\(\int \frac {1+2 x^2}{1+x^2+4 x^4} \, dx\) [46]

Optimal result

Integrand size = 20, antiderivative size = 46 \[ \int \frac {1+2 x^2}{1+x^2+4 x^4} \, dx=-\frac {\arctan \left (\frac {\sqrt {3}-4 x}{\sqrt {5}}\right )}{\sqrt {5}}+\frac {\arctan \left (\frac {\sqrt {3}+4 x}{\sqrt {5}}\right )}{\sqrt {5}} \]

[Out]

-1/5*arctan(1/5*(-4*x+3^(1/2))*5^(1/2))*5^(1/2)+1/5*arctan(1/5*(4*x+3^(1/2))*5^(1/2))*5^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1175, 632, 210} \[ \int \frac {1+2 x^2}{1+x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {4 x+\sqrt {3}}{\sqrt {5}}\right )}{\sqrt {5}}-\frac {\arctan \left (\frac {\sqrt {3}-4 x}{\sqrt {5}}\right )}{\sqrt {5}} \]

[In]

Int[(1 + 2*x^2)/(1 + x^2 + 4*x^4),x]

[Out]

-(ArcTan[(Sqrt[3] - 4*x)/Sqrt[5]]/Sqrt[5]) + ArcTan[(Sqrt[3] + 4*x)/Sqrt[5]]/Sqrt[5]

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1175

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {1}{\frac {1}{2}-\frac {\sqrt {3} x}{2}+x^2} \, dx+\frac {1}{4} \int \frac {1}{\frac {1}{2}+\frac {\sqrt {3} x}{2}+x^2} \, dx \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {5}{4}-x^2} \, dx,x,-\frac {\sqrt {3}}{2}+2 x\right )\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {5}{4}-x^2} \, dx,x,\frac {\sqrt {3}}{2}+2 x\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {\sqrt {3}-4 x}{\sqrt {5}}\right )}{\sqrt {5}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3}+4 x}{\sqrt {5}}\right )}{\sqrt {5}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.11 \[ \int \frac {1+2 x^2}{1+x^2+4 x^4} \, dx=\frac {\left (-3 i+\sqrt {15}\right ) \arctan \left (\frac {2 x}{\sqrt {\frac {1}{2} \left (1-i \sqrt {15}\right )}}\right )}{\sqrt {30-30 i \sqrt {15}}}+\frac {\left (3 i+\sqrt {15}\right ) \arctan \left (\frac {2 x}{\sqrt {\frac {1}{2} \left (1+i \sqrt {15}\right )}}\right )}{\sqrt {30+30 i \sqrt {15}}} \]

[In]

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

[Out]

((-3*I + Sqrt[15])*ArcTan[(2*x)/Sqrt[(1 - I*Sqrt[15])/2]])/Sqrt[30 - (30*I)*Sqrt[15]] + ((3*I + Sqrt[15])*ArcT
an[(2*x)/Sqrt[(1 + I*Sqrt[15])/2]])/Sqrt[30 + (30*I)*Sqrt[15]]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {\sqrt {5}\, \arctan \left (\frac {2 x \sqrt {5}}{5}\right )}{5}+\frac {\sqrt {5}\, \arctan \left (\frac {4 x^{3} \sqrt {5}}{5}+\frac {3 x \sqrt {5}}{5}\right )}{5}\)	\(35\)
default	\(\frac {\arctan \left (\frac {\left (4 x +\sqrt {3}\right ) \sqrt {5}}{5}\right ) \sqrt {5}}{5}+\frac {\sqrt {5}\, \arctan \left (\frac {\left (4 x -\sqrt {3}\right ) \sqrt {5}}{5}\right )}{5}\)	\(40\)

[In]

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

[Out]

1/5*5^(1/2)*arctan(2/5*x*5^(1/2))+1/5*5^(1/2)*arctan(4/5*x^3*5^(1/2)+3/5*x*5^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \frac {1+2 x^2}{1+x^2+4 x^4} \, dx=\frac {1}{5} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} {\left (4 \, x^{3} + 3 \, x\right )}\right ) + \frac {1}{5} \, \sqrt {5} \arctan \left (\frac {2}{5} \, \sqrt {5} x\right ) \]

[In]

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

[Out]

1/5*sqrt(5)*arctan(1/5*sqrt(5)*(4*x^3 + 3*x)) + 1/5*sqrt(5)*arctan(2/5*sqrt(5)*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \frac {1+2 x^2}{1+x^2+4 x^4} \, dx=\frac {\sqrt {5} \cdot \left (2 \operatorname {atan}{\left (\frac {2 \sqrt {5} x}{5} \right )} + 2 \operatorname {atan}{\left (\frac {4 \sqrt {5} x^{3}}{5} + \frac {3 \sqrt {5} x}{5} \right )}\right )}{10} \]

[In]

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

[Out]

sqrt(5)*(2*atan(2*sqrt(5)*x/5) + 2*atan(4*sqrt(5)*x**3/5 + 3*sqrt(5)*x/5))/10

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.13 \[ \int \frac {1+2 x^2}{1+x^2+4 x^4} \, dx=\frac {1}{5} \, \sqrt {5} \arctan \left (\frac {2}{5} \, \sqrt {10} \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (4 \, x + \sqrt {6} \left (\frac {1}{4}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{5} \, \sqrt {5} \arctan \left (\frac {2}{5} \, \sqrt {10} \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (4 \, x - \sqrt {6} \left (\frac {1}{4}\right )^{\frac {1}{4}}\right )}\right ) \]

[In]

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

[Out]

1/5*sqrt(5)*arctan(2/5*sqrt(10)*(1/4)^(3/4)*(4*x + sqrt(6)*(1/4)^(1/4))) + 1/5*sqrt(5)*arctan(2/5*sqrt(10)*(1/
4)^(3/4)*(4*x - sqrt(6)*(1/4)^(1/4)))

Mupad [B] (verification not implemented)

Time = 13.72 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.63 \[ \int \frac {1+2 x^2}{1+x^2+4 x^4} \, dx=\frac {\sqrt {5}\,\left (\mathrm {atan}\left (\frac {4\,\sqrt {5}\,x^3}{5}+\frac {3\,\sqrt {5}\,x}{5}\right )+\mathrm {atan}\left (\frac {2\,\sqrt {5}\,x}{5}\right )\right )}{5} \]

[In]

int((2*x^2 + 1)/(x^2 + 4*x^4 + 1),x)

[Out]

(5^(1/2)*(atan((3*5^(1/2)*x)/5 + (4*5^(1/2)*x^3)/5) + atan((2*5^(1/2)*x)/5)))/5