Integrand size = 12, antiderivative size = 301 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^6} \, dx=-\frac {4 a^5}{7 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^7}+\frac {2 a^5}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^6}-\frac {18 a^5}{5 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^5}+\frac {4 a^5}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^4}-\frac {35 a^5}{12 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {11 a^5}{8 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {a^5}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {a^5}{12 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^5}{8 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {a^5}{4 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )} \]
-4/7*a^5/(1-((-a*x+1)/(a*x+1))^(1/2))^7+2*a^5/(1-((-a*x+1)/(a*x+1))^(1/2)) ^6-18/5*a^5/(1-((-a*x+1)/(a*x+1))^(1/2))^5+4*a^5/(1-((-a*x+1)/(a*x+1))^(1/ 2))^4-35/12*a^5/(1-((-a*x+1)/(a*x+1))^(1/2))^3+11/8*a^5/(1-((-a*x+1)/(a*x+ 1))^(1/2))^2-1/4*a^5/(1-((-a*x+1)/(a*x+1))^(1/2))-1/12*a^5/(1+((-a*x+1)/(a *x+1))^(1/2))^3+1/8*a^5/(1+((-a*x+1)/(a*x+1))^(1/2))^2-1/4*a^5/(1+((-a*x+1 )/(a*x+1))^(1/2))
Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.28 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^6} \, dx=\frac {-30+21 a^2 x^2+2 \sqrt {\frac {1-a x}{1+a x}} (1+a x)^2 \left (-15+15 a x-12 a^2 x^2+12 a^3 x^3-8 a^4 x^4+8 a^5 x^5\right )}{105 a^2 x^7} \]
(-30 + 21*a^2*x^2 + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^2*(-15 + 15*a*x - 12*a^2*x^2 + 12*a^3*x^3 - 8*a^4*x^4 + 8*a^5*x^5))/(105*a^2*x^7)
Time = 0.73 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.92, 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.
\(\displaystyle \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^6} \, dx\) |
\(\Big \downarrow \) 6891 |
\(\displaystyle \int \frac {\left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}\right )^2}{x^6}dx\) |
\(\Big \downarrow \) 7268 |
\(\displaystyle -4 a \int \frac {a^4 \sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^4}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^8 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}d\sqrt {\frac {1-a x}{a x+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -4 a^5 \int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^4}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^8 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}d\sqrt {\frac {1-a x}{a x+1}}\) |
\(\Big \downarrow \) 2115 |
\(\displaystyle -4 a^5 \int \left (-\frac {1}{16 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}+\frac {1}{16 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}-\frac {1}{16 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}+\frac {1}{16 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^2}+\frac {11}{16 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^3}+\frac {35}{16 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^4}+\frac {4}{\left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^5}+\frac {9}{2 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^6}+\frac {3}{\left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^7}+\frac {1}{\left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^8}\right )d\sqrt {\frac {1-a x}{a x+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 a^5 \left (\frac {1}{16 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}-\frac {1}{32 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}+\frac {1}{48 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}+\frac {1}{16 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}-\frac {11}{32 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}+\frac {35}{48 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3}-\frac {1}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^4}+\frac {9}{10 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^5}-\frac {1}{2 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^6}+\frac {1}{7 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^7}\right )\) |
-4*a^5*(1/(7*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^7) - 1/(2*(1 - Sqrt[(1 - a*x) /(1 + a*x)])^6) + 9/(10*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^5) - (1 - Sqrt[(1 - a*x)/(1 + a*x)])^(-4) + 35/(48*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^3) - 11/( 32*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^2) + 1/(16*(1 - Sqrt[(1 - a*x)/(1 + a*x )])) + 1/(48*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^3) - 1/(32*(1 + Sqrt[(1 - a*x )/(1 + a*x)])^2) + 1/(16*(1 + Sqrt[(1 - a*x)/(1 + a*x)])))
3.1.75.3.1 Defintions of rubi rules used
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 = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.31
method | result | size |
default | \(\frac {-\frac {1}{7 x^{7}}+\frac {a^{2}}{5 x^{5}}}{a^{2}}+\frac {2 \sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (a^{2} x^{2}-1\right ) \left (8 a^{4} x^{4}+12 a^{2} x^{2}+15\right )}{105 a \,x^{6}}-\frac {1}{7 a^{2} x^{7}}\) | \(92\) |
1/a^2*(-1/7/x^7+1/5*a^2/x^5)+2/105/a*((a*x+1)/a/x)^(1/2)/x^6*(-(a*x-1)/a/x )^(1/2)*(a^2*x^2-1)*(8*a^4*x^4+12*a^2*x^2+15)-1/7/a^2/x^7
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.26 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^6} \, dx=\frac {21 \, a^{2} x^{2} + 2 \, {\left (8 \, a^{7} x^{7} + 4 \, a^{5} x^{5} + 3 \, a^{3} x^{3} - 15 \, a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 30}{105 \, a^{2} x^{7}} \]
1/105*(21*a^2*x^2 + 2*(8*a^7*x^7 + 4*a^5*x^5 + 3*a^3*x^3 - 15*a*x)*sqrt((a *x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 30)/(a^2*x^7)
\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^6} \, dx=\frac {\int \frac {2}{x^{8}}\, dx + \int \left (- \frac {a^{2}}{x^{6}}\right )\, dx + \int \frac {2 a \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}}{x^{7}}\, dx}{a^{2}} \]
(Integral(2/x**8, x) + Integral(-a**2/x**6, x) + Integral(2*a*sqrt(-1 + 1/ (a*x))*sqrt(1 + 1/(a*x))/x**7, x))/a**2
Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.22 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^6} \, dx=\frac {1}{5 \, x^{5}} + \frac {2 \, {\left (8 \, a^{6} x^{7} + 4 \, a^{4} x^{5} + 3 \, a^{2} x^{3} - 15 \, x\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{105 \, a^{2} x^{8}} - \frac {2}{7 \, a^{2} x^{7}} \]
1/5/x^5 + 2/105*(8*a^6*x^7 + 4*a^4*x^5 + 3*a^2*x^3 - 15*x)*sqrt(a*x + 1)*s qrt(-a*x + 1)/(a^2*x^8) - 2/7/(a^2*x^7)
\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^6} \, dx=\int { \frac {{\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2}}{x^{6}} \,d x } \]
Time = 5.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.35 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^6} \, dx=\frac {\frac {a^2\,x^2}{5}-\frac {2}{7}}{a^2\,x^7}+\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {2\,a\,x^2\,\sqrt {\frac {1}{a\,x}+1}}{35}-\frac {2\,\sqrt {\frac {1}{a\,x}+1}}{7\,a}+\frac {8\,a^3\,x^4\,\sqrt {\frac {1}{a\,x}+1}}{105}+\frac {16\,a^5\,x^6\,\sqrt {\frac {1}{a\,x}+1}}{105}\right )}{x^6} \]