\(\int e^{-\text {sech}^{-1}(a x)} x^m \, dx\) [43]

Optimal result

Integrand size = 12, antiderivative size = 131 \[ \int e^{-\text {sech}^{-1}(a x)} x^m \, dx=-\frac {2^{1+m} \left (\frac {e^{\text {sech}^{-1}(a x)}}{1+e^{2 \text {sech}^{-1}(a x)}}\right )^m \left (1+e^{2 \text {sech}^{-1}(a x)}\right )^m x^m (a x)^{-m} \left (e^{2 \text {sech}^{-1}(a x)} m \operatorname {Hypergeometric2F1}\left (1+\frac {m}{2},2+m,2+\frac {m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )-(2+m) \operatorname {Hypergeometric2F1}\left (\frac {m}{2},2+m,1+\frac {m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )\right )}{a m (2+m)} \] Output:

-2^(1+m)*((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))/(1+(1/a/x+(-1+1/a/x)^(1 
/2)*(1+1/a/x)^(1/2))^2))^m*(1+(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2)^ 
m*x^m*((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*m*hypergeom([2+m, 1+1/2* 
m],[2+1/2*m],-(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2)-(2+m)*hypergeom( 
[2+m, 1/2*m],[1+1/2*m],-(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2))/a/m/( 
2+m)/((a*x)^m)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00 \[ \int e^{-\text {sech}^{-1}(a x)} x^m \, dx=-\frac {2^{1+m} \left (\frac {e^{\text {sech}^{-1}(a x)}}{1+e^{2 \text {sech}^{-1}(a x)}}\right )^m \left (1+e^{2 \text {sech}^{-1}(a x)}\right )^m x^m (a x)^{-m} \left (e^{2 \text {sech}^{-1}(a x)} m \operatorname {Hypergeometric2F1}\left (1+\frac {m}{2},2+m,2+\frac {m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )-(2+m) \operatorname {Hypergeometric2F1}\left (\frac {m}{2},2+m,1+\frac {m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )\right )}{a m (2+m)} \] Input:

Integrate[x^m/E^ArcSech[a*x],x]
 

Output:

-((2^(1 + m)*(E^ArcSech[a*x]/(1 + E^(2*ArcSech[a*x])))^m*(1 + E^(2*ArcSech 
[a*x]))^m*x^m*(E^(2*ArcSech[a*x])*m*Hypergeometric2F1[1 + m/2, 2 + m, 2 + 
m/2, -E^(2*ArcSech[a*x])] - (2 + m)*Hypergeometric2F1[m/2, 2 + m, 1 + m/2, 
 -E^(2*ArcSech[a*x])]))/(a*m*(2 + m)*(a*x)^m))
 

Rubi [F]

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.

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

\(\Big \downarrow \) 6891

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

\(\Big \downarrow \) 7268

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2058

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 \left (1-\frac {1-a x}{a x+1}\right )^{-m} \left (\frac {1-\frac {1-a x}{a x+1}}{a \left (\frac {1-a x}{a x+1}+1\right )}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^m \left (-2 \int \frac {\left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\sqrt {\frac {1-a x}{a x+1}}+1}d\sqrt {\frac {1-a x}{a x+1}}+2 \sqrt {\frac {1-a x}{a x+1}} \operatorname {AppellF1}\left (\frac {1}{2},-m,m+2,\frac {3}{2},\frac {1-a x}{a x+1},-\frac {1-a x}{a x+1}\right )+\frac {\left (1-\frac {1-a x}{a x+1}\right )^{m+1} \left (\frac {1-a x}{a x+1}+1\right )^{-m-1}}{4 (m+1)}\right )}{a}\)

Input:

Int[x^m/E^ArcSech[a*x],x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 25

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

rule 2009

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

rule 2058

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ 
(r_.))^(p_), x_Symbol] :> Simp[Simp[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + 
 b*x^n)^(p*q)*(c + d*x^n)^(p*r))]   Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)^(p* 
r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]
 

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]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \frac {x^{m}}{\frac {1}{a x}+\sqrt {-1+\frac {1}{a x}}\, \sqrt {1+\frac {1}{a x}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(x^m/(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2)),x, algorithm="frica 
s")
 

Output:

integral(-(a*x*x^m*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - x^m)/(a* 
x), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((x**m*x)/(sqrt(a*x + 1)*sqrt( - a*x + 1) + 1),x)*a