\(\int e^{\text {sech}^{-1}(a x^3)} x^m \, dx\) [56]

Optimal result

Integrand size = 12, antiderivative size = 109 \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=-\frac {3 x^{-2+m}}{a \left (2+m-m^2\right )}+\frac {e^{\text {sech}^{-1}\left (a x^3\right )} x^{1+m}}{1+m}-\frac {3 x^{-2+m} \sqrt {\frac {1}{1+a x^3}} \sqrt {1+a x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-2+m),\frac {4+m}{6},a^2 x^6\right )}{a \left (2+m-m^2\right )} \]

[Out]

-3*x^(-2+m)/a/(-m^2+m+2)+(1/a/x^3+(1/a/x^3-1)^(1/2)*(1/a/x^3+1)^(1/2))*x^(1+m)/(1+m)-3*x^(-2+m)*hypergeom([1/2
, -1/3+1/6*m],[2/3+1/6*m],a^2*x^6)*(1/(a*x^3+1))^(1/2)*(a*x^3+1)^(1/2)/a/(-m^2+m+2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6470, 30, 265, 371} \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=-\frac {3 \sqrt {\frac {1}{a x^3+1}} \sqrt {a x^3+1} x^{m-2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m-2}{6},\frac {m+4}{6},a^2 x^6\right )}{a \left (-m^2+m+2\right )}-\frac {3 x^{m-2}}{a \left (-m^2+m+2\right )}+\frac {x^{m+1} e^{\text {sech}^{-1}\left (a x^3\right )}}{m+1} \]

[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 30

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

Rule 265

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 6470

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]/(a*(m + 1)))*Sqrt[1/(1 + a*x^p)], 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*} \text {integral}& = \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} \operatorname {Hypergeometric2F1}\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] (verified)

Time = 0.61 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.72 \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\frac {2^{\frac {1+m}{3}} e^{\text {sech}^{-1}\left (a x^3\right )} \left (\frac {e^{\text {sech}^{-1}\left (a x^3\right )}}{1+e^{2 \text {sech}^{-1}\left (a x^3\right )}}\right )^{\frac {1+m}{3}} \left (1+e^{2 \text {sech}^{-1}\left (a x^3\right )}\right )^{\frac {1+m}{3}} x^{1+m} \left (a x^3\right )^{\frac {1}{3} (-1-m)} \left ((10+m) \operatorname {Hypergeometric2F1}\left (\frac {4+m}{6},\frac {4+m}{3},\frac {10+m}{6},-e^{2 \text {sech}^{-1}\left (a x^3\right )}\right )-e^{2 \text {sech}^{-1}\left (a x^3\right )} (4+m) \operatorname {Hypergeometric2F1}\left (\frac {4+m}{3},\frac {10+m}{6},\frac {16+m}{6},-e^{2 \text {sech}^{-1}\left (a x^3\right )}\right )\right )}{(4+m) (10+m)} \]

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

Maple [F]

\[\int \left (\frac {1}{a \,x^{3}}+\sqrt {\frac {1}{a \,x^{3}}-1}\, \sqrt {\frac {1}{a \,x^{3}}+1}\right ) x^{m}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\int { x^{m} {\left (\sqrt {\frac {1}{a x^{3}} + 1} \sqrt {\frac {1}{a x^{3}} - 1} + \frac {1}{a x^{3}}\right )} \,d x } \]

[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]

\[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\frac {\int \frac {x^{m}}{x^{3}}\, dx + \int a x^{m} \sqrt {-1 + \frac {1}{a x^{3}}} \sqrt {1 + \frac {1}{a x^{3}}}\, dx}{a} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\text {Exception raised: ValueError} \]

[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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(m-3>0)', see `assume?` for mor
e details)Is

Giac [F]

\[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\int { x^{m} {\left (\sqrt {\frac {1}{a x^{3}} + 1} \sqrt {\frac {1}{a x^{3}} - 1} + \frac {1}{a x^{3}}\right )} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int e^{\text {sech}^{-1}\left (a x^3\right )} x^m \, dx=\int x^m\,\left (\sqrt {\frac {1}{a\,x^3}-1}\,\sqrt {\frac {1}{a\,x^3}+1}+\frac {1}{a\,x^3}\right ) \,d x \]

[In]

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

[Out]

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