\(\int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx\) [228]

Optimal result

Integrand size = 20, antiderivative size = 137 \[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\frac {x \sqrt {1+x^2+x^4}}{1+x^2}+\frac {1}{2} \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {1+x^2+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {1+x^2+x^4}} \]

[Out]

1/2*arctan(x/(x^4+x^2+1)^(1/2))+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)+3/4*(x^2+1)*(cos(2*arctan(x)
)^2)^(1/2)/cos(2*arctan(x))*EllipticF(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 = 0.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1222, 1153, 1117, 1209, 1224, 1712, 209} \[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\frac {1}{2} \arctan \left (\frac {x}{\sqrt {x^4+x^2+1}}\right )+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {x^4+x^2+1}}+\frac {\sqrt {x^4+x^2+1} x}{x^2+1} \]

[In]

Int[Sqrt[1 + x^2 + x^4]/(1 + x^2),x]

[Out]

(x*Sqrt[1 + x^2 + x^4])/(1 + x^2) + ArcTan[x/Sqrt[1 + x^2 + x^4]]/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] + (3*(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*Ellipti
cF[2*ArcTan[x], 1/4])/(4*Sqrt[1 + x^2 + x^4])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 1153

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[1/q, Int[1/Sqrt[
a + b*x^2 + c*x^4], x], x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], 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 1222

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

Rule 1224

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

Rule 1712

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[
A, Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B},
x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{\sqrt {1+x^2+x^4}} \, dx+\int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx \\ & = \frac {1}{2} \int \frac {1}{\sqrt {1+x^2+x^4}} \, dx+\frac {1}{2} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx+\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 {3 \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 )}{4 \sqrt {1+x^2+x^4}}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {1+x^2+x^4}}\right ) \\ & = \frac {x \sqrt {1+x^2+x^4}}{1+x^2}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt {1+x^2+x^4}}\right )-\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 {3 \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 )}{4 \sqrt {1+x^2+x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 8.90 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\frac {\sqrt [3]{-1} \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} \left (E\left (i \text {arcsinh}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )+\sqrt [3]{-1} \operatorname {EllipticPi}\left (\sqrt [3]{-1},i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )\right )}{\sqrt {1+x^2+x^4}} \]

[In]

Integrate[Sqrt[1 + x^2 + x^4]/(1 + x^2),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)]
- EllipticF[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] + (-1)^(1/3)*EllipticPi[(-1)^(1/3), 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 = 0.49 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.14

method	result	size
default	\(-\frac {4 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {4 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {\sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \Pi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(293\)
elliptic	\(-\frac {4 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {4 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {\sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \Pi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(293\)

[In]

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

[Out]

-4/(-2+2*I*3^(1/2))^(1/2)*(1+1/2*x^2-1/2*I*x^2*3^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(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))+4/(-2+2*I*3^(1/2))^(1/2
)*(1+1/2*x^2-1/2*I*x^2*3^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(1/2)/(x^4+x^2+1)^(1/2)/(1+I*3^(1/2))*Elli
pticE(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2))+1/(-1/2+1/2*I*3^(1/2))^(1/2)*(1+1/2*x^2-1/2*I*x
^2*3^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(1/2)/(x^4+x^2+1)^(1/2)*EllipticPi((-1/2+1/2*I*3^(1/2))^(1/2)*
x,-1/(-1/2+1/2*I*3^(1/2)),(-1/2-1/2*I*3^(1/2))^(1/2)/(-1/2+1/2*I*3^(1/2))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\frac {2 \, \sqrt {2} {\left (\sqrt {-3} x - x\right )} \sqrt {\sqrt {-3} - 1} E(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) - \sqrt {2} {\left (\sqrt {-3} x - 3 \, x\right )} \sqrt {\sqrt {-3} - 1} F(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) + 4 \, x \arctan \left (\frac {x}{\sqrt {x^{4} + x^{2} + 1}}\right ) + 8 \, \sqrt {x^{4} + x^{2} + 1}}{8 \, x} \]

[In]

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

[Out]

1/8*(2*sqrt(2)*(sqrt(-3)*x - x)*sqrt(sqrt(-3) - 1)*elliptic_e(arcsin(1/2*sqrt(2)*sqrt(sqrt(-3) - 1)/x), 1/2*sq
rt(-3) - 1/2) - sqrt(2)*(sqrt(-3)*x - 3*x)*sqrt(sqrt(-3) - 1)*elliptic_f(arcsin(1/2*sqrt(2)*sqrt(sqrt(-3) - 1)
/x), 1/2*sqrt(-3) - 1/2) + 4*x*arctan(x/sqrt(x^4 + x^2 + 1)) + 8*sqrt(x^4 + x^2 + 1))/x

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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