\(\int \frac {1}{\sqrt {2+5 x^2+4 x^4}} \, dx\) [121]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 90 \[ \int \frac {1}{\sqrt {2+5 x^2+4 x^4}} \, dx=\frac {\left (1+\sqrt {2} x^2\right ) \sqrt {\frac {2+5 x^2+4 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{16} \left (8-5 \sqrt {2}\right )\right )}{2\ 2^{3/4} \sqrt {2+5 x^2+4 x^4}} \]

[Out]

1/4*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticF(sin(2*arctan(2^(1/4)*x)),1/4*(8-5*2^
(1/2))^(1/2))*(1+x^2*2^(1/2))*((4*x^4+5*x^2+2)/(1+x^2*2^(1/2))^2)^(1/2)*2^(1/4)/(4*x^4+5*x^2+2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 90, 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} \[ \int \frac {1}{\sqrt {2+5 x^2+4 x^4}} \, dx=\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {4 x^4+5 x^2+2}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{16} \left (8-5 \sqrt {2}\right )\right )}{2\ 2^{3/4} \sqrt {4 x^4+5 x^2+2}} \]

[In]

Int[1/Sqrt[2 + 5*x^2 + 4*x^4],x]

[Out]

((1 + Sqrt[2]*x^2)*Sqrt[(2 + 5*x^2 + 4*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (8 - 5*Sqrt[2]
)/16])/(2*2^(3/4)*Sqrt[2 + 5*x^2 + 4*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 (1+\sqrt {2} x^2\right ) \sqrt {\frac {2+5 x^2+4 x^4}{\left (1+\sqrt {2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{16} \left (8-5 \sqrt {2}\right )\right )}{2\ 2^{3/4} \sqrt {2+5 x^2+4 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.63 \[ \int \frac {1}{\sqrt {2+5 x^2+4 x^4}} \, dx=-\frac {i \sqrt {1-\frac {8 x^2}{-5-i \sqrt {7}}} \sqrt {1-\frac {8 x^2}{-5+i \sqrt {7}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (2 \sqrt {-\frac {2}{-5-i \sqrt {7}}} x\right ),\frac {-5-i \sqrt {7}}{-5+i \sqrt {7}}\right )}{2 \sqrt {2} \sqrt {-\frac {1}{-5-i \sqrt {7}}} \sqrt {2+5 x^2+4 x^4}} \]

[In]

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

[Out]

((-1/2*I)*Sqrt[1 - (8*x^2)/(-5 - I*Sqrt[7])]*Sqrt[1 - (8*x^2)/(-5 + I*Sqrt[7])]*EllipticF[I*ArcSinh[2*Sqrt[-2/
(-5 - I*Sqrt[7])]*x], (-5 - I*Sqrt[7])/(-5 + I*Sqrt[7])])/(Sqrt[2]*Sqrt[-(-5 - I*Sqrt[7])^(-1)]*Sqrt[2 + 5*x^2
 + 4*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.62 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97

method result size
default \(\frac {2 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {7}}{4}\right ) x^{2}}\, F\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )}{\sqrt {-5+i \sqrt {7}}\, \sqrt {4 x^{4}+5 x^{2}+2}}\) \(87\)
elliptic \(\frac {2 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {7}}{4}\right ) x^{2}}\, F\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )}{\sqrt {-5+i \sqrt {7}}\, \sqrt {4 x^{4}+5 x^{2}+2}}\) \(87\)

[In]

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

[Out]

2/(-5+I*7^(1/2))^(1/2)*(1-(-5/4+1/4*I*7^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*I*7^(1/2))*x^2)^(1/2)/(4*x^4+5*x^2+2)^(
1/2)*EllipticF(1/2*x*(-5+I*7^(1/2))^(1/2),1/4*(9+5*I*7^(1/2))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\sqrt {2+5 x^2+4 x^4}} \, dx=-\frac {1}{32} \, \sqrt {2} {\left (\sqrt {-7} + 5\right )} \sqrt {\sqrt {-7} - 5} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {-7} - 5}\right )\,|\,\frac {5}{16} \, \sqrt {-7} + \frac {9}{16}) \]

[In]

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

[Out]

-1/32*sqrt(2)*(sqrt(-7) + 5)*sqrt(sqrt(-7) - 5)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(-7) - 5)), 5/16*sqrt(-7) + 9
/16)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(5*x^2 + 4*x^4 + 2)^(1/2),x)

[Out]

int(1/(5*x^2 + 4*x^4 + 2)^(1/2), x)