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

Optimal. Leaf size=91 \[ -\frac {\sqrt {\frac {1}{a^2 x^4}+1}}{3 x}-\frac {\sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) F\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{3 \sqrt {a} \sqrt {\frac {1}{a^2 x^4}+1}}-\frac {1}{3 a x^3} \]

[Out]

-1/3/a/x^3-1/3*(1+1/a^2/x^4)^(1/2)/x-1/3*(a+1/x^2)*(cos(2*arccot(x*a^(1/2)))^2)^(1/2)/cos(2*arccot(x*a^(1/2)))
*EllipticF(sin(2*arccot(x*a^(1/2))),1/2*2^(1/2))*((a^2+1/x^4)/(a+1/x^2)^2)^(1/2)/a^(1/2)/(1+1/a^2/x^4)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 30

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

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]

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 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 \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*}

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Mathematica [C]  time = 0.16, size = 96, normalized size = 1.05 \[ -\frac {a x \sqrt {\frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{2 e^{2 \text {csch}^{-1}\left (a x^2\right )}-2}} \left (4 \sqrt {1-e^{2 \text {csch}^{-1}\left (a x^2\right )}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )+e^{2 \text {csch}^{-1}\left (a x^2\right )}-1\right )}{3 \sqrt {a x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a x^{2} \sqrt {\frac {a^{2} x^{4} + 1}{a^{2} x^{4}}} + 1}{a x^{4}}, x\right ) \]

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^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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {1}{a^{2} x^{4}} + 1} + \frac {1}{a x^{2}}}{x^{2}}\,{d x} \]

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)

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maple [C]  time = 0.05, size = 111, normalized size = 1.22 \[ -\frac {\sqrt {\frac {a^{2} x^{4}+1}{a^{2} x^{4}}}\, \left (-2 \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \EllipticF \left (x \sqrt {i a}, i\right ) x^{3} a^{2}+\sqrt {i a}\, x^{4} a^{2}+\sqrt {i a}\right )}{3 x \left (a^{2} x^{4}+1\right ) \sqrt {i a}}-\frac {1}{3 x^{3} a} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {\frac {\Gamma \left (-\frac {3}{4}\right ) \,_2F_1\left (\begin {matrix} -\frac {3}{4},-\frac {1}{2} \\ \frac {1}{4} \end {matrix} ; -a^{2} x^{4} \right )}{4 \, x^{3} \Gamma \left (\frac {1}{4}\right )}}{a} - \frac {1}{3 \, a x^{3}} \]

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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mupad [B]  time = 2.37, size = 27, normalized size = 0.30 \[ -\frac {{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {1}{4};\ \frac {5}{4};\ -\frac {1}{a^2\,x^4}\right )}{x}-\frac {1}{3\,a\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- hypergeom([-1/2, 1/4], 5/4, -1/(a^2*x^4))/x - 1/(3*a*x^3)

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sympy [C]  time = 2.28, size = 42, normalized size = 0.46 \[ - \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 x \Gamma \left (\frac {5}{4}\right )} - \frac {1}{3 a x^{3}} \]

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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