\(\int \frac {1+x^2}{\sqrt {1+x^2+x^4}} \, dx\) [234]

Optimal result

Integrand size = 18, antiderivative size = 115 \[ \int \frac {1+x^2}{\sqrt {1+x^2+x^4}} \, dx=\frac {x \sqrt {1+x^2+x^4}}{1+x^2}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {1+x^2+x^4}}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{\sqrt {1+x^2+x^4}} \]

[Out]

x*(x^4+x^2+1)^(1/2)/(x^2+1)-(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticE(sin(2*arctan(x)),1/2
)*((x^4+x^2+1)/(x^2+1)^2)^(1/2)/(x^4+x^2+1)^(1/2)+(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*Elliptic
F(sin(2*arctan(x)),1/2)*((x^4+x^2+1)/(x^2+1)^2)^(1/2)/(x^4+x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1211, 1117, 1209} \[ \int \frac {1+x^2}{\sqrt {1+x^2+x^4}} \, dx=\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{\sqrt {x^4+x^2+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {x^4+x^2+1}}+\frac {\sqrt {x^4+x^2+1} x}{x^2+1} \]

[In]

Int[(1 + x^2)/Sqrt[1 + x^2 + x^4],x]

[Out]

(x*Sqrt[1 + x^2 + x^4])/(1 + x^2) - ((1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/
Sqrt[1 + x^2 + x^4] + ((1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/4])/Sqrt[1 + x^2 +
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]

Rule 1209

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

Rule 1211

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.82 \[ \int \frac {1+x^2}{\sqrt {1+x^2+x^4}} \, dx=\frac {\sqrt [3]{-1} \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} \left (E\left (i \text {arcsinh}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+\left (-1+\sqrt [3]{-1}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )\right )}{\sqrt {1+x^2+x^4}} \]

[In]

Integrate[(1 + x^2)/Sqrt[1 + x^2 + x^4],x]

[Out]

((-1)^(1/3)*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^(2/3)*x^2]*(EllipticE[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)]
+ (-1 + (-1)^(1/3))*EllipticF[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)]))/Sqrt[1 + x^2 + x^4]

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.78

method	result	size
default	\(\frac {2 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}-\frac {4 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\)	\(205\)
elliptic	\(\frac {2 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}-\frac {4 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\)	\(205\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int \frac {1+x^2}{\sqrt {1+x^2+x^4}} \, dx=\frac {\sqrt {2} {\left (\sqrt {-3} x - x\right )} \sqrt {\sqrt {-3} - 1} E(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) + 2 \, \sqrt {2} x \sqrt {\sqrt {-3} - 1} F(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) + 4 \, \sqrt {x^{4} + x^{2} + 1}}{4 \, x} \]

[In]

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

[Out]

1/4*(sqrt(2)*(sqrt(-3)*x - x)*sqrt(sqrt(-3) - 1)*elliptic_e(arcsin(1/2*sqrt(2)*sqrt(sqrt(-3) - 1)/x), 1/2*sqrt
(-3) - 1/2) + 2*sqrt(2)*x*sqrt(sqrt(-3) - 1)*elliptic_f(arcsin(1/2*sqrt(2)*sqrt(sqrt(-3) - 1)/x), 1/2*sqrt(-3)
- 1/2) + 4*sqrt(x^4 + x^2 + 1))/x

Sympy [F]

\[ \int \frac {1+x^2}{\sqrt {1+x^2+x^4}} \, dx=\int \frac {x^{2} + 1}{\sqrt {\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )}}\, dx \]

[In]

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

[Out]

Integral((x**2 + 1)/sqrt((x**2 - x + 1)*(x**2 + x + 1)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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