Integrand size = 22, antiderivative size = 109 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {x}{c^3}-\frac {1}{16 a c^3 (1-a x)}+\frac {1}{12 a c^3 (1+a x)^3}-\frac {5}{8 a c^3 (1+a x)^2}+\frac {39}{16 a c^3 (1+a x)}-\frac {\log (1-a x)}{4 a c^3}+\frac {9 \log (1+a x)}{4 a c^3} \] Output:
-x/c^3-1/16/a/c^3/(-a*x+1)+1/12/a/c^3/(a*x+1)^3-5/8/a/c^3/(a*x+1)^2+39/16/ a/c^3/(a*x+1)-1/4*ln(-a*x+1)/a/c^3+9/4*ln(a*x+1)/a/c^3
Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {-2 \left (11+7 a x-24 a^2 x^2-15 a^3 x^3+12 a^4 x^4+6 a^5 x^5\right )-3 (-1+a x) (1+a x)^3 \log (1-a x)+27 (-1+a x) (1+a x)^3 \log (1+a x)}{12 a (-1+a x) (c+a c x)^3} \] Input:
Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))^3),x]
Output:
(-2*(11 + 7*a*x - 24*a^2*x^2 - 15*a^3*x^3 + 12*a^4*x^4 + 6*a^5*x^5) - 3*(- 1 + a*x)*(1 + a*x)^3*Log[1 - a*x] + 27*(-1 + a*x)*(1 + a*x)^3*Log[1 + a*x] )/(12*a*(-1 + a*x)*(c + a*c*x)^3)
Time = 0.44 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6707, 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.
\(\displaystyle \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle -\frac {a^6 \int \frac {e^{-2 \text {arctanh}(a x)} x^6}{\left (1-a^2 x^2\right )^3}dx}{c^3}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle -\frac {a^6 \int \frac {x^6}{(1-a x)^2 (a x+1)^4}dx}{c^3}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a^6 \int \left (-\frac {9}{4 a^6 (a x+1)}+\frac {39}{16 a^6 (a x+1)^2}-\frac {5}{4 a^6 (a x+1)^3}+\frac {1}{4 a^6 (a x+1)^4}+\frac {1}{a^6}+\frac {1}{4 a^6 (a x-1)}+\frac {1}{16 a^6 (a x-1)^2}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^6 \left (\frac {1}{16 a^7 (1-a x)}-\frac {39}{16 a^7 (a x+1)}+\frac {5}{8 a^7 (a x+1)^2}-\frac {1}{12 a^7 (a x+1)^3}+\frac {\log (1-a x)}{4 a^7}-\frac {9 \log (a x+1)}{4 a^7}+\frac {x}{a^6}\right )}{c^3}\) |
Input:
Int[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))^3),x]
Output:
-((a^6*(x/a^6 + 1/(16*a^7*(1 - a*x)) - 1/(12*a^7*(1 + a*x)^3) + 5/(8*a^7*( 1 + a*x)^2) - 39/(16*a^7*(1 + a*x)) + Log[1 - a*x]/(4*a^7) - (9*Log[1 + a* x])/(4*a^7)))/c^3)
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])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {a^{6} \left (-\frac {x}{a^{6}}+\frac {9 \ln \left (a x +1\right )}{4 a^{7}}+\frac {1}{12 a^{7} \left (a x +1\right )^{3}}-\frac {5}{8 a^{7} \left (a x +1\right )^{2}}+\frac {39}{16 a^{7} \left (a x +1\right )}+\frac {1}{16 a^{7} \left (a x -1\right )}-\frac {\ln \left (a x -1\right )}{4 a^{7}}\right )}{c^{3}}\) | \(85\) |
risch | \(-\frac {x}{c^{3}}+\frac {\frac {5 a^{2} c^{3} x^{3}}{2}+2 a \,c^{3} x^{2}-\frac {13 c^{3} x}{6}-\frac {11 c^{3}}{6 a}}{c^{6} \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right )}+\frac {9 \ln \left (-a x -1\right )}{4 a \,c^{3}}-\frac {\ln \left (a x -1\right )}{4 a \,c^{3}}\) | \(94\) |
norman | \(\frac {-\frac {a^{6} x^{7}}{c}-\frac {5 x}{2 c}-\frac {4 a \,x^{2}}{c}+\frac {11 a^{2} x^{3}}{3 c}+\frac {47 a^{3} x^{4}}{6 c}-\frac {a^{4} x^{5}}{6 c}-\frac {23 a^{5} x^{6}}{6 c}}{c^{2} \left (a x -1\right )^{2} \left (a x +1\right )^{4}}-\frac {\ln \left (a x -1\right )}{4 a \,c^{3}}+\frac {9 \ln \left (a x +1\right )}{4 a \,c^{3}}\) | \(119\) |
parallelrisch | \(\frac {-12 a^{5} x^{5}-3 \ln \left (a x -1\right ) x^{4} a^{4}+27 \ln \left (a x +1\right ) x^{4} a^{4}-46 a^{4} x^{4}-6 a^{3} \ln \left (a x -1\right ) x^{3}+54 \ln \left (a x +1\right ) x^{3} a^{3}-14 a^{3} x^{3}+48 a^{2} x^{2}+6 a \ln \left (a x -1\right ) x -54 \ln \left (a x +1\right ) x a +30 a x +3 \ln \left (a x -1\right )-27 \ln \left (a x +1\right )}{12 c^{3} \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right ) a}\) | \(156\) |
Input:
int(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^3,x,method=_RETURNVERBOSE)
Output:
a^6/c^3*(-x/a^6+9/4*ln(a*x+1)/a^7+1/12/a^7/(a*x+1)^3-5/8/a^7/(a*x+1)^2+39/ 16/a^7/(a*x+1)+1/16/a^7/(a*x-1)-1/4/a^7*ln(a*x-1))
Time = 0.07 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {12 \, a^{5} x^{5} + 24 \, a^{4} x^{4} - 30 \, a^{3} x^{3} - 48 \, a^{2} x^{2} + 14 \, a x - 27 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 22}{12 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^3,x, algorithm="fricas")
Output:
-1/12*(12*a^5*x^5 + 24*a^4*x^4 - 30*a^3*x^3 - 48*a^2*x^2 + 14*a*x - 27*(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) + 22)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a* c^3)
Time = 0.34 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=- a^{6} \left (\frac {- 15 a^{3} x^{3} - 12 a^{2} x^{2} + 13 a x + 11}{6 a^{11} c^{3} x^{4} + 12 a^{10} c^{3} x^{3} - 12 a^{8} c^{3} x - 6 a^{7} c^{3}} + \frac {x}{a^{6} c^{3}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{4} - \frac {9 \log {\left (x + \frac {1}{a} \right )}}{4}}{a^{7} c^{3}}\right ) \] Input:
integrate(1/(a*x+1)**2*(-a**2*x**2+1)/(c-c/a**2/x**2)**3,x)
Output:
-a**6*((-15*a**3*x**3 - 12*a**2*x**2 + 13*a*x + 11)/(6*a**11*c**3*x**4 + 1 2*a**10*c**3*x**3 - 12*a**8*c**3*x - 6*a**7*c**3) + x/(a**6*c**3) + (log(x - 1/a)/4 - 9*log(x + 1/a)/4)/(a**7*c**3))
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {15 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 13 \, a x - 11}{6 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} - \frac {x}{c^{3}} + \frac {9 \, \log \left (a x + 1\right )}{4 \, a c^{3}} - \frac {\log \left (a x - 1\right )}{4 \, a c^{3}} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^3,x, algorithm="maxima")
Output:
1/6*(15*a^3*x^3 + 12*a^2*x^2 - 13*a*x - 11)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3) - x/c^3 + 9/4*log(a*x + 1)/(a*c^3) - 1/4*log(a*x - 1 )/(a*c^3)
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {{\left (a x + 1\right )} {\left (\frac {65}{a x + 1} - 32\right )}}{32 \, a c^{3} {\left (\frac {2}{a x + 1} - 1\right )}} - \frac {2 \, \log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a c^{3}} - \frac {\log \left ({\left | -\frac {2}{a x + 1} + 1 \right |}\right )}{4 \, a c^{3}} + \frac {\frac {117 \, a^{11} c^{6}}{a x + 1} - \frac {30 \, a^{11} c^{6}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, a^{11} c^{6}}{{\left (a x + 1\right )}^{3}}}{48 \, a^{12} c^{9}} \] Input:
integrate(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^3,x, algorithm="giac")
Output:
-1/32*(a*x + 1)*(65/(a*x + 1) - 32)/(a*c^3*(2/(a*x + 1) - 1)) - 2*log(abs( a*x + 1)/((a*x + 1)^2*abs(a)))/(a*c^3) - 1/4*log(abs(-2/(a*x + 1) + 1))/(a *c^3) + 1/48*(117*a^11*c^6/(a*x + 1) - 30*a^11*c^6/(a*x + 1)^2 + 4*a^11*c^ 6/(a*x + 1)^3)/(a^12*c^9)
Time = 14.38 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {\frac {13\,x}{6}-2\,a\,x^2+\frac {11}{6\,a}-\frac {5\,a^2\,x^3}{2}}{-a^4\,c^3\,x^4-2\,a^3\,c^3\,x^3+2\,a\,c^3\,x+c^3}-\frac {x}{c^3}-\frac {\ln \left (a\,x-1\right )}{4\,a\,c^3}+\frac {9\,\ln \left (a\,x+1\right )}{4\,a\,c^3} \] Input:
int(-(a^2*x^2 - 1)/((c - c/(a^2*x^2))^3*(a*x + 1)^2),x)
Output:
((13*x)/6 - 2*a*x^2 + 11/(6*a) - (5*a^2*x^3)/2)/(c^3 - 2*a^3*c^3*x^3 - a^4 *c^3*x^4 + 2*a*c^3*x) - x/c^3 - log(a*x - 1)/(4*a*c^3) + (9*log(a*x + 1))/ (4*a*c^3)
Time = 0.15 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {-3 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-6 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+6 \,\mathrm {log}\left (a x -1\right ) a x +3 \,\mathrm {log}\left (a x -1\right )+27 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+54 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-54 \,\mathrm {log}\left (a x +1\right ) a x -27 \,\mathrm {log}\left (a x +1\right )-12 a^{5} x^{5}-39 a^{4} x^{4}+48 a^{2} x^{2}+16 a x -7}{12 a \,c^{3} \left (a^{4} x^{4}+2 a^{3} x^{3}-2 a x -1\right )} \] Input:
int(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a^2/x^2)^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) + 27*log(a*x + 1)*a**4*x**4 + 54*log(a*x + 1)*a**3*x** 3 - 54*log(a*x + 1)*a*x - 27*log(a*x + 1) - 12*a**5*x**5 - 39*a**4*x**4 + 48*a**2*x**2 + 16*a*x - 7)/(12*a*c**3*(a**4*x**4 + 2*a**3*x**3 - 2*a*x - 1 ))