\(\int \frac {1}{\sqrt {2+5 x^2+5 x^4}} \, dx\) [120]

Optimal result

Integrand size = 16, antiderivative size = 92 \[ \int \frac {1}{\sqrt {2+5 x^2+5 x^4}} \, dx=\frac {\left (2+\sqrt {10} x^2\right ) \sqrt {\frac {2+5 x^2+5 x^4}{\left (2+\sqrt {10} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {5}{2}} x\right ),\frac {1}{8} \left (4-\sqrt {10}\right )\right )}{2 \sqrt [4]{10} \sqrt {2+5 x^2+5 x^4}} \]

[Out]

1/20*(cos(2*arctan(1/2*5^(1/4)*2^(3/4)*x))^2)^(1/2)/cos(2*arctan(1/2*5^(1/4)*2^(3/4)*x))*EllipticF(sin(2*arcta
n(1/2*5^(1/4)*2^(3/4)*x)),1/4*(8-2*10^(1/2))^(1/2))*(2+x^2*10^(1/2))*((5*x^4+5*x^2+2)/(2+x^2*10^(1/2))^2)^(1/2
)*10^(3/4)/(5*x^4+5*x^2+2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1117} \[ \int \frac {1}{\sqrt {2+5 x^2+5 x^4}} \, dx=\frac {\left (\sqrt {10} x^2+2\right ) \sqrt {\frac {5 x^4+5 x^2+2}{\left (\sqrt {10} x^2+2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {5}{2}} x\right ),\frac {1}{8} \left (4-\sqrt {10}\right )\right )}{2 \sqrt [4]{10} \sqrt {5 x^4+5 x^2+2}} \]

[In]

Int[1/Sqrt[2 + 5*x^2 + 5*x^4],x]

[Out]

((2 + Sqrt[10]*x^2)*Sqrt[(2 + 5*x^2 + 5*x^4)/(2 + Sqrt[10]*x^2)^2]*EllipticF[2*ArcTan[(5/2)^(1/4)*x], (4 - Sqr
t[10])/8])/(2*10^(1/4)*Sqrt[2 + 5*x^2 + 5*x^4])

Rule 1117

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (2+\sqrt {10} x^2\right ) \sqrt {\frac {2+5 x^2+5 x^4}{\left (2+\sqrt {10} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac {5}{2}} x\right )|\frac {1}{8} \left (4-\sqrt {10}\right )\right )}{2 \sqrt [4]{10} \sqrt {2+5 x^2+5 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\sqrt {2+5 x^2+5 x^4}} \, dx=-\frac {i \sqrt {1-\frac {10 x^2}{-5-i \sqrt {15}}} \sqrt {1-\frac {10 x^2}{-5+i \sqrt {15}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {10}{-5-i \sqrt {15}}} x\right ),\frac {-5-i \sqrt {15}}{-5+i \sqrt {15}}\right )}{\sqrt {10} \sqrt {-\frac {1}{-5-i \sqrt {15}}} \sqrt {2+5 x^2+5 x^4}} \]

[In]

Integrate[1/Sqrt[2 + 5*x^2 + 5*x^4],x]

[Out]

((-I)*Sqrt[1 - (10*x^2)/(-5 - I*Sqrt[15])]*Sqrt[1 - (10*x^2)/(-5 + I*Sqrt[15])]*EllipticF[I*ArcSinh[Sqrt[-10/(
-5 - I*Sqrt[15])]*x], (-5 - I*Sqrt[15])/(-5 + I*Sqrt[15])])/(Sqrt[10]*Sqrt[-(-5 - I*Sqrt[15])^(-1)]*Sqrt[2 + 5
*x^2 + 5*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.68 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {2 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, F\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )}{\sqrt {-5+i \sqrt {15}}\, \sqrt {5 x^{4}+5 x^{2}+2}}\)	\(87\)
elliptic	\(\frac {2 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, F\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )}{\sqrt {-5+i \sqrt {15}}\, \sqrt {5 x^{4}+5 x^{2}+2}}\)	\(87\)

[In]

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

[Out]

2/(-5+I*15^(1/2))^(1/2)*(1-(-5/4+1/4*I*15^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*I*15^(1/2))*x^2)^(1/2)/(5*x^4+5*x^2+2
)^(1/2)*EllipticF(1/2*x*(-5+I*15^(1/2))^(1/2),1/2*(1+I*15^(1/2))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\sqrt {2+5 x^2+5 x^4}} \, dx=-\frac {1}{40} \, \sqrt {2} {\left (\sqrt {-15} + 5\right )} \sqrt {\sqrt {-15} - 5} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {-15} - 5}\right )\,|\,\frac {1}{4} \, \sqrt {-15} + \frac {1}{4}) \]

[In]

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

[Out]

-1/40*sqrt(2)*(sqrt(-15) + 5)*sqrt(sqrt(-15) - 5)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(-15) - 5)), 1/4*sqrt(-15)
+ 1/4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(5*x^2 + 5*x^4 + 2)^(1/2),x)

[Out]

int(1/(5*x^2 + 5*x^4 + 2)^(1/2), x)