Integrand size = 22, antiderivative size = 71 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {x}{c^2}+\frac {1}{3 a c^2 (1-a x)^3}-\frac {2}{a c^2 (1-a x)^2}+\frac {6}{a c^2 (1-a x)}+\frac {4 \log (1-a x)}{a c^2} \] Output:
x/c^2+1/3/a/c^2/(-a*x+1)^3-2/a/c^2/(-a*x+1)^2+6/a/c^2/(-a*x+1)+4*ln(-a*x+1 )/a/c^2
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {-13+27 a x-9 a^2 x^2-9 a^3 x^3+3 a^4 x^4+12 (-1+a x)^3 \log (1-a x)}{3 a c^2 (-1+a x)^3} \] Input:
Integrate[E^(4*ArcCoth[a*x])/(c - c/(a^2*x^2))^2,x]
Output:
(-13 + 27*a*x - 9*a^2*x^2 - 9*a^3*x^3 + 3*a^4*x^4 + 12*(-1 + a*x)^3*Log[1 - a*x])/(3*a*c^2*(-1 + a*x)^3)
Time = 0.79 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6717, 27, 6707, 6700, 49, 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.
\(\displaystyle \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle \int \frac {a^4 e^{4 \text {arctanh}(a x)}}{c^2 \left (a^2-\frac {1}{x^2}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^4 \int \frac {e^{4 \text {arctanh}(a x)}}{\left (a^2-\frac {1}{x^2}\right )^2}dx}{c^2}\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle \frac {a^4 \int \frac {e^{4 \text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^2}dx}{c^2}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {a^4 \int \frac {x^4}{(1-a x)^4}dx}{c^2}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a^4 \int \left (\frac {1}{a^4}+\frac {4}{a^4 (a x-1)}+\frac {6}{a^4 (a x-1)^2}+\frac {4}{a^4 (a x-1)^3}+\frac {1}{a^4 (a x-1)^4}\right )dx}{c^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^4 \left (\frac {6}{a^5 (1-a x)}-\frac {2}{a^5 (1-a x)^2}+\frac {1}{3 a^5 (1-a x)^3}+\frac {4 \log (1-a x)}{a^5}+\frac {x}{a^4}\right )}{c^2}\) |
Input:
Int[E^(4*ArcCoth[a*x])/(c - c/(a^2*x^2))^2,x]
Output:
(a^4*(x/a^4 + 1/(3*a^5*(1 - a*x)^3) - 2/(a^5*(1 - a*x)^2) + 6/(a^5*(1 - a* x)) + (4*Log[1 - a*x])/a^5))/c^2
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 [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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]
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 = 56, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {x}{c^{2}}+\frac {-6 a \,c^{2} x^{2}+10 c^{2} x -\frac {13 c^{2}}{3 a}}{c^{4} \left (a x -1\right )^{3}}+\frac {4 \ln \left (a x -1\right )}{a \,c^{2}}\) | \(56\) |
default | \(\frac {a^{4} \left (\frac {x}{a^{4}}-\frac {6}{a^{5} \left (a x -1\right )}+\frac {4 \ln \left (a x -1\right )}{a^{5}}-\frac {2}{a^{5} \left (a x -1\right )^{2}}-\frac {1}{3 a^{5} \left (a x -1\right )^{3}}\right )}{c^{2}}\) | \(61\) |
norman | \(\frac {\frac {a^{4} x^{5}}{c}+\frac {6 a \,x^{2}}{c}-\frac {4 x}{c}+\frac {8 a^{2} x^{3}}{3 c}-\frac {19 a^{3} x^{4}}{3 c}}{c \left (a x +1\right ) \left (a x -1\right )^{3}}+\frac {4 \ln \left (a x -1\right )}{a \,c^{2}}\) | \(82\) |
parallelrisch | \(\frac {3 a^{4} x^{4}+12 a^{3} \ln \left (a x -1\right ) x^{3}-22 a^{3} x^{3}-36 a^{2} \ln \left (a x -1\right ) x^{2}+30 a^{2} x^{2}+36 a \ln \left (a x -1\right ) x -12 a x -12 \ln \left (a x -1\right )}{3 \left (a x -1\right )^{3} c^{2} a}\) | \(91\) |
Input:
int(1/(a*x-1)^2*(a*x+1)^2/(c-c/a^2/x^2)^2,x,method=_RETURNVERBOSE)
Output:
x/c^2+(-6*a*c^2*x^2+10*c^2*x-13/3*c^2/a)/c^4/(a*x-1)^3+4/a/c^2*ln(a*x-1)
Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {3 \, a^{4} x^{4} - 9 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 27 \, a x + 12 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 13}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a^2/x^2)^2,x, algorithm="fricas")
Output:
1/3*(3*a^4*x^4 - 9*a^3*x^3 - 9*a^2*x^2 + 27*a*x + 12*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x - 1) - 13)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)
Time = 0.18 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=a^{4} \left (\frac {- 18 a^{2} x^{2} + 30 a x - 13}{3 a^{8} c^{2} x^{3} - 9 a^{7} c^{2} x^{2} + 9 a^{6} c^{2} x - 3 a^{5} c^{2}} + \frac {x}{a^{4} c^{2}} + \frac {4 \log {\left (a x - 1 \right )}}{a^{5} c^{2}}\right ) \] Input:
integrate(1/(a*x-1)**2*(a*x+1)**2/(c-c/a**2/x**2)**2,x)
Output:
a**4*((-18*a**2*x**2 + 30*a*x - 13)/(3*a**8*c**2*x**3 - 9*a**7*c**2*x**2 + 9*a**6*c**2*x - 3*a**5*c**2) + x/(a**4*c**2) + 4*log(a*x - 1)/(a**5*c**2) )
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=-\frac {18 \, a^{2} x^{2} - 30 \, a x + 13}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} + \frac {x}{c^{2}} + \frac {4 \, \log \left (a x - 1\right )}{a c^{2}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a^2/x^2)^2,x, algorithm="maxima")
Output:
-1/3*(18*a^2*x^2 - 30*a*x + 13)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2) + x/c^2 + 4*log(a*x - 1)/(a*c^2)
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.31 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {a x - 1}{a c^{2}} - \frac {4 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a c^{2}} - \frac {\frac {18 \, a^{5} c^{4}}{a x - 1} + \frac {6 \, a^{5} c^{4}}{{\left (a x - 1\right )}^{2}} + \frac {a^{5} c^{4}}{{\left (a x - 1\right )}^{3}}}{3 \, a^{6} c^{6}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a^2/x^2)^2,x, algorithm="giac")
Output:
(a*x - 1)/(a*c^2) - 4*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/(a*c^2) - 1/3 *(18*a^5*c^4/(a*x - 1) + 6*a^5*c^4/(a*x - 1)^2 + a^5*c^4/(a*x - 1)^3)/(a^6 *c^6)
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {6\,a\,x^2-10\,x+\frac {13}{3\,a}}{-a^3\,c^2\,x^3+3\,a^2\,c^2\,x^2-3\,a\,c^2\,x+c^2}+\frac {x}{c^2}+\frac {4\,\ln \left (a\,x-1\right )}{a\,c^2} \] Input:
int((a*x + 1)^2/((c - c/(a^2*x^2))^2*(a*x - 1)^2),x)
Output:
(6*a*x^2 - 10*x + 13/(3*a))/(c^2 + 3*a^2*c^2*x^2 - a^3*c^2*x^3 - 3*a*c^2*x ) + x/c^2 + (4*log(a*x - 1))/(a*c^2)
Time = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {12 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-36 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+36 \,\mathrm {log}\left (a x -1\right ) a x -12 \,\mathrm {log}\left (a x -1\right )+3 a^{4} x^{4}-12 a^{3} x^{3}+18 a x -10}{3 a \,c^{2} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int(1/(a*x-1)^2*(a*x+1)^2/(c-c/a^2/x^2)^2,x)
Output:
(12*log(a*x - 1)*a**3*x**3 - 36*log(a*x - 1)*a**2*x**2 + 36*log(a*x - 1)*a *x - 12*log(a*x - 1) + 3*a**4*x**4 - 12*a**3*x**3 + 18*a*x - 10)/(3*a*c**2 *(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))