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

Optimal. Leaf size=59 \[ \frac {x^{m+1} \, _2F_1\left (-\frac {1}{2},\frac {1}{4} (-m-1);\frac {3-m}{4};-\frac {1}{a^2 x^4}\right )}{m+1}-\frac {x^{m-1}}{a (1-m)} \]

[Out]

-x^(-1+m)/a/(1-m)+x^(1+m)*hypergeom([-1/2, -1/4-1/4*m],[3/4-1/4*m],-1/a^2/x^4)/(1+m)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, 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 = {6336, 30, 339, 364} \[ \frac {x^{m+1} \, _2F_1\left (-\frac {1}{2},\frac {1}{4} (-m-1);\frac {3-m}{4};-\frac {1}{a^2 x^4}\right )}{m+1}-\frac {x^{m-1}}{a (1-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

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

Rule 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, Subst
[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]

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 6336

Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Dist[1/a, Int[x^(m - p), x], x] + Int[x^m*Sqrt[1 + 1/
(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, x]

Rubi steps

\begin {align*} \int e^{\text {csch}^{-1}\left (a x^2\right )} x^m \, dx &=\frac {\int x^{-2+m} \, dx}{a}+\int \sqrt {1+\frac {1}{a^2 x^4}} x^m \, dx\\ &=-\frac {x^{-1+m}}{a (1-m)}-\left (\left (\frac {1}{x}\right )^m x^m\right ) \operatorname {Subst}\left (\int x^{-2-m} \sqrt {1+\frac {x^4}{a^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {x^{-1+m}}{a (1-m)}+\frac {x^{1+m} \, _2F_1\left (-\frac {1}{2},\frac {1}{4} (-1-m);\frac {3-m}{4};-\frac {1}{a^2 x^4}\right )}{1+m}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 55, normalized size = 0.93 \[ x^{m-1} \left (\frac {x^2 \, _2F_1\left (-\frac {1}{2},-\frac {m}{4}-\frac {1}{4};\frac {3}{4}-\frac {m}{4};-\frac {1}{a^2 x^4}\right )}{m+1}+\frac {1}{a (m-1)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a x^{2} x^{m} \sqrt {\frac {a^{2} x^{4} + 1}{a^{2} x^{4}}} + x^{m}}{a x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^m,x, algorithm="fricas")

[Out]

integral((a*x^2*x^m*sqrt((a^2*x^4 + 1)/(a^2*x^4)) + x^m)/(a*x^2), x)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^m,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \left (\frac {1}{a \,x^{2}}+\sqrt {1+\frac {1}{a^{2} x^{4}}}\right ) x^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^m,x)

[Out]

int((1/a/x^2+(1+1/a^2/x^4)^(1/2))*x^m,x)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x^2+(1+1/a^2/x^4)^(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-2>0)', see `assume?` for mor
e details)Is m-2 equal to -1?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^m\,\left (\sqrt {\frac {1}{a^2\,x^4}+1}+\frac {1}{a\,x^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*((1/(a^2*x^4) + 1)^(1/2) + 1/(a*x^2)),x)

[Out]

int(x^m*((1/(a^2*x^4) + 1)^(1/2) + 1/(a*x^2)), x)

________________________________________________________________________________________

sympy [A]  time = 6.03, size = 66, normalized size = 1.12 \[ - \frac {x x^{m} \Gamma \left (- \frac {m}{4} - \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {m}{4} - \frac {1}{4} \\ \frac {3}{4} - \frac {m}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4} - \frac {m}{4}\right )} + \frac {\begin {cases} \frac {x^{m}}{m x - x} & \text {for}\: m \neq 1 \\\log {\relax (x )} & \text {otherwise} \end {cases}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x**2+(1+1/a**2/x**4)**(1/2))*x**m,x)

[Out]

-x*x**m*gamma(-m/4 - 1/4)*hyper((-1/2, -m/4 - 1/4), (3/4 - m/4,), exp_polar(I*pi)/(a**2*x**4))/(4*gamma(3/4 -
m/4)) + Piecewise((x**m/(m*x - x), Ne(m, 1)), (log(x), True))/a

________________________________________________________________________________________