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

Optimal. Leaf size=91 \[ \frac {\sqrt {\frac {1}{a x+1}} \sqrt {a x+1} x^m \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};a^2 x^2\right )}{a m (m+1)}+\frac {x^{m+1} e^{\text {sech}^{-1}(a x)}}{m+1}+\frac {x^m}{a m (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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])

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]

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*}

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Mathematica [A]  time = 0.31, size = 145, normalized size = 1.59 \[ -\frac {2^{m+1} x^m (a x)^{-m} e^{2 \text {sech}^{-1}(a x)} \left (\frac {e^{\text {sech}^{-1}(a x)}}{e^{2 \text {sech}^{-1}(a x)}+1}\right )^m \left (e^{2 \text {sech}^{-1}(a x)}+1\right )^m \left ((m+2) e^{2 \text {sech}^{-1}(a x)} \, _2F_1\left (\frac {m}{2}+2,m+2;\frac {m}{2}+3;-e^{2 \text {sech}^{-1}(a x)}\right )-(m+4) \, _2F_1\left (\frac {m}{2}+1,m+2;\frac {m}{2}+2;-e^{2 \text {sech}^{-1}(a x)}\right )\right )}{a (m+2) (m+4)} \]

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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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a x x^{m} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + x^{m}}{a x}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}\,{d x} \]

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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right ) x^{m}\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {a x + 1} \sqrt {-a x + 1} x^{m}}{x}\,{d x}}{a} + \frac {x^{m}}{a m} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,\left (\sqrt {\frac {1}{a\,x}-1}\,\sqrt {\frac {1}{a\,x}+1}+\frac {1}{a\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{m}}{x}\, dx + \int a x^{m} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}\, dx}{a} \]

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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