\(\int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx\) [39]

Optimal result

Integrand size = 12, antiderivative size = 353 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx=-\frac {a^6}{10 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^5}+\frac {a^6}{4 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^4}-\frac {5 a^6}{12 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {3 a^6}{8 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {5 a^6}{16 \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}-\frac {2 a^6}{7 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^7}+\frac {a^6}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^6}-\frac {19 a^6}{10 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^5}+\frac {9 a^6}{4 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^4}-\frac {11 a^6}{6 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^3}+\frac {a^6}{\left (1+\sqrt {\frac {1-a x}{1+a x}}\right )^2}-\frac {5 a^6}{16 \left (1+\sqrt {\frac {1-a x}{1+a x}}\right )} \] Output:

-1/10*a^6/(1-((-a*x+1)/(a*x+1))^(1/2))^5+1/4*a^6/(1-((-a*x+1)/(a*x+1))^(1/ 
2))^4-5/12*a^6/(1-((-a*x+1)/(a*x+1))^(1/2))^3+3/8*a^6/(1-((-a*x+1)/(a*x+1) 
)^(1/2))^2-5*a^6/(16-16*((-a*x+1)/(a*x+1))^(1/2))-2/7*a^6/(1+((-a*x+1)/(a* 
x+1))^(1/2))^7+a^6/(1+((-a*x+1)/(a*x+1))^(1/2))^6-19/10*a^6/(1+((-a*x+1)/( 
a*x+1))^(1/2))^5+9/4*a^6/(1+((-a*x+1)/(a*x+1))^(1/2))^4-11/6*a^6/(1+((-a*x 
+1)/(a*x+1))^(1/2))^3+a^6/(1+((-a*x+1)/(a*x+1))^(1/2))^2-5*a^6/(16+16*((-a 
*x+1)/(a*x+1))^(1/2))
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.22 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx=-\frac {15+\sqrt {\frac {1-a x}{1+a x}} (1+a x)^2 \left (-15+15 a x-12 a^2 x^2+12 a^3 x^3-8 a^4 x^4+8 a^5 x^5\right )}{105 a x^7} \] Input:

Integrate[1/(E^ArcSech[a*x]*x^7),x]
 

Output:

-1/105*(15 + 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))/(a*x^7)
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 328, normalized size of antiderivative = 0.93, 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.

\(\displaystyle \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx\)

\(\Big \downarrow \) 6891

\(\displaystyle \int \frac {1}{x^7 \left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}\right )}dx\)

\(\Big \downarrow \) 7268

\(\displaystyle -4 a \int \frac {a^5 \sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^5}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^6 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^8}d\sqrt {\frac {1-a x}{a x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle -4 a^6 \int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (\frac {1-a x}{a x+1}+1\right )^5}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^6 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^8}d\sqrt {\frac {1-a x}{a x+1}}\)

\(\Big \downarrow \) 2115

\(\displaystyle -4 a^6 \int \left (-\frac {5}{64 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}+\frac {1}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}-\frac {11}{8 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}+\frac {9}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5}-\frac {19}{8 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^6}+\frac {3}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^7}-\frac {1}{2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^8}+\frac {5}{64 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^2}+\frac {3}{16 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^3}+\frac {5}{16 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^4}+\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^5}+\frac {1}{8 \left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^6}\right )d\sqrt {\frac {1-a x}{a x+1}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 a^6 \left (\frac {5}{64 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}-\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}+\frac {11}{24 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}-\frac {9}{16 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^4}+\frac {19}{40 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^5}-\frac {1}{4 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^6}+\frac {1}{14 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^7}+\frac {5}{64 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}-\frac {3}{32 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^2}+\frac {5}{48 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^3}-\frac {1}{16 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^4}+\frac {1}{40 \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )^5}\right )\)

Input:

Int[1/(E^ArcSech[a*x]*x^7),x]
 

Output:

-4*a^6*(1/(40*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^5) - 1/(16*(1 - Sqrt[(1 - a* 
x)/(1 + a*x)])^4) + 5/(48*(1 - Sqrt[(1 - a*x)/(1 + a*x)])^3) - 3/(32*(1 - 
Sqrt[(1 - a*x)/(1 + a*x)])^2) + 5/(64*(1 - Sqrt[(1 - a*x)/(1 + a*x)])) + 1 
/(14*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^7) - 1/(4*(1 + Sqrt[(1 - a*x)/(1 + a* 
x)])^6) + 19/(40*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^5) - 9/(16*(1 + Sqrt[(1 - 
 a*x)/(1 + a*x)])^4) + 11/(24*(1 + Sqrt[(1 - a*x)/(1 + a*x)])^3) - 1/(4*(1 
 + Sqrt[(1 - a*x)/(1 + a*x)])^2) + 5/(64*(1 + Sqrt[(1 - a*x)/(1 + a*x)])))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 2115

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]
 

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 = 0.49 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.22

method	result	size
default	\(a \left (-\frac {1}{7 a^{2} x^{7}}-\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \left (a^{2} x^{2}-1\right ) \left (8 a^{4} x^{4}+12 a^{2} x^{2}+15\right )}{105 a \,x^{6}}\right )\)	\(76\)

Input:

int(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^7,x,method=_RETURNVERBOSE 
)
 

Output:

a*(-1/7/a^2/x^7-1/105/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))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.20 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx=-\frac {{\left (8 \, a^{7} x^{7} + 4 \, a^{5} x^{5} + 3 \, a^{3} x^{3} - 15 \, a x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 15}{105 \, a x^{7}} \] Input:

integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^7,x, algorithm="fri 
cas")
 

Output:

-1/105*((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)) + 15)/(a*x^7)
 

Sympy [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx=a \int \frac {1}{a x^{7} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + x^{6}}\, dx \] Input:

integrate(1/(1/a/x+(-1+1/a/x)**(1/2)*(1+1/a/x)**(1/2))/x**7,x)
 

Output:

a*Integral(1/(a*x**7*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)) + x**6), x)
 

Maxima [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx=\int { \frac {1}{x^{7} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \] Input:

integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^7,x, algorithm="max 
ima")
 

Output:

integrate(1/(x^7*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)
 

Giac [F]

\[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx=\int { \frac {1}{x^{7} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}} \,d x } \] Input:

integrate(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^7,x, algorithm="gia 
c")
 

Output:

integrate(1/(x^7*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))), x)
 

Mupad [B] (verification not implemented)

Time = 25.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.26 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx=-\frac {1}{7\,a\,x^7}-\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {a\,x^2}{35}-\frac {x}{7}-\frac {1}{7\,a}+\frac {a^2\,x^3}{35}+\frac {4\,a^3\,x^4}{105}+\frac {4\,a^4\,x^5}{105}+\frac {8\,a^5\,x^6}{105}+\frac {8\,a^6\,x^7}{105}\right )}{x^7\,\sqrt {\frac {1}{a\,x}+1}} \] Input:

int(1/(x^7*((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x))),x)
 

Output:

- 1/(7*a*x^7) - ((1/(a*x) - 1)^(1/2)*((a*x^2)/35 - x/7 - 1/(7*a) + (a^2*x^ 
3)/35 + (4*a^3*x^4)/105 + (4*a^4*x^5)/105 + (8*a^5*x^6)/105 + (8*a^6*x^7)/ 
105))/(x^7*(1/(a*x) + 1)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.39 \[ \int \frac {e^{-\text {sech}^{-1}(a x)}}{x^7} \, dx=\frac {a \left (15 \sqrt {a x +1}\, \sqrt {-a x +1}\, a^{5} x^{5}-8 \sqrt {a x +1}\, \sqrt {-a x +1}\, a^{4} x^{4}-4 \sqrt {a x +1}\, \sqrt {-a x +1}\, a^{2} x^{2}-3 \sqrt {a x +1}\, \sqrt {-a x +1}+8 a^{6} x^{6}+15 a^{5} x^{5}-4 a^{4} x^{4}-a^{2} x^{2}-18\right )}{105 x^{5} \left (\sqrt {a x +1}\, \sqrt {-a x +1}+1\right )} \] Input:

int(1/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/x^7,x)
 

Output:

(a*(15*sqrt(a*x + 1)*sqrt( - a*x + 1)*a**5*x**5 - 8*sqrt(a*x + 1)*sqrt( - 
a*x + 1)*a**4*x**4 - 4*sqrt(a*x + 1)*sqrt( - a*x + 1)*a**2*x**2 - 3*sqrt(a 
*x + 1)*sqrt( - a*x + 1) + 8*a**6*x**6 + 15*a**5*x**5 - 4*a**4*x**4 - a**2 
*x**2 - 18))/(105*x**5*(sqrt(a*x + 1)*sqrt( - a*x + 1) + 1))