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

Optimal. Leaf size=64 $\frac{2 x^m \text{Hypergeometric2F1}\left (-\frac{1}{2},-\frac{m}{2},1-\frac{m}{2},-\frac{1}{a^2 x^2}\right )}{a m}-\frac{2 x^{m-1}}{a^2 (1-m)}+\frac{x^{m+1}}{m+1}$

[Out]

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

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Rubi [A]  time = 0.321878, antiderivative size = 64, normalized size of antiderivative = 1., 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 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]]

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])

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*}

Mathematica [A]  time = 0.0648334, size = 57, normalized size = 0.89 $x^m \left (\frac{2 \text{Hypergeometric2F1}\left (-\frac{1}{2},-\frac{m}{2},1-\frac{m}{2},-\frac{1}{a^2 x^2}\right )}{a m}+\frac{2}{a^2 (m-1) x}+\frac{x}{m+1}\right )$

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

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Maple [F]  time = 0.185, size = 0, normalized size = 0. \begin{align*} \int \left ({\frac{1}{ax}}+\sqrt{1+{\frac{1}{{a}^{2}{x}^{2}}}} \right ) ^{2}{x}^{m}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((1/a/x+(1+1/a^2/x^2)^(1/2))^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+(1+1/a^2/x^2)^(1/2))^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{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 ) \end{align*}

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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Sympy [A]  time = 14.1205, size = 71, normalized size = 1.11 \begin{align*} \begin{cases} \frac{x^{m + 1}}{m + 1} & \text{for}\: m \neq -1 \\\log{\left (x \right )} & \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{\left (x \right )} & \text{otherwise} \end{cases}\right )}{a^{2}} \end{align*}

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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