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

Optimal. Leaf size=64 \[ \frac {2 x^m \, _2F_1\left (-\frac {1}{2},-\frac {m}{2};1-\frac {m}{2};-\frac {1}{a^2 x^2}\right )}{a m}-\frac {2 x^{m-1}}{a^2 (1-m)}+\frac {x^{m+1}}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6338, 6742, 339, 364} \[ \frac {2 x^m \, _2F_1\left (-\frac {1}{2},-\frac {m}{2};1-\frac {m}{2};-\frac {1}{a^2 x^2}\right )}{a m}-\frac {2 x^{m-1}}{a^2 (1-m)}+\frac {x^{m+1}}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 6338

Int[E^(ArcCsch[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[1 + 1/u^2])^n, x] /; FreeQ[m, x] && Int
egerQ[n]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 57, normalized size = 0.89 \[ x^m \left (\frac {2 \, _2F_1\left (-\frac {1}{2},-\frac {m}{2};1-\frac {m}{2};-\frac {1}{a^2 x^2}\right )}{a m}+\frac {2}{a^2 (m-1) x}+\frac {x}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, a x x^{m} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} + {\left (a^{2} x^{2} + 2\right )} x^{m}}{a^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Error: Bad Argument Type

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (\frac {1}{a x}+\sqrt {1+\frac {1}{x^{2} a^{2}}}\right )^{2} x^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+(1+1/a^2/x^2)^(1/2))^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?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.15, size = 71, normalized size = 1.11 \[ \begin {cases} \frac {x^{m + 1}}{m + 1} & \text {for}\: m \neq -1 \\\log {\relax (x )} & \text {otherwise} \end {cases} - \frac {x^{m} \Gamma \left (- \frac {m}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {m}{2} \\ 1 - \frac {m}{2} \end {matrix}\middle | {\frac {e^{i \pi }}{a^{2} x^{2}}} \right )}}{a \Gamma \left (1 - \frac {m}{2}\right )} + \frac {2 \left (\begin {cases} \frac {x^{m}}{m x - x} & \text {for}\: m \neq 1 \\\log {\relax (x )} & \text {otherwise} \end {cases}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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