\(\int e^{\text {sech}^{-1}(a x^2)} x^m \, dx\) [61]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F(-2)]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 12, antiderivative size = 85 \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^m \, dx=-\frac {x^{-1+m}}{a (1-m)}+\frac {\sqrt {-1+\frac {1}{a x^2}} x^{1+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} (-1-m),\frac {3-m}{4},\frac {1}{a^2 x^4}\right )}{(1+m) \sqrt {1-\frac {1}{a x^2}}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 2.20 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.87 \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^m \, dx=\frac {2^{\frac {1+m}{2}} e^{\text {sech}^{-1}\left (a x^2\right )} \left (\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{1+e^{2 \text {sech}^{-1}\left (a x^2\right )}}\right )^{\frac {1+m}{2}} x^{1+m} \left (a x^2\right )^{\frac {1}{2} (-1-m)} \left ((7+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1-m}{4},\frac {7+m}{4},-e^{2 \text {sech}^{-1}\left (a x^2\right )}\right )-e^{2 \text {sech}^{-1}\left (a x^2\right )} (3+m) \operatorname {Hypergeometric2F1}\left (1,\frac {5-m}{4},\frac {11+m}{4},-e^{2 \text {sech}^{-1}\left (a x^2\right )}\right )\right )}{(3+m) (7+m)} \] Input:

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

Output:

(2^((1 + m)/2)*E^ArcSech[a*x^2]*(E^ArcSech[a*x^2]/(1 + E^(2*ArcSech[a*x^2] 
)))^((1 + m)/2)*x^(1 + m)*(a*x^2)^((-1 - m)/2)*((7 + m)*Hypergeometric2F1[ 
1, (1 - m)/4, (7 + m)/4, -E^(2*ArcSech[a*x^2])] - E^(2*ArcSech[a*x^2])*(3 
+ m)*Hypergeometric2F1[1, (5 - m)/4, (11 + m)/4, -E^(2*ArcSech[a*x^2])]))/ 
((3 + m)*(7 + m))
 

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6889, 15, 335, 888}

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.

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

\(\Big \downarrow \) 6889

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 335

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

\(\Big \downarrow \) 888

\(\displaystyle -\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} x^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m-1}{4},\frac {m+3}{4},a^2 x^4\right )}{a (1-m) (m+1)}-\frac {2 x^{m-1}}{a (1-m) (m+1)}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (a x^2\right )}}{m+1}\)

Input:

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

Output:

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

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 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 888
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

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]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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-2>0)', see `assume?` for more 
details)Is
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(x**m*sqrt(a*x**2 + 1)*sqrt( - a*x**2 + 1)*m - x**m*sqrt(a*x**2 + 1)*sqrt( 
 - a*x**2 + 1) + x**m*m + x**m - 2*int((x**m*sqrt(a*x**2 + 1)*sqrt( - a*x* 
*2 + 1))/(a**2*m*x**6 + a**2*x**6 - m*x**2 - x**2),x)*m**2*x + 2*int((x**m 
*sqrt(a*x**2 + 1)*sqrt( - a*x**2 + 1))/(a**2*m*x**6 + a**2*x**6 - m*x**2 - 
 x**2),x)*x)/(a*x*(m**2 - 1))