\(\int \frac {(1+x^2)^2}{\sqrt {1+x^2+x^4}} \, dx\) [233]

Optimal result

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

[Out]

1/3*x*(x^4+x^2+1)^(1/2)+4/3*x*(x^4+x^2+1)^(1/2)/(x^2+1)-4/3*(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))*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 = 0.03 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1220, 1211, 1117, 1209} \[ \int \frac {\left (1+x^2\right )^2}{\sqrt {1+x^2+x^4}} \, dx=\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{\sqrt {x^4+x^2+1}}-\frac {4 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{3 \sqrt {x^4+x^2+1}}+\frac {4 \sqrt {x^4+x^2+1} x}{3 \left (x^2+1\right )}+\frac {1}{3} \sqrt {x^4+x^2+1} x \]

[In]

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

[Out]

(x*Sqrt[1 + x^2 + x^4])/3 + (4*x*Sqrt[1 + x^2 + x^4])/(3*(1 + x^2)) - (4*(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x
^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(3*Sqrt[1 + x^2 + x^4]) + ((1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*Ell
ipticF[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]

Rule 1220

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e^q*x^(2*q - 3)*((
a + b*x^2 + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2*p + 2*q - 1)*e^q*x^(2*q - 2) -
c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && IGtQ[q, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x \sqrt {1+x^2+x^4}+\frac {1}{3} \int \frac {2+4 x^2}{\sqrt {1+x^2+x^4}} \, dx \\ & = \frac {1}{3} x \sqrt {1+x^2+x^4}-\frac {4}{3} \int \frac {1-x^2}{\sqrt {1+x^2+x^4}} \, dx+2 \int \frac {1}{\sqrt {1+x^2+x^4}} \, dx \\ & = \frac {1}{3} x \sqrt {1+x^2+x^4}+\frac {4 x \sqrt {1+x^2+x^4}}{3 \left (1+x^2\right )}-\frac {4 \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 )}{3 \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 = 10.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04 \[ \int \frac {\left (1+x^2\right )^2}{\sqrt {1+x^2+x^4}} \, dx=\frac {x+x^3+x^5+4 \sqrt [3]{-1} \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} E\left (i \text {arcsinh}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+2 \sqrt [3]{-1} \left (-2+\sqrt [3]{-1}\right ) \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )}{3 \sqrt {1+x^2+x^4}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {4 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {x \sqrt {x^{4}+x^{2}+1}}{3}-\frac {16 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\)	\(218\)
risch	\(\frac {4 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {x \sqrt {x^{4}+x^{2}+1}}{3}-\frac {16 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\)	\(218\)
elliptic	\(\frac {4 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {x \sqrt {x^{4}+x^{2}+1}}{3}-\frac {16 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{3 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\)	\(218\)

[In]

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

[Out]

4/3/(-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))+1/3*x*(x^4+x^2+1)^(1/2)-16/3/(-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 = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.82 \[ \int \frac {\left (1+x^2\right )^2}{\sqrt {1+x^2+x^4}} \, dx=\frac {2 \, \sqrt {2} {\left (\sqrt {-3} x - x\right )} \sqrt {\sqrt {-3} - 1} E(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) - \sqrt {2} {\left (\sqrt {-3} x - 3 \, x\right )} \sqrt {\sqrt {-3} - 1} F(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) + 2 \, \sqrt {x^{4} + x^{2} + 1} {\left (x^{2} + 4\right )}}{6 \, x} \]

[In]

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

[Out]

1/6*(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) + 2*sqrt(x^4 + x^2 + 1)*(x^2 + 4))/x

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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