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

Optimal. Leaf size=91 \[ \frac{\sqrt{\frac{1}{a x+1}} \sqrt{a x+1} x^m \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2},\frac{m+2}{2},a^2 x^2\right )}{a m (m+1)}+\frac{x^m}{a m (m+1)}+\frac{x^{m+1} e^{\text{sech}^{-1}(a x)}}{m+1} \]

[Out]

x^m/(a*m*(1 + m)) + (E^ArcSech[a*x]*x^(1 + m))/(1 + m) + (x^m*Sqrt[(1 + a*x)^(-1)]*Sqrt[1 + a*x]*Hypergeometri
c2F1[1/2, m/2, (2 + m)/2, a^2*x^2])/(a*m*(1 + m))

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Rubi [A]  time = 0.0395333, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6335, 30, 125, 364} \[ \frac{\sqrt{\frac{1}{a x+1}} \sqrt{a x+1} x^m \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m+2}{2};a^2 x^2\right )}{a m (m+1)}+\frac{x^m}{a m (m+1)}+\frac{x^{m+1} e^{\text{sech}^{-1}(a x)}}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^m/(a*m*(1 + m)) + (E^ArcSech[a*x]*x^(1 + m))/(1 + m) + (x^m*Sqrt[(1 + a*x)^(-1)]*Sqrt[1 + a*x]*Hypergeometri
c2F1[1/2, m/2, (2 + m)/2, a^2*x^2])/(a*m*(1 + m))

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 125

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[(a*c + b*d*x^2)
^m*(f*x)^p, x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n, 0] && GtQ[a, 0] && GtQ
[c, 0]

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}(a x)} x^m \, dx &=\frac{e^{\text{sech}^{-1}(a x)} x^{1+m}}{1+m}+\frac{\int x^{-1+m} \, dx}{a (1+m)}+\frac{\left (\sqrt{\frac{1}{1+a x}} \sqrt{1+a x}\right ) \int \frac{x^{-1+m}}{\sqrt{1-a x} \sqrt{1+a x}} \, dx}{a (1+m)}\\ &=\frac{x^m}{a m (1+m)}+\frac{e^{\text{sech}^{-1}(a x)} x^{1+m}}{1+m}+\frac{\left (\sqrt{\frac{1}{1+a x}} \sqrt{1+a x}\right ) \int \frac{x^{-1+m}}{\sqrt{1-a^2 x^2}} \, dx}{a (1+m)}\\ &=\frac{x^m}{a m (1+m)}+\frac{e^{\text{sech}^{-1}(a x)} x^{1+m}}{1+m}+\frac{x^m \sqrt{\frac{1}{1+a x}} \sqrt{1+a x} \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{2+m}{2};a^2 x^2\right )}{a m (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.302242, size = 145, normalized size = 1.59 \[ -\frac{2^{m+1} x^m (a x)^{-m} e^{2 \text{sech}^{-1}(a x)} \left (\frac{e^{\text{sech}^{-1}(a x)}}{e^{2 \text{sech}^{-1}(a x)}+1}\right )^m \left (e^{2 \text{sech}^{-1}(a x)}+1\right )^m \left ((m+2) e^{2 \text{sech}^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{m}{2}+2,m+2,\frac{m}{2}+3,-e^{2 \text{sech}^{-1}(a x)}\right )-(m+4) \text{Hypergeometric2F1}\left (\frac{m}{2}+1,m+2,\frac{m}{2}+2,-e^{2 \text{sech}^{-1}(a x)}\right )\right )}{a (m+2) (m+4)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2^(1 + m)*E^(2*ArcSech[a*x])*(E^ArcSech[a*x]/(1 + E^(2*ArcSech[a*x])))^m*(1 + E^(2*ArcSech[a*x]))^m*x^m*(-(
(4 + m)*Hypergeometric2F1[1 + m/2, 2 + m, 2 + m/2, -E^(2*ArcSech[a*x])]) + E^(2*ArcSech[a*x])*(2 + m)*Hypergeo
metric2F1[2 + m/2, 2 + m, 3 + m/2, -E^(2*ArcSech[a*x])]))/(a*(2 + m)*(4 + m)*(a*x)^m))

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Maple [F]  time = 0.174, size = 0, normalized size = 0. \begin{align*} \int \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ){x}^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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