\(\int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx\) [124]

Optimal result

Integrand size = 14, antiderivative size = 108 \[ \int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx=\frac {\sqrt {\frac {4+\left (5-\sqrt {17}\right ) x^2}{4+\left (5+\sqrt {17}\right ) x^2}} \left (4+\left (5+\sqrt {17}\right ) x^2\right ) \operatorname {EllipticF}\left (\arctan \left (\frac {1}{2} \sqrt {5+\sqrt {17}} x\right ),\frac {1}{4} \left (-17+5 \sqrt {17}\right )\right )}{2 \sqrt {5+\sqrt {17}} \sqrt {2+5 x^2+x^4}} \]

[Out]

1/2*(1/(4+x^2*(5+17^(1/2))))^(1/2)*(4+x^2*(5+17^(1/2)))^(3/2)*EllipticF(x*(5+17^(1/2))^(1/2)/(4+x^2*(5+17^(1/2
)))^(1/2),1/2*(-17+5*17^(1/2))^(1/2))*((4+x^2*(5-17^(1/2)))/(4+x^2*(5+17^(1/2))))^(1/2)/(x^4+5*x^2+2)^(1/2)/(5
+17^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1113} \[ \int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx=\frac {\sqrt {\frac {\left (5-\sqrt {17}\right ) x^2+4}{\left (5+\sqrt {17}\right ) x^2+4}} \left (\left (5+\sqrt {17}\right ) x^2+4\right ) \operatorname {EllipticF}\left (\arctan \left (\frac {1}{2} \sqrt {5+\sqrt {17}} x\right ),\frac {1}{4} \left (-17+5 \sqrt {17}\right )\right )}{2 \sqrt {5+\sqrt {17}} \sqrt {x^4+5 x^2+2}} \]

[In]

Int[1/Sqrt[2 + 5*x^2 + x^4],x]

[Out]

(Sqrt[(4 + (5 - Sqrt[17])*x^2)/(4 + (5 + Sqrt[17])*x^2)]*(4 + (5 + Sqrt[17])*x^2)*EllipticF[ArcTan[(Sqrt[5 + S
qrt[17]]*x)/2], (-17 + 5*Sqrt[17])/4])/(2*Sqrt[5 + Sqrt[17]]*Sqrt[2 + 5*x^2 + x^4])

Rule 1113

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(2*a + (b + q
)*x^2)*(Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*Elli
pticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\frac {4+\left (5-\sqrt {17}\right ) x^2}{4+\left (5+\sqrt {17}\right ) x^2}} \left (4+\left (5+\sqrt {17}\right ) x^2\right ) F\left (\tan ^{-1}\left (\frac {1}{2} \sqrt {5+\sqrt {17}} x\right )|\frac {1}{4} \left (-17+5 \sqrt {17}\right )\right )}{2 \sqrt {5+\sqrt {17}} \sqrt {2+5 x^2+x^4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx=-\frac {i \sqrt {5-\sqrt {17}+2 x^2} \sqrt {5+\sqrt {17}+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{5+\sqrt {17}}} x\right ),\frac {21}{4}+\frac {5 \sqrt {17}}{4}\right )}{\sqrt {2 \left (5-\sqrt {17}\right )} \sqrt {2+5 x^2+x^4}} \]

[In]

Integrate[1/Sqrt[2 + 5*x^2 + x^4],x]

[Out]

((-I)*Sqrt[5 - Sqrt[17] + 2*x^2]*Sqrt[5 + Sqrt[17] + 2*x^2]*EllipticF[I*ArcSinh[Sqrt[2/(5 + Sqrt[17])]*x], 21/
4 + (5*Sqrt[17])/4])/(Sqrt[2*(5 - Sqrt[17])]*Sqrt[2 + 5*x^2 + x^4])

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {2 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {17}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {17}}{4}\right ) x^{2}}\, F\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )}{\sqrt {-5+\sqrt {17}}\, \sqrt {x^{4}+5 x^{2}+2}}\)	\(76\)
elliptic	\(\frac {2 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {17}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {17}}{4}\right ) x^{2}}\, F\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )}{\sqrt {-5+\sqrt {17}}\, \sqrt {x^{4}+5 x^{2}+2}}\)	\(76\)

[In]

int(1/(x^4+5*x^2+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/(-5+17^(1/2))^(1/2)*(1-(-5/4+1/4*17^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*17^(1/2))*x^2)^(1/2)/(x^4+5*x^2+2)^(1/2)*
EllipticF(1/2*x*(-5+17^(1/2))^(1/2),5/4*2^(1/2)+1/4*34^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx=-\frac {1}{8} \, {\left (\sqrt {17} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {\sqrt {17} - 5} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {17} - 5}\right )\,|\,\frac {5}{4} \, \sqrt {17} + \frac {21}{4}) \]

[In]

integrate(1/(x^4+5*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

-1/8*(sqrt(17)*sqrt(2) + 5*sqrt(2))*sqrt(sqrt(17) - 5)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(17) - 5)), 5/4*sqrt(1
7) + 21/4)

Sympy [F]

\[ \int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx=\int \frac {1}{\sqrt {x^{4} + 5 x^{2} + 2}}\, dx \]

[In]

integrate(1/(x**4+5*x**2+2)**(1/2),x)

[Out]

Integral(1/sqrt(x**4 + 5*x**2 + 2), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} + 5 \, x^{2} + 2}} \,d x } \]

[In]

integrate(1/(x^4+5*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(x^4 + 5*x^2 + 2), x)

Giac [F]

\[ \int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} + 5 \, x^{2} + 2}} \,d x } \]

[In]

integrate(1/(x^4+5*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(x^4 + 5*x^2 + 2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {2+5 x^2+x^4}} \, dx=\int \frac {1}{\sqrt {x^4+5\,x^2+2}} \,d x \]

[In]

int(1/(5*x^2 + x^4 + 2)^(1/2),x)

[Out]

int(1/(5*x^2 + x^4 + 2)^(1/2), x)