3.56 \(\int e^{\text{sech}^{-1}(a x^3)} x^m \, dx\)

Optimal. Leaf size=109 \[ -\frac{3 \sqrt{\frac{1}{a x^3+1}} \sqrt{a x^3+1} x^{m-2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m-2}{6},\frac{m+4}{6},a^2 x^6\right )}{a \left (-m^2+m+2\right )}-\frac{3 x^{m-2}}{a \left (-m^2+m+2\right )}+\frac{x^{m+1} e^{\text{sech}^{-1}\left (a x^3\right )}}{m+1} \]

[Out]

(-3*x^(-2 + m))/(a*(2 + m - m^2)) + (E^ArcSech[a*x^3]*x^(1 + m))/(1 + m) - (3*x^(-2 + m)*Sqrt[(1 + a*x^3)^(-1)
]*Sqrt[1 + a*x^3]*Hypergeometric2F1[1/2, (-2 + m)/6, (4 + m)/6, a^2*x^6])/(a*(2 + m - m^2))

________________________________________________________________________________________

Rubi [A]  time = 0.0677183, antiderivative size = 109, 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} \[ -\frac{3 \sqrt{\frac{1}{a x^3+1}} \sqrt{a x^3+1} x^{m-2} \, _2F_1\left (\frac{1}{2},\frac{m-2}{6};\frac{m+4}{6};a^2 x^6\right )}{a \left (-m^2+m+2\right )}-\frac{3 x^{m-2}}{a \left (-m^2+m+2\right )}+\frac{x^{m+1} e^{\text{sech}^{-1}\left (a x^3\right )}}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSech[a*x^3]*x^m,x]

[Out]

(-3*x^(-2 + m))/(a*(2 + m - m^2)) + (E^ArcSech[a*x^3]*x^(1 + m))/(1 + m) - (3*x^(-2 + m)*Sqrt[(1 + a*x^3)^(-1)
]*Sqrt[1 + a*x^3]*Hypergeometric2F1[1/2, (-2 + m)/6, (4 + m)/6, a^2*x^6])/(a*(2 + 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 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^3\right )} x^m \, dx &=\frac{e^{\text{sech}^{-1}\left (a x^3\right )} x^{1+m}}{1+m}+\frac{3 \int x^{-3+m} \, dx}{a (1+m)}+\frac{\left (3 \sqrt{\frac{1}{1+a x^3}} \sqrt{1+a x^3}\right ) \int \frac{x^{-3+m}}{\sqrt{1-a x^3} \sqrt{1+a x^3}} \, dx}{a (1+m)}\\ &=-\frac{3 x^{-2+m}}{a \left (2+m-m^2\right )}+\frac{e^{\text{sech}^{-1}\left (a x^3\right )} x^{1+m}}{1+m}+\frac{\left (3 \sqrt{\frac{1}{1+a x^3}} \sqrt{1+a x^3}\right ) \int \frac{x^{-3+m}}{\sqrt{1-a^2 x^6}} \, dx}{a (1+m)}\\ &=-\frac{3 x^{-2+m}}{a \left (2+m-m^2\right )}+\frac{e^{\text{sech}^{-1}\left (a x^3\right )} x^{1+m}}{1+m}-\frac{3 x^{-2+m} \sqrt{\frac{1}{1+a x^3}} \sqrt{1+a x^3} \, _2F_1\left (\frac{1}{2},\frac{1}{6} (-2+m);\frac{4+m}{6};a^2 x^6\right )}{a \left (2+m-m^2\right )}\\ \end{align*}

Mathematica [A]  time = 2.49469, size = 159, normalized size = 1.46 \[ \frac{2^{\frac{m+1}{3}} x^{m+1} \left (a x^3\right )^{\frac{1}{3} (-m-1)} e^{\text{sech}^{-1}\left (a x^3\right )} \left (\frac{e^{\text{sech}^{-1}\left (a x^3\right )}}{e^{2 \text{sech}^{-1}\left (a x^3\right )}+1}\right )^{\frac{m+1}{3}} \left ((m+10) \text{Hypergeometric2F1}\left (1,\frac{2-m}{6},\frac{m+10}{6},-e^{2 \text{sech}^{-1}\left (a x^3\right )}\right )-(m+4) e^{2 \text{sech}^{-1}\left (a x^3\right )} \text{Hypergeometric2F1}\left (1,\frac{8-m}{6},\frac{m+16}{6},-e^{2 \text{sech}^{-1}\left (a x^3\right )}\right )\right )}{(m+4) (m+10)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSech[a*x^3]*x^m,x]

[Out]

(2^((1 + m)/3)*E^ArcSech[a*x^3]*(E^ArcSech[a*x^3]/(1 + E^(2*ArcSech[a*x^3])))^((1 + m)/3)*x^(1 + m)*(a*x^3)^((
-1 - m)/3)*((10 + m)*Hypergeometric2F1[1, (2 - m)/6, (10 + m)/6, -E^(2*ArcSech[a*x^3])] - E^(2*ArcSech[a*x^3])
*(4 + m)*Hypergeometric2F1[1, (8 - m)/6, (16 + m)/6, -E^(2*ArcSech[a*x^3])]))/((4 + m)*(10 + m))

________________________________________________________________________________________

Maple [F]  time = 0.23, size = 0, normalized size = 0. \begin{align*} \int \left ({\frac{1}{{x}^{3}a}}+\sqrt{{\frac{1}{{x}^{3}a}}-1}\sqrt{{\frac{1}{{x}^{3}a}}+1} \right ){x}^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/x^3/a+(1/x^3/a-1)^(1/2)*(1/x^3/a+1)^(1/2))*x^m,x)

[Out]

int((1/x^3/a+(1/x^3/a-1)^(1/2)*(1/x^3/a+1)^(1/2))*x^m,x)

________________________________________________________________________________________

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^3+(1/a/x^3-1)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a x^{3} x^{m} \sqrt{\frac{a x^{3} + 1}{a x^{3}}} \sqrt{-\frac{a x^{3} - 1}{a x^{3}}} + x^{m}}{a x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^3+(1/a/x^3-1)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x, algorithm="fricas")

[Out]

integral((a*x^3*x^m*sqrt((a*x^3 + 1)/(a*x^3))*sqrt(-(a*x^3 - 1)/(a*x^3)) + x^m)/(a*x^3), x)

________________________________________________________________________________________

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**3+(1/a/x**3-1)**(1/2)*(1/a/x**3+1)**(1/2))*x**m,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m}{\left (\sqrt{\frac{1}{a x^{3}} + 1} \sqrt{\frac{1}{a x^{3}} - 1} + \frac{1}{a x^{3}}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^3+(1/a/x^3-1)^(1/2)*(1/a/x^3+1)^(1/2))*x^m,x, algorithm="giac")

[Out]

integrate(x^m*(sqrt(1/(a*x^3) + 1)*sqrt(1/(a*x^3) - 1) + 1/(a*x^3)), x)