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

Optimal. Leaf size=107 \[ -\frac{2 \sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} x^{m-1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m-1}{4},\frac{m+3}{4},a^2 x^4\right )}{a \left (1-m^2\right )}-\frac{2 x^{m-1}}{a \left (1-m^2\right )}+\frac{x^{m+1} e^{\text{sech}^{-1}\left (a x^2\right )}}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0557968, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6335, 30, 259, 364} \[ -\frac{2 \sqrt{\frac{1}{a x^2+1}} \sqrt{a x^2+1} x^{m-1} \, _2F_1\left (\frac{1}{2},\frac{m-1}{4};\frac{m+3}{4};a^2 x^4\right )}{a \left (1-m^2\right )}-\frac{2 x^{m-1}}{a \left (1-m^2\right )}+\frac{x^{m+1} e^{\text{sech}^{-1}\left (a x^2\right )}}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^(-1 + m))/(a*(1 - m^2)) + (E^ArcSech[a*x^2]*x^(1 + m))/(1 + m) - (2*x^(-1 + m)*Sqrt[(1 + a*x^2)^(-1)]*Sq
rt[1 + a*x^2]*Hypergeometric2F1[1/2, (-1 + m)/4, (3 + m)/4, a^2*x^4])/(a*(1 - 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 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 (a x^2\right )} x^m \, dx &=\frac{e^{\text{sech}^{-1}\left (a x^2\right )} x^{1+m}}{1+m}+\frac{2 \int x^{-2+m} \, dx}{a (1+m)}+\frac{\left (2 \sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \int \frac{x^{-2+m}}{\sqrt{1-a x^2} \sqrt{1+a x^2}} \, dx}{a (1+m)}\\ &=-\frac{2 x^{-1+m}}{a \left (1-m^2\right )}+\frac{e^{\text{sech}^{-1}\left (a x^2\right )} x^{1+m}}{1+m}+\frac{\left (2 \sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2}\right ) \int \frac{x^{-2+m}}{\sqrt{1-a^2 x^4}} \, dx}{a (1+m)}\\ &=-\frac{2 x^{-1+m}}{a \left (1-m^2\right )}+\frac{e^{\text{sech}^{-1}\left (a x^2\right )} x^{1+m}}{1+m}-\frac{2 x^{-1+m} \sqrt{\frac{1}{1+a x^2}} \sqrt{1+a x^2} \, _2F_1\left (\frac{1}{2},\frac{1}{4} (-1+m);\frac{3+m}{4};a^2 x^4\right )}{a \left (1-m^2\right )}\\ \end{align*}

Mathematica [A]  time = 2.30329, size = 159, normalized size = 1.49 \[ \frac{2^{\frac{m+1}{2}} x^{m+1} \left (a x^2\right )^{\frac{1}{2} (-m-1)} e^{\text{sech}^{-1}\left (a x^2\right )} \left (\frac{e^{\text{sech}^{-1}\left (a x^2\right )}}{e^{2 \text{sech}^{-1}\left (a x^2\right )}+1}\right )^{\frac{m+1}{2}} \left ((m+7) \text{Hypergeometric2F1}\left (1,\frac{1-m}{4},\frac{m+7}{4},-e^{2 \text{sech}^{-1}\left (a x^2\right )}\right )-(m+3) e^{2 \text{sech}^{-1}\left (a x^2\right )} \text{Hypergeometric2F1}\left (1,\frac{5-m}{4},\frac{m+11}{4},-e^{2 \text{sech}^{-1}\left (a x^2\right )}\right )\right )}{(m+3) (m+7)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(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))

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Maple [F]  time = 0.194, size = 0, normalized size = 0. \begin{align*} \int \left ({\frac{1}{a{x}^{2}}}+\sqrt{{\frac{1}{a{x}^{2}}}-1}\sqrt{{\frac{1}{a{x}^{2}}}+1} \right ){x}^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(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^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(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^{2} x^{m} \sqrt{\frac{a x^{2} + 1}{a x^{2}}} \sqrt{-\frac{a x^{2} - 1}{a x^{2}}} + x^{m}}{a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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