\(\int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx\) [34]

Optimal result

Integrand size = 12, antiderivative size = 72 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx=-\frac {a}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}+\frac {a}{1+\sqrt {\frac {1-a x}{1+a x}}}-a \text {arctanh}\left (\sqrt {\frac {1-a x}{1+a x}}\right ) \] Output:

-a/(1+((-a*x+1)/(a*x+1))^(1/2))^2+a/(1+((-a*x+1)/(a*x+1))^(1/2))-a*arctanh 
(((-a*x+1)/(a*x+1))^(1/2))
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx=\frac {1}{2} \left (-\frac {1}{a x^2}+\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{a x^2}+a \log (x)-a \log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right )\right ) \] Input:

Integrate[1/(E^ArcSech[a*x]*x^2),x]
 

Output:

(-(1/(a*x^2)) + (Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x))/(a*x^2) + a*Log[x] - 
 a*Log[1 + Sqrt[(1 - a*x)/(1 + a*x)] + a*x*Sqrt[(1 - a*x)/(1 + a*x)]])/2
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.10, 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, 86, 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.

\(\displaystyle \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx\)

\(\Big \downarrow \) 6891

\(\displaystyle \int \frac {1}{x^2 \left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}\right )}dx\)

\(\Big \downarrow \) 7268

\(\displaystyle 4 a \int -\frac {\sqrt {\frac {1-a x}{a x+1}}}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle -4 a \int \frac {\sqrt {\frac {1-a x}{a x+1}}}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}\)

\(\Big \downarrow \) 86

\(\displaystyle -4 a \int \left (\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}-\frac {1}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}-\frac {1}{4 \left (\frac {1-a x}{a x+1}-1\right )}\right )d\sqrt {\frac {1-a x}{a x+1}}\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 a \left (-\frac {1}{4} \text {arctanh}\left (\sqrt {\frac {1-a x}{a x+1}}\right )+\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}-\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}\right )\)

Input:

Int[1/(E^ArcSech[a*x]*x^2),x]
 

Output:

4*a*(-1/4*1/(1 + Sqrt[(1 - a*x)/(1 + a*x)])^2 + 1/(4*(1 + Sqrt[(1 - a*x)/( 
1 + a*x)])) - ArcTanh[Sqrt[(1 - a*x)/(1 + a*x)]]/4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6891

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]
 

rule 7268

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]]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.33

method	result	size
default	\(a \left (-\frac {1}{2 a^{2} x^{2}}-\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \left (a^{2} x^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\sqrt {-a^{2} x^{2}+1}\right )}{2 a x \sqrt {-a^{2} x^{2}+1}}\right )\)	\(96\)

Input:

int(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^2,x,method=_RETURNVERBOSE 
)
 

Output:

a*(-1/2/a^2/x^2-1/2/a*(-(a*x-1)/a/x)^(1/2)/x*((a*x+1)/a/x)^(1/2)*(a^2*x^2* 
arctanh(1/(-a^2*x^2+1)^(1/2))-(-a^2*x^2+1)^(1/2))/(-a^2*x^2+1)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.78 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx=-\frac {a^{2} x^{2} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - a^{2} x^{2} \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) - 2 \, a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 2}{4 \, a x^{2}} \] Input:

integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^2,x, algorithm="fri 
cas")
 

Output:

-1/4*(a^2*x^2*log(a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 1) - 
a^2*x^2*log(a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 1) - 2*a*x* 
sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 2)/(a*x^2)
 

Sympy [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx=a \int \frac {1}{a x^{2} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + x}\, dx \] Input:

integrate(1/(1/a/x+(-1+1/a/x)**(1/2)*(1+1/a/x)**(1/2))/x**2,x)
 

Output:

a*Integral(1/(a*x**2*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)) + x), x)
 

Maxima [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx=\int { \frac {1}{x^{2} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \] Input:

integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^2,x, algorithm="max 
ima")
 

Output:

integrate(1/(x^2*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)
 

Giac [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx=\int { \frac {1}{x^{2} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \] Input:

integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^2,x, algorithm="gia 
c")
 

Output:

integrate(1/(x^2*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)
 

Mupad [B] (verification not implemented)

Time = 38.74 (sec) , antiderivative size = 323, normalized size of antiderivative = 4.49 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx=2\,a\,\mathrm {atanh}\left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )-a\,\mathrm {acosh}\left (\frac {1}{a\,x}\right )-\frac {1}{2\,a\,x^2}-\frac {a\,\left (\frac {14\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^3}+\frac {14\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^5}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^5}+\frac {2\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^7}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^7}+\frac {2\,\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}{\sqrt {\frac {1}{a\,x}+1}-1}\right )}{1+\frac {6\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}-\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}-\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}} \] Input:

int(1/(x^2*((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x))),x)
 

Output:

2*a*atanh(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1)) - a*acosh( 
1/(a*x)) - 1/(2*a*x^2) - (a*((14*((1/(a*x) - 1)^(1/2) - 1i)^3)/((1/(a*x) + 
 1)^(1/2) - 1)^3 + (14*((1/(a*x) - 1)^(1/2) - 1i)^5)/((1/(a*x) + 1)^(1/2) 
- 1)^5 + (2*((1/(a*x) - 1)^(1/2) - 1i)^7)/((1/(a*x) + 1)^(1/2) - 1)^7 + (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) - 1 
i)^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)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 424, normalized size of antiderivative = 5.89 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^2} \, dx=\frac {a \left (4 \sqrt {a x +1}\, \sqrt {-a x +1}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )^{2}+1\right )-3 \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 )-\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 )-3 \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 )-\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 )+2 \sqrt {a x +1}\, \sqrt {-a x +1}\, \mathrm {log}\left (\frac {2 \sqrt {a x +1}-2}{\sqrt {2}}\right )+2 \sqrt {a x +1}\, \sqrt {-a x +1}\, \mathrm {log}\left (\frac {2 \sqrt {a x +1}+2}{\sqrt {2}}\right )+4 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )^{2}+1\right )-3 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right )-\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right )-3 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right )-\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right )+2 \,\mathrm {log}\left (\frac {2 \sqrt {a x +1}-2}{\sqrt {2}}\right )+2 \,\mathrm {log}\left (\frac {2 \sqrt {a x +1}+2}{\sqrt {2}}\right )-1\right )}{2 \sqrt {a x +1}\, \sqrt {-a x +1}+2} \] Input:

int(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^2,x)
 

Output:

(a*(4*sqrt(a*x + 1)*sqrt( - a*x + 1)*log(tan(asin(sqrt( - a*x + 1)/sqrt(2) 
)/2)**2 + 1) - 3*sqrt(a*x + 1)*sqrt( - a*x + 1)*log( - sqrt(2) + tan(asin( 
sqrt( - a*x + 1)/sqrt(2))/2) - 1) - sqrt(a*x + 1)*sqrt( - a*x + 1)*log( - 
sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) + 1) - 3*sqrt(a*x + 1)*sqr 
t( - a*x + 1)*log(sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) - 1) - s 
qrt(a*x + 1)*sqrt( - a*x + 1)*log(sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt 
(2))/2) + 1) + 2*sqrt(a*x + 1)*sqrt( - a*x + 1)*log((2*sqrt(a*x + 1) - 2)/ 
sqrt(2)) + 2*sqrt(a*x + 1)*sqrt( - a*x + 1)*log((2*sqrt(a*x + 1) + 2)/sqrt 
(2)) + 4*log(tan(asin(sqrt( - a*x + 1)/sqrt(2))/2)**2 + 1) - 3*log( - sqrt 
(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) - 1) - log( - sqrt(2) + tan(as 
in(sqrt( - a*x + 1)/sqrt(2))/2) + 1) - 3*log(sqrt(2) + tan(asin(sqrt( - a* 
x + 1)/sqrt(2))/2) - 1) - log(sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2)) 
/2) + 1) + 2*log((2*sqrt(a*x + 1) - 2)/sqrt(2)) + 2*log((2*sqrt(a*x + 1) + 
 2)/sqrt(2)) - 1))/(2*(sqrt(a*x + 1)*sqrt( - a*x + 1) + 1))