Integrand size = 22, antiderivative size = 84 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {1}{16 a c^3 (1-a x)}+\frac {1}{12 a c^3 (1+a x)^3}+\frac {1}{8 a c^3 (1+a x)^2}+\frac {3}{16 a c^3 (1+a x)}-\frac {\text {arctanh}(a x)}{4 a c^3} \] Output:
-1/16/a/c^3/(-a*x+1)+1/12/a/c^3/(a*x+1)^3+1/8/a/c^3/(a*x+1)^2+3/16/a/c^3/( a*x+1)-1/4*arctanh(a*x)/a/c^3
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {-4+a x+6 a^2 x^2+3 a^3 x^3-3 (-1+a x) (1+a x)^3 \text {arctanh}(a x)}{12 a (-1+a x) (c+a c x)^3} \] Input:
Integrate[1/(E^(2*ArcCoth[a*x])*(c - a^2*c*x^2)^3),x]
Output:
(-4 + a*x + 6*a^2*x^2 + 3*a^3*x^3 - 3*(-1 + a*x)*(1 + a*x)^3*ArcTanh[a*x]) /(12*a*(-1 + a*x)*(c + a*c*x)^3)
Time = 0.60 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6717, 27, 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.
\(\displaystyle \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {e^{-2 \text {arctanh}(a x)}}{c^3 \left (1-a^2 x^2\right )^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {e^{-2 \text {arctanh}(a x)}}{\left (1-a^2 x^2\right )^3}dx}{c^3}\) |
\(\Big \downarrow \) 6690 |
\(\displaystyle -\frac {\int \frac {1}{(1-a x)^2 (a x+1)^4}dx}{c^3}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -\frac {\int \left (\frac {1}{16 (a x-1)^2}+\frac {3}{16 (a x+1)^2}+\frac {1}{4 (a x+1)^3}+\frac {1}{4 (a x+1)^4}-\frac {1}{4 \left (a^2 x^2-1\right )}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\text {arctanh}(a x)}{4 a}+\frac {1}{16 a (1-a x)}-\frac {3}{16 a (a x+1)}-\frac {1}{8 a (a x+1)^2}-\frac {1}{12 a (a x+1)^3}}{c^3}\) |
Input:
Int[1/(E^(2*ArcCoth[a*x])*(c - a^2*c*x^2)^3),x]
Output:
-((1/(16*a*(1 - a*x)) - 1/(12*a*(1 + a*x)^3) - 1/(8*a*(1 + a*x)^2) - 3/(16 *a*(1 + a*x)) + ArcTanh[a*x]/(4*a))/c^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2) Int[ u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {\frac {1}{12 a \left (a x +1\right )^{3}}+\frac {1}{8 a \left (a x +1\right )^{2}}+\frac {3}{16 a \left (a x +1\right )}-\frac {\ln \left (a x +1\right )}{8 a}+\frac {1}{16 \left (a x -1\right ) a}+\frac {\ln \left (a x -1\right )}{8 a}}{c^{3}}\) | \(76\) |
risch | \(\frac {\frac {a^{2} x^{3}}{4}+\frac {a \,x^{2}}{2}+\frac {x}{12}-\frac {1}{3 a}}{\left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right ) c^{3}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{3}}+\frac {\ln \left (-a x +1\right )}{8 a \,c^{3}}\) | \(76\) |
norman | \(\frac {-\frac {3 x}{4 c}+\frac {a \,x^{2}}{4 c}+\frac {11 a^{2} x^{3}}{12 c}-\frac {a^{3} x^{4}}{12 c}-\frac {a^{4} x^{5}}{3 c}}{\left (a x +1\right )^{3} c^{2} \left (a x -1\right )^{2}}+\frac {\ln \left (a x -1\right )}{8 a \,c^{3}}-\frac {\ln \left (a x +1\right )}{8 a \,c^{3}}\) | \(97\) |
parallelrisch | \(\frac {3 \ln \left (a x -1\right ) x^{4} a^{4}-3 \ln \left (a x +1\right ) x^{4} a^{4}-8 a^{4} x^{4}+6 a^{3} \ln \left (a x -1\right ) x^{3}-6 \ln \left (a x +1\right ) x^{3} a^{3}-10 a^{3} x^{3}+12 a^{2} x^{2}-6 a \ln \left (a x -1\right ) x +6 \ln \left (a x +1\right ) x a +18 a x -3 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )}{24 c^{3} \left (a x +1\right )^{2} \left (a^{2} x^{2}-1\right ) a}\) | \(148\) |
Input:
int((a*x-1)/(a*x+1)/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)
Output:
1/c^3*(1/12/a/(a*x+1)^3+1/8/a/(a*x+1)^2+3/16/a/(a*x+1)-1/8*ln(a*x+1)/a+1/1 6/(a*x-1)/a+1/8/a*ln(a*x-1))
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {6 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 2 \, a x - 3 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \log \left (a x - 1\right ) - 8}{24 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a^2*c*x^2+c)^3,x, algorithm="fricas")
Output:
1/24*(6*a^3*x^3 + 12*a^2*x^2 + 2*a*x - 3*(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) - 8)/(a^5 *c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3)
Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=- \frac {- 3 a^{3} x^{3} - 6 a^{2} x^{2} - a x + 4}{12 a^{5} c^{3} x^{4} + 24 a^{4} c^{3} x^{3} - 24 a^{2} c^{3} x - 12 a c^{3}} - \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{8} + \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a c^{3}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a**2*c*x**2+c)**3,x)
Output:
-(-3*a**3*x**3 - 6*a**2*x**2 - a*x + 4)/(12*a**5*c**3*x**4 + 24*a**4*c**3* x**3 - 24*a**2*c**3*x - 12*a*c**3) - (-log(x - 1/a)/8 + log(x + 1/a)/8)/(a *c**3)
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + a x - 4}{12 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}} - \frac {\log \left (a x + 1\right )}{8 \, a c^{3}} + \frac {\log \left (a x - 1\right )}{8 \, a c^{3}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a^2*c*x^2+c)^3,x, algorithm="maxima")
Output:
1/12*(3*a^3*x^3 + 6*a^2*x^2 + a*x - 4)/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^ 2*c^3*x - a*c^3) - 1/8*log(a*x + 1)/(a*c^3) + 1/8*log(a*x - 1)/(a*c^3)
Time = 0.13 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{3}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{3}} + \frac {3 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + a x - 4}{12 \, {\left (a x + 1\right )}^{3} {\left (a x - 1\right )} a c^{3}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a^2*c*x^2+c)^3,x, algorithm="giac")
Output:
-1/8*log(abs(a*x + 1))/(a*c^3) + 1/8*log(abs(a*x - 1))/(a*c^3) + 1/12*(3*a ^3*x^3 + 6*a^2*x^2 + a*x - 4)/((a*x + 1)^3*(a*x - 1)*a*c^3)
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=-\frac {\frac {x}{12}+\frac {a\,x^2}{2}-\frac {1}{3\,a}+\frac {a^2\,x^3}{4}}{-a^4\,c^3\,x^4-2\,a^3\,c^3\,x^3+2\,a\,c^3\,x+c^3}-\frac {\mathrm {atanh}\left (a\,x\right )}{4\,a\,c^3} \] Input:
int((a*x - 1)/((c - a^2*c*x^2)^3*(a*x + 1)),x)
Output:
- (x/12 + (a*x^2)/2 - 1/(3*a) + (a^2*x^3)/4)/(c^3 - 2*a^3*c^3*x^3 - a^4*c^ 3*x^4 + 2*a*c^3*x) - atanh(a*x)/(4*a*c^3)
Time = 0.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.73 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {3 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+6 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-6 \,\mathrm {log}\left (a x -1\right ) a x -3 \,\mathrm {log}\left (a x -1\right )-3 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}-6 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+6 \,\mathrm {log}\left (a x +1\right ) a x +3 \,\mathrm {log}\left (a x +1\right )-3 a^{4} x^{4}+12 a^{2} x^{2}+8 a x -5}{24 a \,c^{3} \left (a^{4} x^{4}+2 a^{3} x^{3}-2 a x -1\right )} \] Input:
int((a*x-1)/(a*x+1)/(-a^2*c*x^2+c)^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) - 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) - 3*a**4*x**4 + 12*a**2*x**2 + 8*a*x - 5)/(24*a*c**3*(a**4*x**4 + 2*a**3*x**3 - 2*a*x - 1))