\(\int e^{3 \text {sech}^{-1}(a x)} x^m \, dx\) [40]

Optimal result

Integrand size = 12, antiderivative size = 146 \[ \int e^{3 \text {sech}^{-1}(a x)} x^m \, dx=-\frac {2^{1+m} e^{4 \text {sech}^{-1}(a x)} \left (\frac {e^{\text {sech}^{-1}(a x)}}{1+e^{2 \text {sech}^{-1}(a x)}}\right )^m \left (1+e^{2 \text {sech}^{-1}(a x)}\right )^m x^m (a x)^{-m} \left ((-6-m) \operatorname {Hypergeometric2F1}\left (2+\frac {m}{2},2+m,3+\frac {m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )+e^{2 \text {sech}^{-1}(a x)} (4+m) \operatorname {Hypergeometric2F1}\left (3+\frac {m}{2},2+m,4+\frac {m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )\right )}{a (4+m) (6+m)} \] Output:

-2^(1+m)*(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^4*((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*((-6-m)*hypergeom([2+m, 2 
+1/2*m],[3+1/2*m],-(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2)+(1/a/x+(-1+ 
1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*(4+m)*hypergeom([2+m, 3+1/2*m],[4+1/2*m],- 
(1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2))/a/(4+m)/(6+m)/((a*x)^m)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.99 \[ \int e^{3 \text {sech}^{-1}(a x)} x^m \, dx=-\frac {2^{1+m} e^{4 \text {sech}^{-1}(a x)} \left (\frac {e^{\text {sech}^{-1}(a x)}}{1+e^{2 \text {sech}^{-1}(a x)}}\right )^m \left (1+e^{2 \text {sech}^{-1}(a x)}\right )^m x^m (a x)^{-m} \left (-\left ((6+m) \operatorname {Hypergeometric2F1}\left (2+\frac {m}{2},2+m,3+\frac {m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )\right )+e^{2 \text {sech}^{-1}(a x)} (4+m) \operatorname {Hypergeometric2F1}\left (3+\frac {m}{2},2+m,4+\frac {m}{2},-e^{2 \text {sech}^{-1}(a x)}\right )\right )}{a (4+m) (6+m)} \] Input:

Integrate[E^(3*ArcSech[a*x])*x^m,x]
 

Output:

-((2^(1 + m)*E^(4*ArcSech[a*x])*(E^ArcSech[a*x]/(1 + E^(2*ArcSech[a*x])))^ 
m*(1 + E^(2*ArcSech[a*x]))^m*x^m*(-((6 + m)*Hypergeometric2F1[2 + m/2, 2 + 
 m, 3 + m/2, -E^(2*ArcSech[a*x])]) + E^(2*ArcSech[a*x])*(4 + m)*Hypergeome 
tric2F1[3 + m/2, 2 + m, 4 + m/2, -E^(2*ArcSech[a*x])]))/(a*(4 + m)*(6 + 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.

\(\displaystyle \int x^m e^{3 \text {sech}^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 6891

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

\(\Big \downarrow \) 7268

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2058

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

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {4 \left (1-\frac {1-a x}{a x+1}\right )^{-m} \left (\frac {1-\frac {1-a x}{a x+1}}{a \left (\frac {1-a x}{a x+1}+1\right )}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^m \int \left (-\sqrt {\frac {1-a x}{a x+1}} \left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}-\frac {18 \left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\sqrt {\frac {1-a x}{a x+1}}-1}-\frac {20 \left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^2}-\frac {8 \left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^3}-6 \left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}\right )d\sqrt {\frac {1-a x}{a x+1}}}{a}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 \left (1-\frac {1-a x}{a x+1}\right )^{-m} \left (\frac {1-\frac {1-a x}{a x+1}}{a \left (\frac {1-a x}{a x+1}+1\right )}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^m \left (-8 \int \frac {\left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}-20 \int \frac {\left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\left (\sqrt {\frac {1-a x}{a x+1}}-1\right )^2}d\sqrt {\frac {1-a x}{a x+1}}-18 \int \frac {\left (1-\frac {1-a x}{a x+1}\right )^m \left (\frac {1-a x}{a x+1}+1\right )^{-m-2}}{\sqrt {\frac {1-a x}{a x+1}}-1}d\sqrt {\frac {1-a x}{a x+1}}-6 \sqrt {\frac {1-a x}{a x+1}} \operatorname {AppellF1}\left (\frac {1}{2},-m,m+2,\frac {3}{2},\frac {1-a x}{a x+1},-\frac {1-a x}{a x+1}\right )+\frac {\left (1-\frac {1-a x}{a x+1}\right )^{m+1} \left (\frac {1-a x}{a x+1}+1\right )^{-m-1}}{4 (m+1)}\right )}{a}\)

Input:

Int[E^(3*ArcSech[a*x])*x^m,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]

\[\int \left (\frac {1}{a x}+\sqrt {-1+\frac {1}{a x}}\, \sqrt {1+\frac {1}{a x}}\right )^{3} x^{m}d x\]

Input:

int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^3*x^m,x)
 

Output:

int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^3*x^m,x)
 

Fricas [F]

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

integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^3*x^m,x, algorithm="fri 
cas")
 

Output:

integral(-((a^3*x^3 - 4*a*x)*x^m*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a* 
x)) + (3*a^2*x^2 - 4)*x^m)/(a^3*x^3), x)
 

Sympy [F]

\[ \int e^{3 \text {sech}^{-1}(a x)} x^m \, dx=\frac {\int \frac {4 x^{m}}{x^{3}}\, dx + \int \left (- \frac {3 a^{2} x^{m}}{x}\right )\, dx + \int \left (- a^{3} x^{m} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}\right )\, dx + \int \frac {4 a x^{m} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}}{x^{2}}\, dx}{a^{3}} \] Input:

integrate((1/a/x+(-1+1/a/x)**(1/2)*(1+1/a/x)**(1/2))**3*x**m,x)
 

Output:

(Integral(4*x**m/x**3, x) + Integral(-3*a**2*x**m/x, x) + Integral(-a**3*x 
**m*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)), x) + Integral(4*a*x**m*sqrt(-1 + 
 1/(a*x))*sqrt(1 + 1/(a*x))/x**2, x))/a**3
 

Maxima [F(-2)]

Exception generated. \[ \int e^{3 \text {sech}^{-1}(a x)} x^m \, dx=\text {Exception raised: ValueError} \] Input:

integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^3*x^m,x, algorithm="max 
ima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(m-3>0)', see `assume?` for more 
details)Is
 

Giac [F]

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

integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^3*x^m,x, algorithm="gia 
c")
 

Output:

integrate(x^m*(sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{3 \text {sech}^{-1}(a x)} x^m \, dx=\int x^m\,{\left (\sqrt {\frac {1}{a\,x}-1}\,\sqrt {\frac {1}{a\,x}+1}+\frac {1}{a\,x}\right )}^3 \,d x \] Input:

int(x^m*((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x))^3,x)
 

Output:

int(x^m*((1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2) + 1/(a*x))^3, x)
 

Reduce [F]

\[ \int e^{3 \text {sech}^{-1}(a x)} x^m \, dx=\frac {-3 x^{m} a^{2} m \,x^{2}+6 x^{m} a^{2} x^{2}+4 x^{m} m +4 \left (\int \frac {x^{m} \sqrt {a x +1}\, \sqrt {-a x +1}}{x^{3}}d x \right ) m^{2} x^{2}-8 \left (\int \frac {x^{m} \sqrt {a x +1}\, \sqrt {-a x +1}}{x^{3}}d x \right ) m \,x^{2}-\left (\int \frac {x^{m} \sqrt {a x +1}\, \sqrt {-a x +1}}{x}d x \right ) a^{2} m^{2} x^{2}+2 \left (\int \frac {x^{m} \sqrt {a x +1}\, \sqrt {-a x +1}}{x}d x \right ) a^{2} m \,x^{2}}{a^{3} m \,x^{2} \left (m -2\right )} \] Input:

int((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^3*x^m,x)
                                                                                    
                                                                                    
 

Output:

( - 3*x**m*a**2*m*x**2 + 6*x**m*a**2*x**2 + 4*x**m*m + 4*int((x**m*sqrt(a* 
x + 1)*sqrt( - a*x + 1))/x**3,x)*m**2*x**2 - 8*int((x**m*sqrt(a*x + 1)*sqr 
t( - a*x + 1))/x**3,x)*m*x**2 - int((x**m*sqrt(a*x + 1)*sqrt( - a*x + 1))/ 
x,x)*a**2*m**2*x**2 + 2*int((x**m*sqrt(a*x + 1)*sqrt( - a*x + 1))/x,x)*a** 
2*m*x**2)/(a**3*m*x**2*(m - 2))