3.1.55 \(\int \frac {e^{\text {sech}^{-1}(a x^2)}}{x^3} \, dx\) [55]

3.1.55.1 Optimal result
3.1.55.2 Mathematica [A] (verified)
3.1.55.3 Rubi [A] (verified)
3.1.55.4 Maple [C] (verified)
3.1.55.5 Fricas [A] (verification not implemented)
3.1.55.6 Sympy [A] (verification not implemented)
3.1.55.7 Maxima [F]
3.1.55.8 Giac [F(-2)]
3.1.55.9 Mupad [B] (verification not implemented)

3.1.55.1 Optimal result

Integrand size = 12, antiderivative size = 118 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^3} \, dx=\frac {1}{4 a x^4}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{2 x^2}+\frac {\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \sqrt {1-a^2 x^4}}{4 a x^4}+\frac {1}{4} a \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \text {arctanh}\left (\sqrt {1-a^2 x^4}\right ) \]

output
1/4/a/x^4-1/2*(1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))/x^2+1/4*a*arct 
anh((-a^2*x^4+1)^(1/2))*(1/(a*x^2+1))^(1/2)*(a*x^2+1)^(1/2)+1/4*(1/(a*x^2+ 
1))^(1/2)*(a*x^2+1)^(1/2)*(-a^2*x^4+1)^(1/2)/a/x^4
 
3.1.55.2 Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^3} \, dx=-\frac {\frac {1}{x^4}+\frac {\sqrt {\frac {1-a x^2}{1+a x^2}} \left (1+a x^2\right )}{x^4}-\frac {a^2 \sqrt {\frac {1-a x^2}{1+a x^2}} \left (1+a x^2\right ) \arctan \left (\sqrt {-1+a^2 x^4}\right )}{\sqrt {-1+a^2 x^4}}}{4 a} \]

input
Integrate[E^ArcSech[a*x^2]/x^3,x]
 
output
-1/4*(x^(-4) + (Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(1 + a*x^2))/x^4 - (a^2*Sqrt 
[(1 - a*x^2)/(1 + a*x^2)]*(1 + a*x^2)*ArcTan[Sqrt[-1 + a^2*x^4]])/Sqrt[-1 
+ a^2*x^4])/a
 
3.1.55.3 Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6889, 15, 335, 798, 52, 73, 221}

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.

\(\displaystyle \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^3} \, dx\)

\(\Big \downarrow \) 6889

\(\displaystyle -\frac {\int \frac {1}{x^5}dx}{a}-\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^5 \sqrt {1-a x^2} \sqrt {a x^2+1}}dx}{a}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{2 x^2}\)

\(\Big \downarrow \) 15

\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^5 \sqrt {1-a x^2} \sqrt {a x^2+1}}dx}{a}+\frac {1}{4 a x^4}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{2 x^2}\)

\(\Big \downarrow \) 335

\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^5 \sqrt {1-a^2 x^4}}dx}{a}+\frac {1}{4 a x^4}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{2 x^2}\)

\(\Big \downarrow \) 798

\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {1}{x^8 \sqrt {1-a^2 x^4}}dx^4}{4 a}+\frac {1}{4 a x^4}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{2 x^2}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {1}{2} a^2 \int \frac {1}{x^4 \sqrt {1-a^2 x^4}}dx^4-\frac {\sqrt {1-a^2 x^4}}{x^4}\right )}{4 a}+\frac {1}{4 a x^4}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{2 x^2}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (-\int \frac {1}{\frac {1}{a^2}-\frac {x^8}{a^2}}d\sqrt {1-a^2 x^4}-\frac {\sqrt {1-a^2 x^4}}{x^4}\right )}{4 a}+\frac {1}{4 a x^4}-\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{2 x^2}\)

\(\Big \downarrow \) 221

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

input
Int[E^ArcSech[a*x^2]/x^3,x]
 
output
1/(4*a*x^4) - E^ArcSech[a*x^2]/(2*x^2) - (Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + 
a*x^2]*(-(Sqrt[1 - a^2*x^4]/x^4) - a^2*ArcTanh[Sqrt[1 - a^2*x^4]]))/(4*a)
 

3.1.55.3.1 Defintions of rubi rules used

rule 15
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 52
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 335
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p 
_.), x_Symbol] :> Int[(e*x)^m*(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, e 
, m, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] 
))
 

rule 798
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 6889
Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(E^ 
ArcSech[a*x^p]/(m + 1)), x] + (Simp[p/(a*(m + 1))   Int[x^(m - p), x], x] + 
 Simp[p*(Sqrt[1 + a*x^p]/(a*(m + 1)))*Sqrt[1/(1 + a*x^p)]   Int[x^(m - p)/( 
Sqrt[1 + a*x^p]*Sqrt[1 - a*x^p]), x], x]) /; FreeQ[{a, m, p}, x] && NeQ[m, 
-1]
 
3.1.55.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.09

method result size
default \(\frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (\ln \left (\frac {2 \,\operatorname {csgn}\left (\frac {1}{a}\right ) a \sqrt {-\frac {x^{4} a^{2}-1}{a^{2}}}+2}{a^{2} x^{2}}\right ) x^{4} a -\sqrt {-\frac {x^{4} a^{2}-1}{a^{2}}}\, \operatorname {csgn}\left (\frac {1}{a}\right )\right ) \operatorname {csgn}\left (\frac {1}{a}\right )}{4 x^{2} \sqrt {-\frac {x^{4} a^{2}-1}{a^{2}}}}-\frac {1}{4 a \,x^{4}}\) \(129\)

input
int((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))/x^3,x,method=_RETURNVERB 
OSE)
 
output
1/4*(-(a*x^2-1)/a/x^2)^(1/2)/x^2*((a*x^2+1)/a/x^2)^(1/2)*(ln(2*(csgn(1/a)* 
a*(-(a^2*x^4-1)/a^2)^(1/2)+1)/a^2/x^2)*x^4*a-(-(a^2*x^4-1)/a^2)^(1/2)*csgn 
(1/a))*csgn(1/a)/(-(a^2*x^4-1)/a^2)^(1/2)-1/4/a/x^4
 
3.1.55.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^3} \, dx=\frac {a^{2} x^{4} \log \left (a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} + 1\right ) - a^{2} x^{4} \log \left (a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - 1\right ) - 2 \, a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - 2}{8 \, a x^{4}} \]

input
integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))/x^3,x, algorithm=" 
fricas")
 
output
1/8*(a^2*x^4*log(a*x^2*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2) 
) + 1) - a^2*x^4*log(a*x^2*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a* 
x^2)) - 1) - 2*a*x^2*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) 
- 2)/(a*x^4)
 
3.1.55.6 Sympy [A] (verification not implemented)

Time = 4.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^3} \, dx=- \frac {a \left (2 \sqrt {-1 + \frac {1}{a x^{2}}} \left (\frac {\left (1 + \frac {1}{a x^{2}}\right )^{\frac {3}{2}}}{4} - \frac {\sqrt {1 + \frac {1}{a x^{2}}}}{4}\right ) - \log {\left (2 \sqrt {-1 + \frac {1}{a x^{2}}} + 2 \sqrt {1 + \frac {1}{a x^{2}}} \right )}\right )}{2} - \frac {1}{4 a x^{4}} \]

input
integrate((1/a/x**2+(1/a/x**2-1)**(1/2)*(1/a/x**2+1)**(1/2))/x**3,x)
 
output
-a*(2*sqrt(-1 + 1/(a*x**2))*((1 + 1/(a*x**2))**(3/2)/4 - sqrt(1 + 1/(a*x** 
2))/4) - log(2*sqrt(-1 + 1/(a*x**2)) + 2*sqrt(1 + 1/(a*x**2))))/2 - 1/(4*a 
*x**4)
 
3.1.55.7 Maxima [F]

\[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^3} \, dx=\int { \frac {\sqrt {\frac {1}{a x^{2}} + 1} \sqrt {\frac {1}{a x^{2}} - 1} + \frac {1}{a x^{2}}}{x^{3}} \,d x } \]

input
integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))/x^3,x, algorithm=" 
maxima")
 
output
integrate(sqrt(a*x^2 + 1)*sqrt(-a*x^2 + 1)/x^5, x)/a - 1/4/(a*x^4)
 
3.1.55.8 Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^3} \, dx=\text {Exception raised: TypeError} \]

input
integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))/x^3,x, algorithm=" 
giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 
3.1.55.9 Mupad [B] (verification not implemented)

Time = 5.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.60 \[ \int \frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{x^3} \, dx=\frac {a\,\ln \left (\sqrt {\frac {1}{a\,x^2}-1}\,\sqrt {\frac {1}{a\,x^2}+1}+\frac {1}{a\,x^2}\right )}{4}-\frac {1}{4\,a\,x^4}-\frac {\sqrt {\frac {1}{a\,x^2}-1}\,\sqrt {\frac {1}{a\,x^2}+1}}{4\,x^2} \]

input
int(((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2) + 1/(a*x^2))/x^3,x)
 
output
(a*log((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2) + 1/(a*x^2)))/4 - 1/(4* 
a*x^4) - ((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2))/(4*x^2)