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

Optimal. Leaf size=109 \[ -\frac{\sqrt{\frac{1}{\frac{a}{x}+1}} \sqrt{\frac{a}{x}+1} x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-m-2),-\frac{m}{2},\frac{a^2}{x^2}\right )}{a \left (m^2+3 m+2\right )}-\frac{x^{m+2}}{a \left (m^2+3 m+2\right )}+\frac{x^{m+1} e^{\text{sech}^{-1}\left (\frac{a}{x}\right )}}{m+1} \]

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

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Rubi [A]  time = 0.0751257, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6335, 30, 259, 339, 364} \[ -\frac{\sqrt{\frac{1}{\frac{a}{x}+1}} \sqrt{\frac{a}{x}+1} x^{m+2} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-m-2);-\frac{m}{2};\frac{a^2}{x^2}\right )}{a \left (m^2+3 m+2\right )}-\frac{x^{m+2}}{a \left (m^2+3 m+2\right )}+\frac{x^{m+1} e^{\text{sech}^{-1}\left (\frac{a}{x}\right )}}{m+1} \]

Antiderivative was successfully verified.

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

Rule 6335

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]*Sqrt[1/(1 + a*x^p)])/(a*(m + 1)), Int[x^(m - p
)/(Sqrt[1 + a*x^p]*Sqrt[1 - a*x^p]), x], x]) /; FreeQ[{a, m, p}, x] && NeQ[m, -1]

Rule 30

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

Rule 259

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 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, 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 364

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

Rubi steps

\begin{align*} \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{\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 ) \operatorname{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} \, _2F_1\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]  time = 0.825083, size = 139, normalized size = 1.28 \[ -\frac{a 2^{-m-1} x^m \left (\frac{a}{x}\right )^m e^{\text{sech}^{-1}\left (\frac{a}{x}\right )} \left (\frac{e^{\text{sech}^{-1}\left (\frac{a}{x}\right )}}{e^{2 \text{sech}^{-1}\left (\frac{a}{x}\right )}+1}\right )^{-m-1} \left (m e^{2 \text{sech}^{-1}\left (\frac{a}{x}\right )} \text{Hypergeometric2F1}\left (1,\frac{m}{2}+2,2-\frac{m}{2},-e^{2 \text{sech}^{-1}\left (\frac{a}{x}\right )}\right )-(m-2) \text{Hypergeometric2F1}\left (1,\frac{m}{2}+1,1-\frac{m}{2},-e^{2 \text{sech}^{-1}\left (\frac{a}{x}\right )}\right )\right )}{(m-2) m} \]

Warning: Unable to verify antiderivative.

[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))

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Maple [F]  time = 0.643, size = 0, normalized size = 0. \begin{align*} \int \left ({\frac{x}{a}}+\sqrt{-1+{\frac{x}{a}}}\sqrt{1+{\frac{x}{a}}} \right ){x}^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x^{2} x^{m}}{a{\left (m + 2\right )}} + \frac{\int \sqrt{a + x} \sqrt{-a + x} x^{m}\,{d x}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a x^{m} \sqrt{\frac{a + x}{a}} \sqrt{-\frac{a - x}{a}} + x x^{m}}{a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int x x^{m}\, dx + \int a x^{m} \sqrt{-1 + \frac{x}{a}} \sqrt{1 + \frac{x}{a}}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m}{\left (\sqrt{\frac{x}{a} + 1} \sqrt{\frac{x}{a} - 1} + \frac{x}{a}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)