\(\int e^{-2 \text {sech}^{-1}(a x)} x^m \, dx\) [44]

Optimal result

Integrand size = 12, antiderivative size = 155 \[ \int e^{-2 \text {sech}^{-1}(a x)} x^m \, dx=-\frac {2^{1+m} e^{-m \text {sech}^{-1}(a x)} \left (\frac {e^{\text {sech}^{-1}(a x)}}{1+e^{2 \text {sech}^{-1}(a x)}}\right )^m \left (1+e^{2 \text {sech}^{-1}(a x)}\right )^m x^m (a x)^{-m} \left (-e^{(-1+m) \text {sech}^{-1}(a x)} (1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),2+m,\frac {1+m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )+e^{(1+m) \text {sech}^{-1}(a x)} (-1+m) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},2+m,\frac {3+m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )\right )}{a \left (-1+m^2\right )} \] 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*(-exp((-1+m)*arcsech(a*x))*(1+m)*hypergeom([2+m, -1/2+1/2*m],[1/2+1/ 
2*m],-(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2)+exp((1+m)*arcsech(a*x))* 
(-1+m)*hypergeom([2+m, 1/2+1/2*m],[3/2+1/2*m],-(1/a/x+(-1+1/a/x)^(1/2)*(1+ 
1/a/x)^(1/2))^2))/a/exp(m*arcsech(a*x))/(m^2-1)/((a*x)^m)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.88 \[ \int e^{-2 \text {sech}^{-1}(a x)} x^m \, dx=\frac {2^{1+m} \left (\frac {e^{\text {sech}^{-1}(a x)}}{1+e^{2 \text {sech}^{-1}(a x)}}\right )^{-1+m} \left (1+e^{2 \text {sech}^{-1}(a x)}\right )^{-1+m} x^m (a x)^{-m} \left ((1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),2+m,\frac {1+m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )-e^{2 \text {sech}^{-1}(a x)} (-1+m) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},2+m,\frac {3+m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )\right )}{a \left (-1+m^2\right )} \] Input:

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

Output:

(2^(1 + m)*(E^ArcSech[a*x]/(1 + E^(2*ArcSech[a*x])))^(-1 + m)*(1 + E^(2*Ar 
cSech[a*x]))^(-1 + m)*x^m*((1 + m)*Hypergeometric2F1[(-1 + m)/2, 2 + m, (1 
 + m)/2, -E^(2*ArcSech[a*x])] - E^(2*ArcSech[a*x])*(-1 + m)*Hypergeometric 
2F1[(1 + m)/2, 2 + m, (3 + m)/2, -E^(2*ArcSech[a*x])]))/(a*(-1 + m^2)*(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.

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

\(\Big \downarrow \) 6891

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

\(\Big \downarrow \) 7268

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

\(\Big \downarrow \) 2058

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 \left (1-\frac {1-a x}{a x+1}\right )^{-m} \left (\frac {1-\frac {1-a x}{a x+1}}{a \left (\frac {1-a x}{a x+1}+1\right )}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^m \left (-4 \int \frac {\left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^2}d\sqrt {\frac {1-a x}{a x+1}}+8 \int \frac {\left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\sqrt {\frac {1-a x}{a x+1}}+1}d\sqrt {\frac {1-a x}{a x+1}}-4 \sqrt {\frac {1-a x}{a x+1}} \operatorname {AppellF1}\left (\frac {1}{2},-m,m+2,\frac {3}{2},\frac {1-a x}{a x+1},-\frac {1-a x}{a x+1}\right )-\frac {\left (1-\frac {1-a x}{a x+1}\right )^{m+1} \left (\frac {1-a x}{a x+1}+1\right )^{-m-1}}{4 (m+1)}\right )}{a}\)

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

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]

\[\int \frac {x^{m}}{\left (\frac {1}{a x}+\sqrt {-1+\frac {1}{a x}}\, \sqrt {1+\frac {1}{a x}}\right )^{2}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-%%%{2,[2,2]%%%}/%%%{-8,[3,3]%%%}+%%%{8,[2,2]%%%}*%%%{8,[3 
,3]%%%}+%
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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