3.46 \(\int \frac{1+2 x^2}{1+x^2+4 x^4} \, dx\)

Optimal. Leaf size=46 \[ \frac{\tan ^{-1}\left (\frac{4 x+\sqrt{3}}{\sqrt{5}}\right )}{\sqrt{5}}-\frac{\tan ^{-1}\left (\frac{\sqrt{3}-4 x}{\sqrt{5}}\right )}{\sqrt{5}} \]

[Out]

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

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Rubi [A]  time = 0.0431827, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {1161, 618, 204} \[ \frac{\tan ^{-1}\left (\frac{4 x+\sqrt{3}}{\sqrt{5}}\right )}{\sqrt{5}}-\frac{\tan ^{-1}\left (\frac{\sqrt{3}-4 x}{\sqrt{5}}\right )}{\sqrt{5}} \]

Antiderivative was successfully verified.

[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 1161

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]))

Rule 618

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 204

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

Rubi steps

\begin{align*} \int \frac{1+2 x^2}{1+x^2+4 x^4} \, dx &=\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} \operatorname{Subst}\left (\int \frac{1}{-\frac{5}{4}-x^2} \, dx,x,-\frac{\sqrt{3}}{2}+2 x\right )\right )-\frac{1}{2} \operatorname{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]  time = 0.225133, size = 97, normalized size = 2.11 \[ \frac{\left (\sqrt{15}-3 i\right ) \tan ^{-1}\left (\frac{2 x}{\sqrt{\frac{1}{2} \left (1-i \sqrt{15}\right )}}\right )}{\sqrt{30-30 i \sqrt{15}}}+\frac{\left (\sqrt{15}+3 i\right ) \tan ^{-1}\left (\frac{2 x}{\sqrt{\frac{1}{2} \left (1+i \sqrt{15}\right )}}\right )}{\sqrt{30+30 i \sqrt{15}}} \]

Antiderivative was successfully verified.

[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]]

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Maple [A]  time = 0.061, size = 40, normalized size = 0.9 \begin{align*}{\frac{\sqrt{5}}{5}\arctan \left ({\frac{ \left ( 4\,x-\sqrt{3} \right ) \sqrt{5}}{5}} \right ) }+{\frac{\sqrt{5}}{5}\arctan \left ({\frac{ \left ( 4\,x+\sqrt{3} \right ) \sqrt{5}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{2 \, x^{2} + 1}{4 \, x^{4} + x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.34345, size = 112, normalized size = 2.43 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.112677, size = 44, normalized size = 0.96 \begin{align*} \frac{\sqrt{5} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{2 \, x^{2} + 1}{4 \, x^{4} + x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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