\(\int e^{\text {sech}^{-1}(\frac {a}{x})} x^m \, dx\) [59]

Optimal result

Integrand size = 12, antiderivative size = 109 \[ \int e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} x^m \, dx=\frac {e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} x^{1+m}}{1+m}-\frac {x^{2+m}}{a \left (2+3 m+m^2\right )}-\frac {\sqrt {\frac {1}{1+\frac {a}{x}}} \sqrt {1+\frac {a}{x}} x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2-m),-\frac {m}{2},\frac {a^2}{x^2}\right )}{a \left (2+3 m+m^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6470, 30, 265, 346, 371} \[ \int e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} x^m \, dx=-\frac {\sqrt {\frac {1}{\frac {a}{x}+1}} \sqrt {\frac {a}{x}+1} x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-2),-\frac {m}{2},\frac {a^2}{x^2}\right )}{a \left (m^2+3 m+2\right )}-\frac {x^{m+2}}{a \left (m^2+3 m+2\right )}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (\frac {a}{x}\right )}}{m+1} \]

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 265

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[(c*x)
^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (Intege
rQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 346

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(-c^(-1))*(c*x)^(m + 1)*(1/x)^(m + 1),
Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m
]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 6470

Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(E^ArcSech[a*x^p]/(m + 1)), x] + (Dist
[p/(a*(m + 1)), Int[x^(m - p), x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} x^{1+m}}{1+m}-\frac {\int x^{1+m} \, dx}{a (1+m)}-\frac {\left (\sqrt {\frac {1}{1+\frac {a}{x}}} \sqrt {1+\frac {a}{x}}\right ) \int \frac {x^{1+m}}{\sqrt {1-\frac {a}{x}} \sqrt {1+\frac {a}{x}}} \, dx}{a (1+m)} \\ & = \frac {e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} x^{1+m}}{1+m}-\frac {x^{2+m}}{a \left (2+3 m+m^2\right )}-\frac {\left (\sqrt {\frac {1}{1+\frac {a}{x}}} \sqrt {1+\frac {a}{x}}\right ) \int \frac {x^{1+m}}{\sqrt {1-\frac {a^2}{x^2}}} \, dx}{a (1+m)} \\ & = \frac {e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} x^{1+m}}{1+m}-\frac {x^{2+m}}{a \left (2+3 m+m^2\right )}+\frac {\left (\sqrt {\frac {1}{1+\frac {a}{x}}} \sqrt {1+\frac {a}{x}} \left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-3-m}}{\sqrt {1-a^2 x^2}} \, dx,x,\frac {1}{x}\right )}{a (1+m)} \\ & = \frac {e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} x^{1+m}}{1+m}-\frac {x^{2+m}}{a \left (2+3 m+m^2\right )}-\frac {\sqrt {\frac {1}{1+\frac {a}{x}}} \sqrt {1+\frac {a}{x}} x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2-m),-\frac {m}{2},\frac {a^2}{x^2}\right )}{a \left (2+3 m+m^2\right )} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.71 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.28 \[ \int e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} x^m \, dx=-\frac {2^{-1-m} a e^{\text {sech}^{-1}\left (\frac {a}{x}\right )} \left (\frac {e^{\text {sech}^{-1}\left (\frac {a}{x}\right )}}{1+e^{2 \text {sech}^{-1}\left (\frac {a}{x}\right )}}\right )^{-1-m} \left (\frac {a}{x}\right )^m x^m \left (-\left ((-2+m) \operatorname {Hypergeometric2F1}\left (1,1+\frac {m}{2},1-\frac {m}{2},-e^{2 \text {sech}^{-1}\left (\frac {a}{x}\right )}\right )\right )+e^{2 \text {sech}^{-1}\left (\frac {a}{x}\right )} m \operatorname {Hypergeometric2F1}\left (1,2+\frac {m}{2},2-\frac {m}{2},-e^{2 \text {sech}^{-1}\left (\frac {a}{x}\right )}\right )\right )}{(-2+m) m} \]

[In]

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

[Out]

-((2^(-1 - m)*a*E^ArcSech[a/x]*(E^ArcSech[a/x]/(1 + E^(2*ArcSech[a/x])))^(-1 - m)*(a/x)^m*x^m*(-((-2 + m)*Hype
rgeometric2F1[1, 1 + m/2, 1 - m/2, -E^(2*ArcSech[a/x])]) + E^(2*ArcSech[a/x])*m*Hypergeometric2F1[1, 2 + m/2,
2 - m/2, -E^(2*ArcSech[a/x])]))/((-2 + m)*m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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