Integrand size = 12, antiderivative size = 147 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx=-\frac {a^2}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^4}+\frac {2 a^2}{\left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^3}-\frac {3 a^2}{2 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {a^2}{2 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}+\frac {1}{2} a^2 \text {arctanh}\left (\sqrt {\frac {1-a x}{1+a x}}\right ) \] Output:
-a^2/(1-((-a*x+1)/(a*x+1))^(1/2))^4+2*a^2/(1-((-a*x+1)/(a*x+1))^(1/2))^3-3 /2*a^2/(1-((-a*x+1)/(a*x+1))^(1/2))^2+a^2/(2-2*((-a*x+1)/(a*x+1))^(1/2))+1 /2*a^2*arctanh(((-a*x+1)/(a*x+1))^(1/2))
Time = 0.16 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx=\frac {\frac {(1+a x) \left (-2+2 a x-2 \sqrt {\frac {1-a x}{1+a x}}+a^2 x^2 \sqrt {\frac {1-a x}{1+a x}}\right )}{x^4}-a^4 \log (x)+a^4 \log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right )}{4 a^2} \] Input:
Integrate[E^(2*ArcSech[a*x])/x^3,x]
Output:
(((1 + a*x)*(-2 + 2*a*x - 2*Sqrt[(1 - a*x)/(1 + a*x)] + a^2*x^2*Sqrt[(1 - a*x)/(1 + a*x)]))/x^4 - a^4*Log[x] + a^4*Log[1 + Sqrt[(1 - a*x)/(1 + a*x)] + a*x*Sqrt[(1 - a*x)/(1 + a*x)]])/(4*a^2)
Time = 1.15 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6891, 7268, 25, 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.
| \(\displaystyle \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx\) |
| \(\Big \downarrow \) 6891 |
| \(\displaystyle \int \frac {\left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}\right )^2}{x^3}dx\) |
| \(\Big \downarrow \) 7268 |
| \(\displaystyle 4 a^2 \int -\frac {\sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^5 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}d\sqrt {\frac {1-a x}{a x+1}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -4 a^2 \int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^5 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}d\sqrt {\frac {1-a x}{a x+1}}\) |
| \(\Big \downarrow \) 2115 |
| \(\displaystyle -4 a^2 \int \left (-\frac {1}{8 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^2}-\frac {3}{4 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^3}-\frac {3}{2 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^4}-\frac {1}{\left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^5}+\frac {1}{8 \left (\frac {1-a x}{a x+1}-1\right )}\right )d\sqrt {\frac {1-a x}{a x+1}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 a^2 \left (\frac {1}{8} \text {arctanh}\left (\sqrt {\frac {1-a x}{a x+1}}\right )+\frac {1}{8 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}-\frac {3}{8 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}+\frac {1}{2 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3}-\frac {1}{4 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^4}\right )\) |
Input:
Int[E^(2*ArcSech[a*x])/x^3,x]
Output:
4*a^2*(-1/4*1/(1 - Sqrt[(1 - a*x)/(1 + a*x)])^4 + 1/(2*(1 - Sqrt[(1 - a*x) /(1 + a*x)])^3) - 3/(8*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^2) + 1/(8*(1 - Sqrt [(1 - a*x)/(1 + a*x)])) + ArcTanh[Sqrt[(1 - a*x)/(1 + a*x)]]/8)
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 = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.89
| method | result | size |
| default | \(\frac {\frac {a^{2}}{2 x^{2}}-\frac {1}{4 x^{4}}}{a^{2}}+\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \left (\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right ) a^{4} x^{4}+\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\right )}{4 a \,x^{3} \sqrt {-a^{2} x^{2}+1}}-\frac {1}{4 a^{2} x^{4}}\) | \(131\) |
Input:
int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2/x^3,x,method=_RETURNVERBOSE )
Output:
1/a^2*(1/2*a^2/x^2-1/4/x^4)+1/4/a*(-(a*x-1)/a/x)^(1/2)/x^3*((a*x+1)/a/x)^( 1/2)*(arctanh(1/(-a^2*x^2+1)^(1/2))*a^4*x^4+(-a^2*x^2+1)^(1/2)*a^2*x^2-2*( -a^2*x^2+1)^(1/2))/(-a^2*x^2+1)^(1/2)-1/4/a^2/x^4
Time = 0.11 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.99 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx=\frac {a^{4} x^{4} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - a^{4} x^{4} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) + 4 \, a^{2} x^{2} + 2 \, {\left (a^{3} x^{3} - 2 \, a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 4}{8 \, a^{2} x^{4}} \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2/x^3,x, algorithm="fri cas")
Output:
1/8*(a^4*x^4*log(a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 1) - a ^4*x^4*log(a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 1) + 4*a^2*x ^2 + 2*(a^3*x^3 - 2*a*x)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 4) /(a^2*x^4)
\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx=\frac {\int \frac {2}{x^{5}}\, dx + \int \left (- \frac {a^{2}}{x^{3}}\right )\, dx + \int \frac {2 a \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}}{x^{4}}\, dx}{a^{2}} \] Input:
integrate((1/a/x+(-1+1/a/x)**(1/2)*(1+1/a/x)**(1/2))**2/x**3,x)
Output:
(Integral(2/x**5, x) + Integral(-a**2/x**3, x) + Integral(2*a*sqrt(-1 + 1/ (a*x))*sqrt(1 + 1/(a*x))/x**4, x))/a**2
\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx=\int { \frac {{\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2}}{x^{3}} \,d x } \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2/x^3,x, algorithm="max ima")
Output:
2*integrate(sqrt(a*x + 1)*sqrt(-a*x + 1)/x^5, x)/a^2 - 1/2/(a^2*x^4) - int egrate(x^(-3), x)
\[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx=\int { \frac {{\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2}}{x^{3}} \,d x } \] Input:
integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2/x^3,x, algorithm="gia c")
Output:
integrate((sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))^2/x^3, x)
Time = 81.04 (sec) , antiderivative size = 885, normalized size of antiderivative = 6.02 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx =\text {Too large to display} \] Input:
int(((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x))^2/x^3,x)
Output:
a^2*atanh(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1)) - ((28*a^2 *((1/(a*x) - 1)^(1/2) - 1i)^3)/((1/(a*x) + 1)^(1/2) - 1)^3 + (28*a^2*((1/( a*x) - 1)^(1/2) - 1i)^5)/((1/(a*x) + 1)^(1/2) - 1)^5 + (4*a^2*((1/(a*x) - 1)^(1/2) - 1i)^7)/((1/(a*x) + 1)^(1/2) - 1)^7 + (4*a^2*((1/(a*x) - 1)^(1/2 ) - 1i))/((1/(a*x) + 1)^(1/2) - 1))/((6*((1/(a*x) - 1)^(1/2) - 1i)^4)/((1/ (a*x) + 1)^(1/2) - 1)^4 - (4*((1/(a*x) - 1)^(1/2) - 1i)^2)/((1/(a*x) + 1)^ (1/2) - 1)^2 - (4*((1/(a*x) - 1)^(1/2) - 1i)^6)/((1/(a*x) + 1)^(1/2) - 1)^ 6 + ((1/(a*x) - 1)^(1/2) - 1i)^8/((1/(a*x) + 1)^(1/2) - 1)^8 + 1) - ((23*a ^2*((1/(a*x) - 1)^(1/2) - 1i)^3)/((1/(a*x) + 1)^(1/2) - 1)^3 + (333*a^2*(( 1/(a*x) - 1)^(1/2) - 1i)^5)/((1/(a*x) + 1)^(1/2) - 1)^5 + (671*a^2*((1/(a* x) - 1)^(1/2) - 1i)^7)/((1/(a*x) + 1)^(1/2) - 1)^7 + (671*a^2*((1/(a*x) - 1)^(1/2) - 1i)^9)/((1/(a*x) + 1)^(1/2) - 1)^9 + (333*a^2*((1/(a*x) - 1)^(1 /2) - 1i)^11)/((1/(a*x) + 1)^(1/2) - 1)^11 + (23*a^2*((1/(a*x) - 1)^(1/2) - 1i)^13)/((1/(a*x) + 1)^(1/2) - 1)^13 - (3*a^2*((1/(a*x) - 1)^(1/2) - 1i) ^15)/((1/(a*x) + 1)^(1/2) - 1)^15 - (3*a^2*((1/(a*x) - 1)^(1/2) - 1i))/((1 /(a*x) + 1)^(1/2) - 1))/((28*((1/(a*x) - 1)^(1/2) - 1i)^4)/((1/(a*x) + 1)^ (1/2) - 1)^4 - (8*((1/(a*x) - 1)^(1/2) - 1i)^2)/((1/(a*x) + 1)^(1/2) - 1)^ 2 - (56*((1/(a*x) - 1)^(1/2) - 1i)^6)/((1/(a*x) + 1)^(1/2) - 1)^6 + (70*(( 1/(a*x) - 1)^(1/2) - 1i)^8)/((1/(a*x) + 1)^(1/2) - 1)^8 - (56*((1/(a*x) - 1)^(1/2) - 1i)^10)/((1/(a*x) + 1)^(1/2) - 1)^10 + (28*((1/(a*x) - 1)^(1...
Time = 0.16 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.16 \[ \int \frac {e^{2 \text {sech}^{-1}(a x)}}{x^3} \, dx=\frac {\sqrt {a x +1}\, \sqrt {-a x +1}\, a^{2} x^{2}-2 \sqrt {a x +1}\, \sqrt {-a x +1}+\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) a^{4} x^{4}-\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) a^{4} x^{4}+\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) a^{4} x^{4}-\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) a^{4} x^{4}+2 a^{2} x^{2}-2}{4 a^{2} x^{4}} \] Input:
int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2/x^3,x)
Output:
(sqrt(a*x + 1)*sqrt( - a*x + 1)*a**2*x**2 - 2*sqrt(a*x + 1)*sqrt( - a*x + 1) + log( - sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) - 1)*a**4*x**4 - log( - sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) + 1)*a**4*x**4 + log(sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) - 1)*a**4*x**4 - log( sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) + 1)*a**4*x**4 + 2*a**2*x* *2 - 2)/(4*a**2*x**4)