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

Optimal. Leaf size=176 \[ -\frac{\log \left (x^2-\sqrt{2 \left (\sqrt{2}-1\right )} x+\sqrt{2}\right )}{8 \sqrt{\sqrt{2}-1}}+\frac{\log \left (x^2+\sqrt{2 \left (\sqrt{2}-1\right )} x+\sqrt{2}\right )}{8 \sqrt{\sqrt{2}-1}}-\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{2}-1\right )}-2 x}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right )+\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (\sqrt{2}-1\right )}}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right ) \]

[Out]

-(Sqrt[-1 + Sqrt[2]]*ArcTan[(Sqrt[2*(-1 + Sqrt[2])] - 2*x)/Sqrt[2*(1 + Sqrt[2])]])/4 + (Sqrt[-1 + Sqrt[2]]*Arc
Tan[(Sqrt[2*(-1 + Sqrt[2])] + 2*x)/Sqrt[2*(1 + Sqrt[2])]])/4 - Log[Sqrt[2] - Sqrt[2*(-1 + Sqrt[2])]*x + x^2]/(
8*Sqrt[-1 + Sqrt[2]]) + Log[Sqrt[2] + Sqrt[2*(-1 + Sqrt[2])]*x + x^2]/(8*Sqrt[-1 + Sqrt[2]])

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Rubi [A]  time = 0.161705, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {1094, 634, 618, 204, 628} \[ -\frac{\log \left (x^2-\sqrt{2 \left (\sqrt{2}-1\right )} x+\sqrt{2}\right )}{8 \sqrt{\sqrt{2}-1}}+\frac{\log \left (x^2+\sqrt{2 \left (\sqrt{2}-1\right )} x+\sqrt{2}\right )}{8 \sqrt{\sqrt{2}-1}}-\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{2}-1\right )}-2 x}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right )+\frac{1}{4} \sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (\sqrt{2}-1\right )}}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 + 2*x^2 + x^4)^(-1),x]

[Out]

-(Sqrt[-1 + Sqrt[2]]*ArcTan[(Sqrt[2*(-1 + Sqrt[2])] - 2*x)/Sqrt[2*(1 + Sqrt[2])]])/4 + (Sqrt[-1 + Sqrt[2]]*Arc
Tan[(Sqrt[2*(-1 + Sqrt[2])] + 2*x)/Sqrt[2*(1 + Sqrt[2])]])/4 - Log[Sqrt[2] - Sqrt[2*(-1 + Sqrt[2])]*x + x^2]/(
8*Sqrt[-1 + Sqrt[2]]) + Log[Sqrt[2] + Sqrt[2*(-1 + Sqrt[2])]*x + x^2]/(8*Sqrt[-1 + Sqrt[2]])

Rule 1094

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

Rule 634

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

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])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{1}{2+2 x^2+x^4} \, dx &=\frac{\int \frac{\sqrt{2 \left (-1+\sqrt{2}\right )}-x}{\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx}{4 \sqrt{-1+\sqrt{2}}}+\frac{\int \frac{\sqrt{2 \left (-1+\sqrt{2}\right )}+x}{\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx}{4 \sqrt{-1+\sqrt{2}}}\\ &=\frac{\int \frac{1}{\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx}{4 \sqrt{2}}+\frac{\int \frac{1}{\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx}{4 \sqrt{2}}-\frac{\int \frac{-\sqrt{2 \left (-1+\sqrt{2}\right )}+2 x}{\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx}{8 \sqrt{-1+\sqrt{2}}}+\frac{\int \frac{\sqrt{2 \left (-1+\sqrt{2}\right )}+2 x}{\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2} \, dx}{8 \sqrt{-1+\sqrt{2}}}\\ &=-\frac{\log \left (\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2\right )}{8 \sqrt{-1+\sqrt{2}}}+\frac{\log \left (\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2\right )}{8 \sqrt{-1+\sqrt{2}}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-2 \left (1+\sqrt{2}\right )-x^2} \, dx,x,-\sqrt{2 \left (-1+\sqrt{2}\right )}+2 x\right )}{2 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-2 \left (1+\sqrt{2}\right )-x^2} \, dx,x,\sqrt{2 \left (-1+\sqrt{2}\right )}+2 x\right )}{2 \sqrt{2}}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{2 \left (-1+\sqrt{2}\right )}-2 x}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right )}{4 \sqrt{1+\sqrt{2}}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2 \left (-1+\sqrt{2}\right )}+2 x}{\sqrt{2 \left (1+\sqrt{2}\right )}}\right )}{4 \sqrt{1+\sqrt{2}}}-\frac{\log \left (\sqrt{2}-\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2\right )}{8 \sqrt{-1+\sqrt{2}}}+\frac{\log \left (\sqrt{2}+\sqrt{2 \left (-1+\sqrt{2}\right )} x+x^2\right )}{8 \sqrt{-1+\sqrt{2}}}\\ \end{align*}

Mathematica [C]  time = 0.0351169, size = 41, normalized size = 0.23 \[ \frac{1}{4} \left ((1-i)^{3/2} \tan ^{-1}\left (\frac{x}{\sqrt{1-i}}\right )+(1+i)^{3/2} \tan ^{-1}\left (\frac{x}{\sqrt{1+i}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 2*x^2 + x^4)^(-1),x]

[Out]

((1 - I)^(3/2)*ArcTan[x/Sqrt[1 - I]] + (1 + I)^(3/2)*ArcTan[x/Sqrt[1 + I]])/4

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Maple [B]  time = 0.076, size = 386, normalized size = 2.2 \begin{align*}{\frac{\ln \left ({x}^{2}+\sqrt{2}+x\sqrt{-2+2\,\sqrt{2}} \right ) \sqrt{2}\sqrt{-2+2\,\sqrt{2}}}{16}}+{\frac{\ln \left ({x}^{2}+\sqrt{2}+x\sqrt{-2+2\,\sqrt{2}} \right ) \sqrt{-2+2\,\sqrt{2}}}{8}}-{\frac{\sqrt{2} \left ( -2+2\,\sqrt{2} \right ) }{8\,\sqrt{2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{2}}}{\sqrt{2+2\,\sqrt{2}}}} \right ) }-{\frac{-2+2\,\sqrt{2}}{4\,\sqrt{2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{2}}}{\sqrt{2+2\,\sqrt{2}}}} \right ) }+{\frac{\sqrt{2}}{2\,\sqrt{2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{2}}}{\sqrt{2+2\,\sqrt{2}}}} \right ) }-{\frac{\ln \left ({x}^{2}+\sqrt{2}-x\sqrt{-2+2\,\sqrt{2}} \right ) \sqrt{2}\sqrt{-2+2\,\sqrt{2}}}{16}}-{\frac{\ln \left ({x}^{2}+\sqrt{2}-x\sqrt{-2+2\,\sqrt{2}} \right ) \sqrt{-2+2\,\sqrt{2}}}{8}}-{\frac{\sqrt{2} \left ( -2+2\,\sqrt{2} \right ) }{8\,\sqrt{2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{2}}}{\sqrt{2+2\,\sqrt{2}}}} \right ) }-{\frac{-2+2\,\sqrt{2}}{4\,\sqrt{2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{2}}}{\sqrt{2+2\,\sqrt{2}}}} \right ) }+{\frac{\sqrt{2}}{2\,\sqrt{2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{2}}}{\sqrt{2+2\,\sqrt{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*ln(x^2+2^(1/2)+x*(-2+2*2^(1/2))^(1/2))*2^(1/2)*(-2+2*2^(1/2))^(1/2)+1/8*ln(x^2+2^(1/2)+x*(-2+2*2^(1/2))^(
1/2))*(-2+2*2^(1/2))^(1/2)-1/8/(2+2*2^(1/2))^(1/2)*arctan((2*x+(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))*2^(1
/2)*(-2+2*2^(1/2))-1/4/(2+2*2^(1/2))^(1/2)*arctan((2*x+(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))*(-2+2*2^(1/2
))+1/2/(2+2*2^(1/2))^(1/2)*arctan((2*x+(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))*2^(1/2)-1/16*ln(x^2+2^(1/2)-
x*(-2+2*2^(1/2))^(1/2))*2^(1/2)*(-2+2*2^(1/2))^(1/2)-1/8*ln(x^2+2^(1/2)-x*(-2+2*2^(1/2))^(1/2))*(-2+2*2^(1/2))
^(1/2)-1/8/(2+2*2^(1/2))^(1/2)*arctan((2*x-(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))*2^(1/2)*(-2+2*2^(1/2))-1
/4/(2+2*2^(1/2))^(1/2)*arctan((2*x-(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))*(-2+2*2^(1/2))+1/2/(2+2*2^(1/2))
^(1/2)*arctan((2*x-(-2+2*2^(1/2))^(1/2))/(2+2*2^(1/2))^(1/2))*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^4 + 2*x^2 + 2), x)

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Fricas [A]  time = 1.37879, size = 787, normalized size = 4.47 \begin{align*} \frac{1}{16} \cdot 2^{\frac{1}{4}}{\left (\sqrt{2} + 1\right )} \sqrt{-2 \, \sqrt{2} + 4} \log \left (2^{\frac{3}{4}} x \sqrt{-2 \, \sqrt{2} + 4} + 2 \, x^{2} + 2 \, \sqrt{2}\right ) - \frac{1}{16} \cdot 2^{\frac{1}{4}}{\left (\sqrt{2} + 1\right )} \sqrt{-2 \, \sqrt{2} + 4} \log \left (-2^{\frac{3}{4}} x \sqrt{-2 \, \sqrt{2} + 4} + 2 \, x^{2} + 2 \, \sqrt{2}\right ) - \frac{1}{4} \cdot 2^{\frac{1}{4}} \sqrt{-2 \, \sqrt{2} + 4} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{3}{4}} x \sqrt{-2 \, \sqrt{2} + 4} + \frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{2^{\frac{3}{4}} x \sqrt{-2 \, \sqrt{2} + 4} + 2 \, x^{2} + 2 \, \sqrt{2}} \sqrt{-2 \, \sqrt{2} + 4} - \sqrt{2} + 1\right ) - \frac{1}{4} \cdot 2^{\frac{1}{4}} \sqrt{-2 \, \sqrt{2} + 4} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{3}{4}} x \sqrt{-2 \, \sqrt{2} + 4} + \frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{-2^{\frac{3}{4}} x \sqrt{-2 \, \sqrt{2} + 4} + 2 \, x^{2} + 2 \, \sqrt{2}} \sqrt{-2 \, \sqrt{2} + 4} + \sqrt{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*2^(1/4)*(sqrt(2) + 1)*sqrt(-2*sqrt(2) + 4)*log(2^(3/4)*x*sqrt(-2*sqrt(2) + 4) + 2*x^2 + 2*sqrt(2)) - 1/16
*2^(1/4)*(sqrt(2) + 1)*sqrt(-2*sqrt(2) + 4)*log(-2^(3/4)*x*sqrt(-2*sqrt(2) + 4) + 2*x^2 + 2*sqrt(2)) - 1/4*2^(
1/4)*sqrt(-2*sqrt(2) + 4)*arctan(-1/2*2^(3/4)*x*sqrt(-2*sqrt(2) + 4) + 1/2*2^(1/4)*sqrt(2^(3/4)*x*sqrt(-2*sqrt
(2) + 4) + 2*x^2 + 2*sqrt(2))*sqrt(-2*sqrt(2) + 4) - sqrt(2) + 1) - 1/4*2^(1/4)*sqrt(-2*sqrt(2) + 4)*arctan(-1
/2*2^(3/4)*x*sqrt(-2*sqrt(2) + 4) + 1/2*2^(1/4)*sqrt(-2^(3/4)*x*sqrt(-2*sqrt(2) + 4) + 2*x^2 + 2*sqrt(2))*sqrt
(-2*sqrt(2) + 4) + sqrt(2) - 1)

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Sympy [A]  time = 0.552499, size = 20, normalized size = 0.11 \begin{align*} \operatorname{RootSum}{\left (512 t^{4} - 32 t^{2} + 1, \left ( t \mapsto t \log{\left (128 t^{3} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(512*_t**4 - 32*_t**2 + 1, Lambda(_t, _t*log(128*_t**3 + x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^4 + 2*x^2 + 2), x)