3.44 \(\int \frac{1+2 x^2}{1+3 x^2+4 x^4} \, dx\)

Optimal. Leaf size=38 \[ \frac{\tan ^{-1}\left (\frac{4 x+1}{\sqrt{7}}\right )}{\sqrt{7}}-\frac{\tan ^{-1}\left (\frac{1-4 x}{\sqrt{7}}\right )}{\sqrt{7}} \]

[Out]

-(ArcTan[(1 - 4*x)/Sqrt[7]]/Sqrt[7]) + ArcTan[(1 + 4*x)/Sqrt[7]]/Sqrt[7]

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Rubi [A]  time = 0.0352245, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1161, 618, 204} \[ \frac{\tan ^{-1}\left (\frac{4 x+1}{\sqrt{7}}\right )}{\sqrt{7}}-\frac{\tan ^{-1}\left (\frac{1-4 x}{\sqrt{7}}\right )}{\sqrt{7}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x^2)/(1 + 3*x^2 + 4*x^4),x]

[Out]

-(ArcTan[(1 - 4*x)/Sqrt[7]]/Sqrt[7]) + ArcTan[(1 + 4*x)/Sqrt[7]]/Sqrt[7]

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+3 x^2+4 x^4} \, dx &=\frac{1}{4} \int \frac{1}{\frac{1}{2}-\frac{x}{2}+x^2} \, dx+\frac{1}{4} \int \frac{1}{\frac{1}{2}+\frac{x}{2}+x^2} \, dx\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-\frac{7}{4}-x^2} \, dx,x,-\frac{1}{2}+2 x\right )\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-\frac{7}{4}-x^2} \, dx,x,\frac{1}{2}+2 x\right )\\ &=-\frac{\tan ^{-1}\left (\frac{1-4 x}{\sqrt{7}}\right )}{\sqrt{7}}+\frac{\tan ^{-1}\left (\frac{1+4 x}{\sqrt{7}}\right )}{\sqrt{7}}\\ \end{align*}

Mathematica [C]  time = 0.185003, size = 97, normalized size = 2.55 \[ \frac{\left (\sqrt{7}-i\right ) \tan ^{-1}\left (\frac{2 x}{\sqrt{\frac{1}{2} \left (3-i \sqrt{7}\right )}}\right )}{\sqrt{42-14 i \sqrt{7}}}+\frac{\left (\sqrt{7}+i\right ) \tan ^{-1}\left (\frac{2 x}{\sqrt{\frac{1}{2} \left (3+i \sqrt{7}\right )}}\right )}{\sqrt{42+14 i \sqrt{7}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x^2)/(1 + 3*x^2 + 4*x^4),x]

[Out]

((-I + Sqrt[7])*ArcTan[(2*x)/Sqrt[(3 - I*Sqrt[7])/2]])/Sqrt[42 - (14*I)*Sqrt[7]] + ((I + Sqrt[7])*ArcTan[(2*x)
/Sqrt[(3 + I*Sqrt[7])/2]])/Sqrt[42 + (14*I)*Sqrt[7]]

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Maple [A]  time = 0.043, size = 34, normalized size = 0.9 \begin{align*}{\frac{\sqrt{7}}{7}\arctan \left ({\frac{ \left ( -1+4\,x \right ) \sqrt{7}}{7}} \right ) }+{\frac{\sqrt{7}}{7}\arctan \left ({\frac{ \left ( 4\,x+1 \right ) \sqrt{7}}{7}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+1)/(4*x^4+3*x^2+1),x)

[Out]

1/7*7^(1/2)*arctan(1/7*(-1+4*x)*7^(1/2))+1/7*arctan(1/7*(4*x+1)*7^(1/2))*7^(1/2)

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Maxima [A]  time = 1.44026, size = 45, normalized size = 1.18 \begin{align*} \frac{1}{7} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x + 1\right )}\right ) + \frac{1}{7} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x - 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4+3*x^2+1),x, algorithm="maxima")

[Out]

1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x + 1)) + 1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x - 1))

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Fricas [A]  time = 1.30687, size = 112, normalized size = 2.95 \begin{align*} \frac{1}{7} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x^{3} + 5 \, x\right )}\right ) + \frac{1}{7} \, \sqrt{7} \arctan \left (\frac{2}{7} \, \sqrt{7} x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4+3*x^2+1),x, algorithm="fricas")

[Out]

1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x^3 + 5*x)) + 1/7*sqrt(7)*arctan(2/7*sqrt(7)*x)

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Sympy [A]  time = 0.115235, size = 44, normalized size = 1.16 \begin{align*} \frac{\sqrt{7} \left (2 \operatorname{atan}{\left (\frac{2 \sqrt{7} x}{7} \right )} + 2 \operatorname{atan}{\left (\frac{4 \sqrt{7} x^{3}}{7} + \frac{5 \sqrt{7} x}{7} \right )}\right )}{14} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+1)/(4*x**4+3*x**2+1),x)

[Out]

sqrt(7)*(2*atan(2*sqrt(7)*x/7) + 2*atan(4*sqrt(7)*x**3/7 + 5*sqrt(7)*x/7))/14

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Giac [A]  time = 1.13606, size = 45, normalized size = 1.18 \begin{align*} \frac{1}{7} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x + 1\right )}\right ) + \frac{1}{7} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x - 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+1)/(4*x^4+3*x^2+1),x, algorithm="giac")

[Out]

1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x + 1)) + 1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x - 1))