^5}-\frac {15}{64 \left (a^2 x^2-1\right )}\right )dx}{c^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {15 \text {arctanh}(a x)}{64 a}+\frac {5}{64 a (1-a x)}-\frac {5}{32 a (a x+1)}+\frac {1}{64 a (1-a x)^2}-\frac {3}{32 a (a x+1)^2}-\frac {1}{16 a (a x+1)^3}-\frac {1}{32 a (a x+1)^4}}{c^4}\) |
Input:
Int[1/(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^4),x]
Output:
(1/(64*a*(1 - a*x)^2) + 5/(64*a*(1 - a*x)) - 1/(32*a*(1 + a*x)^4) - 1/(16* a*(1 + a*x)^3) - 3/(32*a*(1 + a*x)^2) - 5/(32*a*(1 + a*x)) + (15*ArcTanh[a *x])/(64*a))/c^4
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 = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {-\frac {15 x^{5} a^{4}}{64}-\frac {15 a^{3} x^{4}}{32}+\frac {5 a^{2} x^{3}}{32}+\frac {25 a \,x^{2}}{32}+\frac {17 x}{64}-\frac {1}{4 a}}{\left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right )^{2} c^{4}}+\frac {15 \ln \left (-a x -1\right )}{128 a \,c^{4}}-\frac {15 \ln \left (a x -1\right )}{128 a \,c^{4}}\) | \(92\) |
default | \(\frac {-\frac {1}{32 a \left (a x +1\right )^{4}}-\frac {1}{16 a \left (a x +1\right )^{3}}-\frac {3}{32 a \left (a x +1\right )^{2}}-\frac {5}{32 a \left (a x +1\right )}+\frac {15 \ln \left (a x +1\right )}{128 a}+\frac {1}{64 a \left (a x -1\right )^{2}}-\frac {5}{64 \left (a x -1\right ) a}-\frac {15 \ln \left (a x -1\right )}{128 a}}{c^{4}}\) | \(100\) |
norman | \(\frac {\frac {5 a^{3} x^{4}}{4 c}-\frac {49 x}{64 c}-\frac {17 a \,x^{2}}{32 c}+\frac {103 a^{2} x^{3}}{64 c}-\frac {71 a^{4} x^{5}}{64 c}-\frac {31 a^{5} x^{6}}{32 c}+\frac {17 a^{6} x^{7}}{64 c}+\frac {a^{7} x^{8}}{4 c}}{c^{3} \left (a x +1\right )^{5} \left (a x -1\right )^{3}}-\frac {15 \ln \left (a x -1\right )}{128 a \,c^{4}}+\frac {15 \ln \left (a x +1\right )}{128 a \,c^{4}}\) | \(130\) |
parallelrisch | \(\frac {-108 a^{3} x^{3}+98 a x -92 a^{4} x^{4}+34 a^{5} x^{5}+32 x^{6} a^{6}+30 \ln \left (a x +1\right ) x a +15 \ln \left (a x +1\right )+68 a^{2} x^{2}-15 \ln \left (a x -1\right )-15 \ln \left (a x +1\right ) x^{2} a^{2}+15 \ln \left (a x -1\right ) x^{4} a^{4}-15 \ln \left (a x +1\right ) x^{4} a^{4}-60 \ln \left (a x +1\right ) x^{3} a^{3}-30 a \ln \left (a x -1\right ) x +60 a^{3} \ln \left (a x -1\right ) x^{3}-30 \ln \left (a x -1\right ) x^{5} a^{5}+30 \ln \left (a x +1\right ) x^{5} a^{5}+15 a^{2} \ln \left (a x -1\right ) x^{2}-15 \ln \left (a x -1\right ) x^{6} a^{6}+15 \ln \left (a x +1\right ) x^{6} a^{6}}{128 c^{4} \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right )^{2} a}\) | \(248\) |
Input:
int(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^4,x,method=_RETURNVERBOSE)
Output:
(-15/64*x^5*a^4-15/32*a^3*x^4+5/32*a^2*x^3+25/32*a*x^2+17/64*x-1/4/a)/(a*x +1)^2/(a^2*x^2-1)^2/c^4+15/128/a/c^4*ln(-a*x-1)-15/128/a/c^4*ln(a*x-1)
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (103) = 206\).
Time = 0.09 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.82 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {30 \, a^{5} x^{5} + 60 \, a^{4} x^{4} - 20 \, a^{3} x^{3} - 100 \, a^{2} x^{2} - 34 \, a x - 15 \, {\left (a^{6} x^{6} + 2 \, a^{5} x^{5} - a^{4} x^{4} - 4 \, a^{3} x^{3} - a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 15 \, {\left (a^{6} x^{6} + 2 \, a^{5} x^{5} - a^{4} x^{4} - 4 \, a^{3} x^{3} - a^{2} x^{2} + 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 32}{128 \, {\left (a^{7} c^{4} x^{6} + 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} + 2 \, a^{2} c^{4} x + a c^{4}\right )}} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^4,x, algorithm="fricas")
Output:
-1/128*(30*a^5*x^5 + 60*a^4*x^4 - 20*a^3*x^3 - 100*a^2*x^2 - 34*a*x - 15*( a^6*x^6 + 2*a^5*x^5 - a^4*x^4 - 4*a^3*x^3 - a^2*x^2 + 2*a*x + 1)*log(a*x + 1) + 15*(a^6*x^6 + 2*a^5*x^5 - a^4*x^4 - 4*a^3*x^3 - a^2*x^2 + 2*a*x + 1) *log(a*x - 1) + 32)/(a^7*c^4*x^6 + 2*a^6*c^4*x^5 - a^5*c^4*x^4 - 4*a^4*c^4 *x^3 - a^3*c^4*x^2 + 2*a^2*c^4*x + a*c^4)
Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=- \frac {15 a^{5} x^{5} + 30 a^{4} x^{4} - 10 a^{3} x^{3} - 50 a^{2} x^{2} - 17 a x + 16}{64 a^{7} c^{4} x^{6} + 128 a^{6} c^{4} x^{5} - 64 a^{5} c^{4} x^{4} - 256 a^{4} c^{4} x^{3} - 64 a^{3} c^{4} x^{2} + 128 a^{2} c^{4} x + 64 a c^{4}} - \frac {\frac {15 \log {\left (x - \frac {1}{a} \right )}}{128} - \frac {15 \log {\left (x + \frac {1}{a} \right )}}{128}}{a c^{4}} \] Input:
integrate(1/(a*x+1)**2*(-a**2*x**2+1)/(-a**2*c*x**2+c)**4,x)
Output:
-(15*a**5*x**5 + 30*a**4*x**4 - 10*a**3*x**3 - 50*a**2*x**2 - 17*a*x + 16) /(64*a**7*c**4*x**6 + 128*a**6*c**4*x**5 - 64*a**5*c**4*x**4 - 256*a**4*c* *4*x**3 - 64*a**3*c**4*x**2 + 128*a**2*c**4*x + 64*a*c**4) - (15*log(x - 1 /a)/128 - 15*log(x + 1/a)/128)/(a*c**4)
Time = 0.03 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {15 \, a^{5} x^{5} + 30 \, a^{4} x^{4} - 10 \, a^{3} x^{3} - 50 \, a^{2} x^{2} - 17 \, a x + 16}{64 \, {\left (a^{7} c^{4} x^{6} + 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} + 2 \, a^{2} c^{4} x + a c^{4}\right )}} + \frac {15 \, \log \left (a x + 1\right )}{128 \, a c^{4}} - \frac {15 \, \log \left (a x - 1\right )}{128 \, a c^{4}} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^4,x, algorithm="maxima")
Output:
-1/64*(15*a^5*x^5 + 30*a^4*x^4 - 10*a^3*x^3 - 50*a^2*x^2 - 17*a*x + 16)/(a ^7*c^4*x^6 + 2*a^6*c^4*x^5 - a^5*c^4*x^4 - 4*a^4*c^4*x^3 - a^3*c^4*x^2 + 2 *a^2*c^4*x + a*c^4) + 15/128*log(a*x + 1)/(a*c^4) - 15/128*log(a*x - 1)/(a *c^4)
Time = 0.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {15 \, \log \left ({\left | -\frac {2}{a x + 1} + 1 \right |}\right )}{128 \, a c^{4}} + \frac {\frac {24}{a x + 1} - 11}{256 \, a c^{4} {\left (\frac {2}{a x + 1} - 1\right )}^{2}} - \frac {\frac {5 \, a^{11} c^{12}}{a x + 1} + \frac {3 \, a^{11} c^{12}}{{\left (a x + 1\right )}^{2}} + \frac {2 \, a^{11} c^{12}}{{\left (a x + 1\right )}^{3}} + \frac {a^{11} c^{12}}{{\left (a x + 1\right )}^{4}}}{32 \, a^{12} c^{16}} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^4,x, algorithm="giac")
Output:
-15/128*log(abs(-2/(a*x + 1) + 1))/(a*c^4) + 1/256*(24/(a*x + 1) - 11)/(a* c^4*(2/(a*x + 1) - 1)^2) - 1/32*(5*a^11*c^12/(a*x + 1) + 3*a^11*c^12/(a*x + 1)^2 + 2*a^11*c^12/(a*x + 1)^3 + a^11*c^12/(a*x + 1)^4)/(a^12*c^16)
Time = 27.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\frac {17\,x}{64}+\frac {25\,a\,x^2}{32}-\frac {1}{4\,a}+\frac {5\,a^2\,x^3}{32}-\frac {15\,a^3\,x^4}{32}-\frac {15\,a^4\,x^5}{64}}{a^6\,c^4\,x^6+2\,a^5\,c^4\,x^5-a^4\,c^4\,x^4-4\,a^3\,c^4\,x^3-a^2\,c^4\,x^2+2\,a\,c^4\,x+c^4}+\frac {15\,\mathrm {atanh}\left (a\,x\right )}{64\,a\,c^4} \] Input:
int(-(a^2*x^2 - 1)/((c - a^2*c*x^2)^4*(a*x + 1)^2),x)
Output:
((17*x)/64 + (25*a*x^2)/32 - 1/(4*a) + (5*a^2*x^3)/32 - (15*a^3*x^4)/32 - (15*a^4*x^5)/64)/(c^4 - a^2*c^4*x^2 - 4*a^3*c^4*x^3 - a^4*c^4*x^4 + 2*a^5* c^4*x^5 + a^6*c^4*x^6 + 2*a*c^4*x) + (15*atanh(a*x))/(64*a*c^4)
Time = 0.14 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.26 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {-15 \,\mathrm {log}\left (a x -1\right ) a^{6} x^{6}-30 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}+15 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+60 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+15 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-30 \,\mathrm {log}\left (a x -1\right ) a x -15 \,\mathrm {log}\left (a x -1\right )+15 \,\mathrm {log}\left (a x +1\right ) a^{6} x^{6}+30 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}-15 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}-60 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-15 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+30 \,\mathrm {log}\left (a x +1\right ) a x +15 \,\mathrm {log}\left (a x +1\right )+15 a^{6} x^{6}-75 a^{4} x^{4}-40 a^{3} x^{3}+85 a^{2} x^{2}+64 a x -17}{128 a \,c^{4} \left (a^{6} x^{6}+2 a^{5} x^{5}-a^{4} x^{4}-4 a^{3} x^{3}-a^{2} x^{2}+2 a x +1\right )} \] Input:
int(1/(a*x+1)^2*(-a^2*x^2+1)/(-a^2*c*x^2+c)^4,x)
Output:
( - 15*log(a*x - 1)*a**6*x**6 - 30*log(a*x - 1)*a**5*x**5 + 15*log(a*x - 1 )*a**4*x**4 + 60*log(a*x - 1)*a**3*x**3 + 15*log(a*x - 1)*a**2*x**2 - 30*l og(a*x - 1)*a*x - 15*log(a*x - 1) + 15*log(a*x + 1)*a**6*x**6 + 30*log(a*x + 1)*a**5*x**5 - 15*log(a*x + 1)*a**4*x**4 - 60*log(a*x + 1)*a**3*x**3 - 15*log(a*x + 1)*a**2*x**2 + 30*log(a*x + 1)*a*x + 15*log(a*x + 1) + 15*a** 6*x**6 - 75*a**4*x**4 - 40*a**3*x**3 + 85*a**2*x**2 + 64*a*x - 17)/(128*a* c**4*(a**6*x**6 + 2*a**5*x**5 - a**4*x**4 - 4*a**3*x**3 - a**2*x**2 + 2*a* x + 1))