### 3.55 $$\int \frac{e^{2 \text{csch}^{-1}(a x)}}{x^2} \, dx$$

Optimal. Leaf size=34 $-\frac{2}{3} a \left (\frac{1}{a^2 x^2}+1\right )^{3/2}-\frac{2}{3 a^2 x^3}-\frac{1}{x}$

[Out]

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

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Rubi [A]  time = 0.203554, antiderivative size = 54, normalized size of antiderivative = 1.59, number of steps used = 4, number of rules used = 3, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {6338, 6715, 2117} $-\frac{1}{6} a \left (\sqrt{\frac{1}{a^2 x^2}+1}+\frac{1}{a x}\right )^3-\frac{1}{2} a \sqrt{\frac{1}{a^2 x^2}+1}-\frac{1}{2 x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(a*Sqrt[1 + 1/(a^2*x^2)])/2 - (a*(Sqrt[1 + 1/(a^2*x^2)] + 1/(a*x))^3)/6 - 1/(2*x)

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 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{e^{2 \text{csch}^{-1}(a x)}}{x^2} \, dx &=\int \frac{\left (\sqrt{1+\frac{1}{a^2 x^2}}+\frac{1}{a x}\right )^2}{x^2} \, dx\\ &=-\operatorname{Subst}\left (\int \left (\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}}\right )^2 \, dx,x,\frac{1}{x}\right )\\ &=-\left (\frac{1}{2} a \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\sqrt{1+\frac{1}{a^2 x^2}}+\frac{1}{a x}\right )\right )\\ &=-\frac{1}{2} a \sqrt{1+\frac{1}{a^2 x^2}}-\frac{1}{6} a \left (\sqrt{1+\frac{1}{a^2 x^2}}+\frac{1}{a x}\right )^3-\frac{1}{2 x}\\ \end{align*}

Mathematica [A]  time = 0.04226, size = 46, normalized size = 1.35 $-\frac{3 a^2 x^2+2 a x \sqrt{\frac{1}{a^2 x^2}+1} \left (a^2 x^2+1\right )+2}{3 a^2 x^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(2 + 3*a^2*x^2 + 2*a*Sqrt[1 + 1/(a^2*x^2)]*x*(1 + a^2*x^2))/(3*a^2*x^3)

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Maple [B]  time = 0.19, size = 63, normalized size = 1.9 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{{a}^{2}}{x}}-{\frac{1}{3\,{x}^{3}}} \right ) }-{\frac{2\,{a}^{2}{x}^{2}+2}{3\,a{x}^{2}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}}}-{\frac{1}{3\,{x}^{3}{a}^{2}}} \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^2,x)

[Out]

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

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Maxima [A]  time = 0.99973, size = 38, normalized size = 1.12 \begin{align*} -\frac{2}{3} \, a{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{3}{2}} - \frac{1}{x} - \frac{2}{3 \, a^{2} x^{3}} \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^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.4791, size = 127, normalized size = 3.74 \begin{align*} -\frac{2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 \,{\left (a^{3} x^{3} + a x\right )} \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} + 2}{3 \, a^{2} x^{3}} \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^2,x, algorithm="fricas")

[Out]

-1/3*(2*a^3*x^3 + 3*a^2*x^2 + 2*(a^3*x^3 + a*x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) + 2)/(a^2*x^3)

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Sympy [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**2,x)

[Out]

Exception raised: TypeError

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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^2,x, algorithm="giac")

[Out]

Exception raised: TypeError