Integrand size = 12, antiderivative size = 233 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {a^4}{6 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^4}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {3 a^4}{8 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {2 a^4}{5 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^5}+\frac {a^4}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^4}-\frac {4 a^4}{3 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^4}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {3 a^4}{8 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )} \] Output:
-1/6*a^4/(1-((-a*x+1)/(a*x+1))^(1/2))^3+1/4*a^4/(1-((-a*x+1)/(a*x+1))^(1/2 ))^2-3*a^4/(8-8*((-a*x+1)/(a*x+1))^(1/2))-2/5*a^4/(1+((-a*x+1)/(a*x+1))^(1 /2))^5+a^4/(1+((-a*x+1)/(a*x+1))^(1/2))^4-4/3*a^4/(1+((-a*x+1)/(a*x+1))^(1 /2))^3+a^4/(1+((-a*x+1)/(a*x+1))^(1/2))^2-3*a^4/(8+8*((-a*x+1)/(a*x+1))^(1 /2))
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.26 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {3+\sqrt {\frac {1-a x}{1+a x}} (1+a x)^2 \left (-3+3 a x-2 a^2 x^2+2 a^3 x^3\right )}{15 a x^5} \] Input:
Integrate[1/(E^ArcSech[a*x]*x^5),x]
Output:
-1/15*(3 + Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^2*(-3 + 3*a*x - 2*a^2*x^2 + 2*a^3*x^3))/(a*x^5)
Time = 1.19 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6891, 7268, 27, 2115, 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.
\(\displaystyle \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx\) |
\(\Big \downarrow \) 6891 |
\(\displaystyle \int \frac {1}{x^5 \left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}\right )}dx\) |
\(\Big \downarrow \) 7268 |
\(\displaystyle -4 a \int \frac {a^3 \sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^3}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^6}d\sqrt {\frac {1-a x}{a x+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -4 a^4 \int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^3}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^6}d\sqrt {\frac {1-a x}{a x+1}}\) |
\(\Big \downarrow \) 2115 |
\(\displaystyle -4 a^4 \int \left (-\frac {3}{32 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}+\frac {1}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}-\frac {1}{\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}+\frac {1}{\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5}-\frac {1}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^6}+\frac {3}{32 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^2}+\frac {1}{8 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^3}+\frac {1}{8 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^4}\right )d\sqrt {\frac {1-a x}{a x+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 a^4 \left (\frac {3}{32 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}-\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}+\frac {1}{3 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}-\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}+\frac {1}{10 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5}+\frac {3}{32 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}-\frac {1}{16 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}+\frac {1}{24 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3}\right )\) |
Input:
Int[1/(E^ArcSech[a*x]*x^5),x]
Output:
-4*a^4*(1/(24*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^3) - 1/(16*(1 - Sqrt[(1 - a* x)/(1 + a*x)])^2) + 3/(32*(1 - Sqrt[(1 - a*x)/(1 + a*x)])) + 1/(10*(1 + Sq rt[(1 - a*x)/(1 + a*x)])^5) - 1/(4*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^4) + 1/ (3*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^3) - 1/(4*(1 + Sqrt[(1 - a*x)/(1 + a*x) ])^2) + 3/(32*(1 + Sqrt[(1 - a*x)/(1 + a*x)])))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f _.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^ n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && IntegersQ[m, n]
Int[E^(ArcSech[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[(1 - u)/(1 + u)] + (1/u)*Sqrt[(1 - u)/(1 + u)])^n, x] /; FreeQ[m, x] && Integer Q[n]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfQuotientOfLinears [u, x]}, Simp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/ls t[[2]])], x] /; !FalseQ[lst]]
Time = 0.45 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.29
method | result | size |
default | \(a \left (-\frac {1}{5 a^{2} x^{5}}-\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \left (a^{2} x^{2}-1\right ) \left (2 a^{2} x^{2}+3\right )}{15 a \,x^{4}}\right )\) | \(68\) |
Input:
int(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^5,x,method=_RETURNVERBOSE )
Output:
a*(-1/5/a^2/x^5-1/15/a*(-(a*x-1)/a/x)^(1/2)/x^4*((a*x+1)/a/x)^(1/2)*(a^2*x ^2-1)*(2*a^2*x^2+3))
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.26 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {{\left (2 \, a^{5} x^{5} + a^{3} x^{3} - 3 \, a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 3}{15 \, a x^{5}} \] Input:
integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^5,x, algorithm="fri cas")
Output:
-1/15*((2*a^5*x^5 + a^3*x^3 - 3*a*x)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1) /(a*x)) + 3)/(a*x^5)
\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=a \int \frac {1}{a x^{5} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + x^{4}}\, dx \] Input:
integrate(1/(1/a/x+(-1+1/a/x)**(1/2)*(1+1/a/x)**(1/2))/x**5,x)
Output:
a*Integral(1/(a*x**5*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)) + x**4), x)
\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=\int { \frac {1}{x^{5} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \] Input:
integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^5,x, algorithm="max ima")
Output:
integrate(1/(x^5*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)
\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=\int { \frac {1}{x^{5} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \] Input:
integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^5,x, algorithm="gia c")
Output:
integrate(1/(x^5*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)
Time = 24.57 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.32 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=-\frac {1}{5\,a\,x^5}-\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {a\,x^2}{15}-\frac {x}{5}-\frac {1}{5\,a}+\frac {a^2\,x^3}{15}+\frac {2\,a^3\,x^4}{15}+\frac {2\,a^4\,x^5}{15}\right )}{x^5\,\sqrt {\frac {1}{a\,x}+1}} \] Input:
int(1/(x^5*((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x))),x)
Output:
- 1/(5*a*x^5) - ((1/(a*x) - 1)^(1/2)*((a*x^2)/15 - x/5 - 1/(5*a) + (a^2*x^ 3)/15 + (2*a^3*x^4)/15 + (2*a^4*x^5)/15))/(x^5*(1/(a*x) + 1)^(1/2))
Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.46 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^5} \, dx=\frac {a \left (3 \sqrt {a x +1}\, \sqrt {-a x +1}\, a^{3} x^{3}-2 \sqrt {a x +1}\, \sqrt {-a x +1}\, a^{2} x^{2}-\sqrt {a x +1}\, \sqrt {-a x +1}+2 a^{4} x^{4}+3 a^{3} x^{3}-a^{2} x^{2}-4\right )}{15 x^{3} \left (\sqrt {a x +1}\, \sqrt {-a x +1}+1\right )} \] Input:
int(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^5,x)
Output:
(a*(3*sqrt(a*x + 1)*sqrt( - a*x + 1)*a**3*x**3 - 2*sqrt(a*x + 1)*sqrt( - a *x + 1)*a**2*x**2 - sqrt(a*x + 1)*sqrt( - a*x + 1) + 2*a**4*x**4 + 3*a**3* x**3 - a**2*x**2 - 4))/(15*x**3*(sqrt(a*x + 1)*sqrt( - a*x + 1) + 1))