Integrand size = 18, antiderivative size = 69 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {1}{6 a c^5 (1-a x)^3}-\frac {1}{8 a c^5 (1-a x)^2}-\frac {1}{8 a c^5 (1-a x)}-\frac {\text {arctanh}(a x)}{8 a c^5} \] Output:
-1/6/a/c^5/(-a*x+1)^3-1/8/a/c^5/(-a*x+1)^2-1/8/a/c^5/(-a*x+1)-1/8*arctanh( a*x)/a/c^5
Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {10-9 a x+3 a^2 x^2-3 (-1+a x)^3 \text {arctanh}(a x)}{24 a c^5 (-1+a x)^3} \] Input:
Integrate[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^5),x]
Output:
(10 - 9*a*x + 3*a^2*x^2 - 3*(-1 + a*x)^3*ArcTanh[a*x])/(24*a*c^5*(-1 + a*x )^3)
Time = 0.53 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6717, 27, 6679, 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.
\(\displaystyle \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {e^{-2 \text {arctanh}(a x)}}{c^5 (1-a x)^5}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {e^{-2 \text {arctanh}(a x)}}{(1-a x)^5}dx}{c^5}\) |
\(\Big \downarrow \) 6679 |
\(\displaystyle -\frac {\int \frac {1}{(1-a x)^4 (a x+1)}dx}{c^5}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -\frac {\int \left (\frac {1}{8 (a x-1)^2}-\frac {1}{4 (a x-1)^3}+\frac {1}{2 (a x-1)^4}-\frac {1}{8 \left (a^2 x^2-1\right )}\right )dx}{c^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\text {arctanh}(a x)}{8 a}+\frac {1}{8 a (1-a x)}+\frac {1}{8 a (1-a x)^2}+\frac {1}{6 a (1-a x)^3}}{c^5}\) |
Input:
Int[1/(E^(2*ArcCoth[a*x])*(c - a*c*x)^5),x]
Output:
-((1/(6*a*(1 - a*x)^3) + 1/(8*a*(1 - a*x)^2) + 1/(8*a*(1 - a*x)) + ArcTanh [a*x]/(8*a))/c^5)
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_.))*(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^(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 = 0.14 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {\frac {a \,x^{2}}{8}-\frac {3 x}{8}+\frac {5}{12 a}}{\left (a x -1\right )^{3} c^{5}}+\frac {\ln \left (-a x +1\right )}{16 c^{5} a}-\frac {\ln \left (a x +1\right )}{16 c^{5} a}\) | \(57\) |
default | \(\frac {-\frac {\ln \left (a x +1\right )}{16 a}+\frac {1}{6 a \left (a x -1\right )^{3}}-\frac {1}{8 a \left (a x -1\right )^{2}}+\frac {1}{8 \left (a x -1\right ) a}+\frac {\ln \left (a x -1\right )}{16 a}}{c^{5}}\) | \(64\) |
norman | \(\frac {-\frac {7 x}{8 c}+\frac {2 a \,x^{2}}{c}-\frac {37 a^{2} x^{3}}{24 c}+\frac {5 a^{3} x^{4}}{12 c}}{c^{4} \left (a x -1\right )^{4}}+\frac {\ln \left (a x -1\right )}{16 a \,c^{5}}-\frac {\ln \left (a x +1\right )}{16 c^{5} a}\) | \(79\) |
parallelrisch | \(\frac {3 a^{3} \ln \left (a x -1\right ) x^{3}-3 \ln \left (a x +1\right ) x^{3} a^{3}+20 a^{3} x^{3}-9 a^{2} \ln \left (a x -1\right ) x^{2}+9 \ln \left (a x +1\right ) x^{2} a^{2}-54 a^{2} x^{2}+9 a \ln \left (a x -1\right ) x -9 \ln \left (a x +1\right ) x a +42 a x -3 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )}{48 c^{5} \left (a x -1\right )^{3} a}\) | \(129\) |
Input:
int((a*x-1)/(a*x+1)/(-a*c*x+c)^5,x,method=_RETURNVERBOSE)
Output:
(1/8*a*x^2-3/8*x+5/12/a)/(a*x-1)^3/c^5+1/16/c^5/a*ln(-a*x+1)-1/16/c^5/a*ln (a*x+1)
Time = 0.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.64 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {6 \, a^{2} x^{2} - 18 \, a x - 3 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) + 20}{48 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^5,x, algorithm="fricas")
Output:
1/48*(6*a^2*x^2 - 18*a*x - 3*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x + 1 ) + 3*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x - 1) + 20)/(a^4*c^5*x^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x - a*c^5)
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=- \frac {- 3 a^{2} x^{2} + 9 a x - 10}{24 a^{4} c^{5} x^{3} - 72 a^{3} c^{5} x^{2} + 72 a^{2} c^{5} x - 24 a c^{5}} - \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{16} + \frac {\log {\left (x + \frac {1}{a} \right )}}{16}}{a c^{5}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a*c*x+c)**5,x)
Output:
-(-3*a**2*x**2 + 9*a*x - 10)/(24*a**4*c**5*x**3 - 72*a**3*c**5*x**2 + 72*a **2*c**5*x - 24*a*c**5) - (-log(x - 1/a)/16 + log(x + 1/a)/16)/(a*c**5)
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {3 \, a^{2} x^{2} - 9 \, a x + 10}{24 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}} - \frac {\log \left (a x + 1\right )}{16 \, a c^{5}} + \frac {\log \left (a x - 1\right )}{16 \, a c^{5}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^5,x, algorithm="maxima")
Output:
1/24*(3*a^2*x^2 - 9*a*x + 10)/(a^4*c^5*x^3 - 3*a^3*c^5*x^2 + 3*a^2*c^5*x - a*c^5) - 1/16*log(a*x + 1)/(a*c^5) + 1/16*log(a*x - 1)/(a*c^5)
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\log \left ({\left | -\frac {2 \, c}{a c x - c} - 1 \right |}\right )}{16 \, a c^{5}} + \frac {\frac {3 \, a^{2} c^{2}}{a c x - c} - \frac {3 \, a^{2} c^{3}}{{\left (a c x - c\right )}^{2}} + \frac {4 \, a^{2} c^{4}}{{\left (a c x - c\right )}^{3}}}{24 \, a^{3} c^{6}} \] Input:
integrate((a*x-1)/(a*x+1)/(-a*c*x+c)^5,x, algorithm="giac")
Output:
-1/16*log(abs(-2*c/(a*c*x - c) - 1))/(a*c^5) + 1/24*(3*a^2*c^2/(a*c*x - c) - 3*a^2*c^3/(a*c*x - c)^2 + 4*a^2*c^4/(a*c*x - c)^3)/(a^3*c^6)
Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\frac {a\,x^2}{8}-\frac {3\,x}{8}+\frac {5}{12\,a}}{-a^3\,c^5\,x^3+3\,a^2\,c^5\,x^2-3\,a\,c^5\,x+c^5}-\frac {\mathrm {atanh}\left (a\,x\right )}{8\,a\,c^5} \] Input:
int((a*x - 1)/((c - a*c*x)^5*(a*x + 1)),x)
Output:
- ((a*x^2)/8 - (3*x)/8 + 5/(12*a))/(c^5 + 3*a^2*c^5*x^2 - a^3*c^5*x^3 - 3* a*c^5*x) - atanh(a*x)/(8*a*c^5)
Time = 0.14 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.99 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {3 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-9 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+9 \,\mathrm {log}\left (a x -1\right ) a x -3 \,\mathrm {log}\left (a x -1\right )-3 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+9 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-9 \,\mathrm {log}\left (a x +1\right ) a x +3 \,\mathrm {log}\left (a x +1\right )+2 a^{3} x^{3}-12 a x +18}{48 a \,c^{5} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:
int((a*x-1)/(a*x+1)/(-a*c*x+c)^5,x)
Output:
(3*log(a*x - 1)*a**3*x**3 - 9*log(a*x - 1)*a**2*x**2 + 9*log(a*x - 1)*a*x - 3*log(a*x - 1) - 3*log(a*x + 1)*a**3*x**3 + 9*log(a*x + 1)*a**2*x**2 - 9 *log(a*x + 1)*a*x + 3*log(a*x + 1) + 2*a**3*x**3 - 12*a*x + 18)/(48*a*c**5 *(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))