### 3.44 $$\int \frac{e^{\text{csch}^{-1}(a x^2)}}{x^2} \, dx$$

Optimal. Leaf size=91 $-\frac{\sqrt{\frac{a^2+\frac{1}{x^4}}{\left (a+\frac{1}{x^2}\right )^2}} \left (a+\frac{1}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\sqrt{a} x\right ),\frac{1}{2}\right )}{3 \sqrt{a} \sqrt{\frac{1}{a^2 x^4}+1}}-\frac{\sqrt{\frac{1}{a^2 x^4}+1}}{3 x}-\frac{1}{3 a x^3}$

[Out]

-1/(3*a*x^3) - Sqrt[1 + 1/(a^2*x^4)]/(3*x) - (Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))*EllipticF[2*Arc
Cot[Sqrt[a]*x], 1/2])/(3*Sqrt[a]*Sqrt[1 + 1/(a^2*x^4)])

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Rubi [A]  time = 0.0472524, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.417, Rules used = {6336, 30, 335, 195, 220} $-\frac{\sqrt{\frac{1}{a^2 x^4}+1}}{3 x}-\frac{\sqrt{\frac{a^2+\frac{1}{x^4}}{\left (a+\frac{1}{x^2}\right )^2}} \left (a+\frac{1}{x^2}\right ) F\left (2 \cot ^{-1}\left (\sqrt{a} x\right )|\frac{1}{2}\right )}{3 \sqrt{a} \sqrt{\frac{1}{a^2 x^4}+1}}-\frac{1}{3 a x^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(3*a*x^3) - Sqrt[1 + 1/(a^2*x^4)]/(3*x) - (Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))*EllipticF[2*Arc
Cot[Sqrt[a]*x], 1/2])/(3*Sqrt[a]*Sqrt[1 + 1/(a^2*x^4)])

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]

Rule 30

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

Rule 335

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

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

Mathematica [C]  time = 0.160232, size = 96, normalized size = 1.05 $-\frac{a x \sqrt{\frac{e^{\text{csch}^{-1}\left (a x^2\right )}}{2 e^{2 \text{csch}^{-1}\left (a x^2\right )}-2}} \left (4 \sqrt{1-e^{2 \text{csch}^{-1}\left (a x^2\right )}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},e^{2 \text{csch}^{-1}\left (a x^2\right )}\right )+e^{2 \text{csch}^{-1}\left (a x^2\right )}-1\right )}{3 \sqrt{a x^2}}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(a*Sqrt[E^ArcCsch[a*x^2]/(-2 + 2*E^(2*ArcCsch[a*x^2]))]*x*(-1 + E^(2*ArcCsch[a*x^2]) + 4*Sqrt[1 - E^(2*ArcCsc
h[a*x^2])]*Hypergeometric2F1[1/4, 1/2, 5/4, E^(2*ArcCsch[a*x^2])]))/(3*Sqrt[a*x^2])

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Maple [C]  time = 0.18, size = 111, normalized size = 1.2 \begin{align*} -{\frac{1}{3\,x \left ({a}^{2}{x}^{4}+1 \right ) }\sqrt{{\frac{{a}^{2}{x}^{4}+1}{{a}^{2}{x}^{4}}}} \left ( -2\,\sqrt{1-ia{x}^{2}}\sqrt{1+ia{x}^{2}}{\it EllipticF} \left ( x\sqrt{ia},i \right ){x}^{3}{a}^{2}+\sqrt{ia}{x}^{4}{a}^{2}+\sqrt{ia} \right ){\frac{1}{\sqrt{ia}}}}-{\frac{1}{3\,{x}^{3}a}} \end{align*}

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^2,x)

[Out]

-1/3*((a^2*x^4+1)/a^2/x^4)^(1/2)*(-2*(1-I*a*x^2)^(1/2)*(1+I*a*x^2)^(1/2)*EllipticF(x*(I*a)^(1/2),I)*x^3*a^2+(I
*a)^(1/2)*x^4*a^2+(I*a)^(1/2))/x/(a^2*x^4+1)/(I*a)^(1/2)-1/3/x^3/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\frac{\Gamma \left (-\frac{3}{4}\right ) \,_2F_1\left (\begin{matrix} -\frac{3}{4},-\frac{1}{2} \\ \frac{1}{4} \end{matrix} ; -a^{2} x^{4} \right )}{4 \, x^{3} \Gamma \left (\frac{1}{4}\right )}}{a} - \frac{1}{3 \, a x^{3}} \end{align*}

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^2,x, algorithm="maxima")

[Out]

integrate(sqrt(a^2*x^4 + 1)/x^4, x)/a - 1/3/(a*x^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a x^{2} \sqrt{\frac{a^{2} x^{4} + 1}{a^{2} x^{4}}} + 1}{a x^{4}}, x\right ) \end{align*}

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^2,x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 2.60444, size = 42, normalized size = 0.46 \begin{align*} - \frac{\Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 x \Gamma \left (\frac{5}{4}\right )} - \frac{1}{3 a x^{3}} \end{align*}

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**2,x)

[Out]

-gamma(1/4)*hyper((-1/2, 1/4), (5/4,), exp_polar(I*pi)/(a**2*x**4))/(4*x*gamma(5/4)) - 1/(3*a*x**3)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{1}{a^{2} x^{4}} + 1} + \frac{1}{a x^{2}}}{x^{2}}\,{d x} \end{align*}

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^2,x, algorithm="giac")

[Out]

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