Integrand size = 22, antiderivative size = 89 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {x}{c^3}-\frac {1}{a c^3 (1-a x)^4}+\frac {16}{3 a c^3 (1-a x)^3}-\frac {25}{2 a c^3 (1-a x)^2}+\frac {19}{a c^3 (1-a x)}+\frac {7 \log (1-a x)}{a c^3} \] Output:
x/c^3-1/a/c^3/(-a*x+1)^4+16/3/a/c^3/(-a*x+1)^3-25/2/a/c^3/(-a*x+1)^2+19/a/ c^3/(-a*x+1)+7*ln(-a*x+1)/a/c^3
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.80 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {65-218 a x+243 a^2 x^2-78 a^3 x^3-24 a^4 x^4+6 a^5 x^5+42 (-1+a x)^4 \log (1-a x)}{6 a c^3 (-1+a x)^4} \] Input:
Integrate[E^(4*ArcCoth[a*x])/(c - c/(a*x))^3,x]
Output:
(65 - 218*a*x + 243*a^2*x^2 - 78*a^3*x^3 - 24*a^4*x^4 + 6*a^5*x^5 + 42*(-1 + a*x)^4*Log[1 - a*x])/(6*a*c^3*(-1 + a*x)^4)
Time = 0.79 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6717, 27, 6681, 6679, 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.
\(\displaystyle \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle \int \frac {a^3 e^{4 \text {arctanh}(a x)}}{c^3 \left (a-\frac {1}{x}\right )^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 \int \frac {e^{4 \text {arctanh}(a x)}}{\left (a-\frac {1}{x}\right )^3}dx}{c^3}\) |
\(\Big \downarrow \) 6681 |
\(\displaystyle -\frac {a^3 \int \frac {e^{4 \text {arctanh}(a x)} x^3}{(1-a x)^3}dx}{c^3}\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle -\frac {a^3 \int \frac {x^3 (a x+1)^2}{(1-a x)^5}dx}{c^3}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a^3 \int \left (-\frac {1}{a^3}-\frac {7}{a^3 (a x-1)}-\frac {19}{a^3 (a x-1)^2}-\frac {25}{a^3 (a x-1)^3}-\frac {16}{a^3 (a x-1)^4}-\frac {4}{a^3 (a x-1)^5}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \left (-\frac {19}{a^4 (1-a x)}+\frac {25}{2 a^4 (1-a x)^2}-\frac {16}{3 a^4 (1-a x)^3}+\frac {1}{a^4 (1-a x)^4}-\frac {7 \log (1-a x)}{a^4}-\frac {x}{a^3}\right )}{c^3}\) |
Input:
Int[E^(4*ArcCoth[a*x])/(c - c/(a*x))^3,x]
Output:
-((a^3*(-(x/a^3) + 1/(a^4*(1 - a*x)^4) - 16/(3*a^4*(1 - a*x)^3) + 25/(2*a^ 4*(1 - a*x)^2) - 19/(a^4*(1 - a*x)) - (7*Log[1 - a*x])/a^4))/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_)*((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_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol ] :> Simp[c^p Int[u*(1 + d*(x/c))^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] , x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol ] :> Simp[d^p Int[u*(1 + c*(x/d))^p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; F reeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 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 = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {x}{c^{3}}+\frac {-19 a^{2} c^{3} x^{3}+\frac {89 a \,c^{3} x^{2}}{2}-\frac {112 c^{3} x}{3}+\frac {65 c^{3}}{6 a}}{c^{6} \left (a x -1\right )^{4}}+\frac {7 \ln \left (a x -1\right )}{a \,c^{3}}\) | \(67\) |
default | \(\frac {a^{3} \left (\frac {x}{a^{3}}-\frac {19}{a^{4} \left (a x -1\right )}+\frac {7 \ln \left (a x -1\right )}{a^{4}}-\frac {1}{a^{4} \left (a x -1\right )^{4}}-\frac {25}{2 a^{4} \left (a x -1\right )^{2}}-\frac {16}{3 a^{4} \left (a x -1\right )^{3}}\right )}{c^{3}}\) | \(73\) |
norman | \(\frac {\frac {a^{4} x^{5}}{c}+\frac {7 x}{c}-\frac {49 a \,x^{2}}{2 c}+\frac {91 a^{2} x^{3}}{3 c}-\frac {89 a^{3} x^{4}}{6 c}}{\left (a x -1\right )^{4} c^{2}}+\frac {7 \ln \left (a x -1\right )}{a \,c^{3}}\) | \(75\) |
parallelrisch | \(\frac {6 a^{5} x^{5}+42 \ln \left (a x -1\right ) x^{4} a^{4}-89 a^{4} x^{4}-168 a^{3} \ln \left (a x -1\right ) x^{3}+182 a^{3} x^{3}+252 a^{2} \ln \left (a x -1\right ) x^{2}-147 a^{2} x^{2}-168 a \ln \left (a x -1\right ) x +42 a x +42 \ln \left (a x -1\right )}{6 \left (a x -1\right )^{4} c^{3} a}\) | \(113\) |
Input:
int(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^3,x,method=_RETURNVERBOSE)
Output:
x/c^3+(-19*a^2*c^3*x^3+89/2*a*c^3*x^2-112/3*c^3*x+65/6*c^3/a)/c^6/(a*x-1)^ 4+7/a/c^3*ln(a*x-1)
Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.42 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {6 \, a^{5} x^{5} - 24 \, a^{4} x^{4} - 78 \, a^{3} x^{3} + 243 \, a^{2} x^{2} - 218 \, a x + 42 \, {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x + 1\right )} \log \left (a x - 1\right ) + 65}{6 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^3,x, algorithm="fricas")
Output:
1/6*(6*a^5*x^5 - 24*a^4*x^4 - 78*a^3*x^3 + 243*a^2*x^2 - 218*a*x + 42*(a^4 *x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x + 1)*log(a*x - 1) + 65)/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {- 114 a^{3} x^{3} + 267 a^{2} x^{2} - 224 a x + 65}{6 a^{5} c^{3} x^{4} - 24 a^{4} c^{3} x^{3} + 36 a^{3} c^{3} x^{2} - 24 a^{2} c^{3} x + 6 a c^{3}} + \frac {x}{c^{3}} + \frac {7 \log {\left (a x - 1 \right )}}{a c^{3}} \] Input:
integrate(1/(a*x-1)**2*(a*x+1)**2/(c-c/a/x)**3,x)
Output:
(-114*a**3*x**3 + 267*a**2*x**2 - 224*a*x + 65)/(6*a**5*c**3*x**4 - 24*a** 4*c**3*x**3 + 36*a**3*c**3*x**2 - 24*a**2*c**3*x + 6*a*c**3) + x/c**3 + 7* log(a*x - 1)/(a*c**3)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=-\frac {114 \, a^{3} x^{3} - 267 \, a^{2} x^{2} + 224 \, a x - 65}{6 \, {\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, a^{2} c^{3} x + a c^{3}\right )}} + \frac {x}{c^{3}} + \frac {7 \, \log \left (a x - 1\right )}{a c^{3}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^3,x, algorithm="maxima")
Output:
-1/6*(114*a^3*x^3 - 267*a^2*x^2 + 224*a*x - 65)/(a^5*c^3*x^4 - 4*a^4*c^3*x ^3 + 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3) + x/c^3 + 7*log(a*x - 1)/(a*c^3)
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.22 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {a x - 1}{a c^{3}} - \frac {7 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a c^{3}} - \frac {\frac {114 \, a^{7} c^{9}}{a x - 1} + \frac {75 \, a^{7} c^{9}}{{\left (a x - 1\right )}^{2}} + \frac {32 \, a^{7} c^{9}}{{\left (a x - 1\right )}^{3}} + \frac {6 \, a^{7} c^{9}}{{\left (a x - 1\right )}^{4}}}{6 \, a^{8} c^{12}} \] Input:
integrate(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^3,x, algorithm="giac")
Output:
(a*x - 1)/(a*c^3) - 7*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/(a*c^3) - 1/6 *(114*a^7*c^9/(a*x - 1) + 75*a^7*c^9/(a*x - 1)^2 + 32*a^7*c^9/(a*x - 1)^3 + 6*a^7*c^9/(a*x - 1)^4)/(a^8*c^12)
Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {x}{c^3}-\frac {\frac {112\,x}{3}-\frac {89\,a\,x^2}{2}-\frac {65}{6\,a}+19\,a^2\,x^3}{a^4\,c^3\,x^4-4\,a^3\,c^3\,x^3+6\,a^2\,c^3\,x^2-4\,a\,c^3\,x+c^3}+\frac {7\,\ln \left (a\,x-1\right )}{a\,c^3} \] Input:
int((a*x + 1)^2/((c - c/(a*x))^3*(a*x - 1)^2),x)
Output:
x/c^3 - ((112*x)/3 - (89*a*x^2)/2 - 65/(6*a) + 19*a^2*x^3)/(c^3 + 6*a^2*c^ 3*x^2 - 4*a^3*c^3*x^3 + a^4*c^3*x^4 - 4*a*c^3*x) + (7*log(a*x - 1))/(a*c^3 )
Time = 0.17 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.45 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {84 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-336 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+504 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-336 \,\mathrm {log}\left (a x -1\right ) a x +84 \,\mathrm {log}\left (a x -1\right )+12 a^{5} x^{5}-87 a^{4} x^{4}+252 a^{2} x^{2}-280 a x +91}{12 a \,c^{3} \left (a^{4} x^{4}-4 a^{3} x^{3}+6 a^{2} x^{2}-4 a x +1\right )} \] Input:
int(1/(a*x-1)^2*(a*x+1)^2/(c-c/a/x)^3,x)
Output:
(84*log(a*x - 1)*a**4*x**4 - 336*log(a*x - 1)*a**3*x**3 + 504*log(a*x - 1) *a**2*x**2 - 336*log(a*x - 1)*a*x + 84*log(a*x - 1) + 12*a**5*x**5 - 87*a* *4*x**4 + 252*a**2*x**2 - 280*a*x + 91)/(12*a*c**3*(a**4*x**4 - 4*a**3*x** 3 + 6*a**2*x**2 - 4*a*x + 1))