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

Optimal. Leaf size=196 \[ -\frac{\log \left (x^2-\sqrt{2 \sqrt{2}-1} x+\sqrt{2}\right )}{4 \sqrt{2 \left (2 \sqrt{2}-1\right )}}+\frac{\log \left (x^2+\sqrt{2 \sqrt{2}-1} x+\sqrt{2}\right )}{4 \sqrt{2 \left (2 \sqrt{2}-1\right )}}-\frac{1}{2} \sqrt{\frac{1}{14} \left (2 \sqrt{2}-1\right )} \tan ^{-1}\left (\frac{\sqrt{2 \sqrt{2}-1}-2 x}{\sqrt{1+2 \sqrt{2}}}\right )+\frac{1}{2} \sqrt{\frac{1}{14} \left (2 \sqrt{2}-1\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \sqrt{2}-1}}{\sqrt{1+2 \sqrt{2}}}\right ) \]

[Out]

-(Sqrt[(-1 + 2*Sqrt[2])/14]*ArcTan[(Sqrt[-1 + 2*Sqrt[2]] - 2*x)/Sqrt[1 + 2*Sqrt[2]]])/2 + (Sqrt[(-1 + 2*Sqrt[2
])/14]*ArcTan[(Sqrt[-1 + 2*Sqrt[2]] + 2*x)/Sqrt[1 + 2*Sqrt[2]]])/2 - Log[Sqrt[2] - Sqrt[-1 + 2*Sqrt[2]]*x + x^
2]/(4*Sqrt[2*(-1 + 2*Sqrt[2])]) + Log[Sqrt[2] + Sqrt[-1 + 2*Sqrt[2]]*x + x^2]/(4*Sqrt[2*(-1 + 2*Sqrt[2])])

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Rubi [A]  time = 0.148191, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {1094, 634, 618, 204, 628} \[ -\frac{\log \left (x^2-\sqrt{2 \sqrt{2}-1} x+\sqrt{2}\right )}{4 \sqrt{2 \left (2 \sqrt{2}-1\right )}}+\frac{\log \left (x^2+\sqrt{2 \sqrt{2}-1} x+\sqrt{2}\right )}{4 \sqrt{2 \left (2 \sqrt{2}-1\right )}}-\frac{1}{2} \sqrt{\frac{1}{14} \left (2 \sqrt{2}-1\right )} \tan ^{-1}\left (\frac{\sqrt{2 \sqrt{2}-1}-2 x}{\sqrt{1+2 \sqrt{2}}}\right )+\frac{1}{2} \sqrt{\frac{1}{14} \left (2 \sqrt{2}-1\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \sqrt{2}-1}}{\sqrt{1+2 \sqrt{2}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 + x^2 + x^4)^(-1),x]

[Out]

-(Sqrt[(-1 + 2*Sqrt[2])/14]*ArcTan[(Sqrt[-1 + 2*Sqrt[2]] - 2*x)/Sqrt[1 + 2*Sqrt[2]]])/2 + (Sqrt[(-1 + 2*Sqrt[2
])/14]*ArcTan[(Sqrt[-1 + 2*Sqrt[2]] + 2*x)/Sqrt[1 + 2*Sqrt[2]]])/2 - Log[Sqrt[2] - Sqrt[-1 + 2*Sqrt[2]]*x + x^
2]/(4*Sqrt[2*(-1 + 2*Sqrt[2])]) + Log[Sqrt[2] + Sqrt[-1 + 2*Sqrt[2]]*x + x^2]/(4*Sqrt[2*(-1 + 2*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+x^2+x^4} \, dx &=\frac{\int \frac{\sqrt{-1+2 \sqrt{2}}-x}{\sqrt{2}-\sqrt{-1+2 \sqrt{2}} x+x^2} \, dx}{2 \sqrt{2 \left (-1+2 \sqrt{2}\right )}}+\frac{\int \frac{\sqrt{-1+2 \sqrt{2}}+x}{\sqrt{2}+\sqrt{-1+2 \sqrt{2}} x+x^2} \, dx}{2 \sqrt{2 \left (-1+2 \sqrt{2}\right )}}\\ &=\frac{\int \frac{1}{\sqrt{2}-\sqrt{-1+2 \sqrt{2}} x+x^2} \, dx}{4 \sqrt{2}}+\frac{\int \frac{1}{\sqrt{2}+\sqrt{-1+2 \sqrt{2}} x+x^2} \, dx}{4 \sqrt{2}}-\frac{\int \frac{-\sqrt{-1+2 \sqrt{2}}+2 x}{\sqrt{2}-\sqrt{-1+2 \sqrt{2}} x+x^2} \, dx}{4 \sqrt{2 \left (-1+2 \sqrt{2}\right )}}+\frac{\int \frac{\sqrt{-1+2 \sqrt{2}}+2 x}{\sqrt{2}+\sqrt{-1+2 \sqrt{2}} x+x^2} \, dx}{4 \sqrt{2 \left (-1+2 \sqrt{2}\right )}}\\ &=-\frac{\log \left (\sqrt{2}-\sqrt{-1+2 \sqrt{2}} x+x^2\right )}{4 \sqrt{2 \left (-1+2 \sqrt{2}\right )}}+\frac{\log \left (\sqrt{2}+\sqrt{-1+2 \sqrt{2}} x+x^2\right )}{4 \sqrt{2 \left (-1+2 \sqrt{2}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-2 \sqrt{2}-x^2} \, dx,x,-\sqrt{-1+2 \sqrt{2}}+2 x\right )}{2 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-2 \sqrt{2}-x^2} \, dx,x,\sqrt{-1+2 \sqrt{2}}+2 x\right )}{2 \sqrt{2}}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{-1+2 \sqrt{2}}-2 x}{\sqrt{1+2 \sqrt{2}}}\right )}{2 \sqrt{2 \left (1+2 \sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{-1+2 \sqrt{2}}+2 x}{\sqrt{1+2 \sqrt{2}}}\right )}{2 \sqrt{2 \left (1+2 \sqrt{2}\right )}}-\frac{\log \left (\sqrt{2}-\sqrt{-1+2 \sqrt{2}} x+x^2\right )}{4 \sqrt{2 \left (-1+2 \sqrt{2}\right )}}+\frac{\log \left (\sqrt{2}+\sqrt{-1+2 \sqrt{2}} x+x^2\right )}{4 \sqrt{2 \left (-1+2 \sqrt{2}\right )}}\\ \end{align*}

Mathematica [C]  time = 0.0616323, size = 91, normalized size = 0.46 \[ \frac{i \tan ^{-1}\left (\frac{x}{\sqrt{\frac{1}{2} \left (1+i \sqrt{7}\right )}}\right )}{\sqrt{\frac{7}{2} \left (1+i \sqrt{7}\right )}}-\frac{i \tan ^{-1}\left (\frac{x}{\sqrt{\frac{1}{2} \left (1-i \sqrt{7}\right )}}\right )}{\sqrt{\frac{7}{2} \left (1-i \sqrt{7}\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + x^2 + x^4)^(-1),x]

[Out]

((-I)*ArcTan[x/Sqrt[(1 - I*Sqrt[7])/2]])/Sqrt[(7*(1 - I*Sqrt[7]))/2] + (I*ArcTan[x/Sqrt[(1 + I*Sqrt[7])/2]])/S
qrt[(7*(1 + I*Sqrt[7]))/2]

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Maple [B]  time = 0.064, size = 386, normalized size = 2. \begin{align*}{\frac{\ln \left ({x}^{2}+\sqrt{2}+x\sqrt{-1+2\,\sqrt{2}} \right ) \sqrt{-1+2\,\sqrt{2}}\sqrt{2}}{56}}+{\frac{\ln \left ({x}^{2}+\sqrt{2}+x\sqrt{-1+2\,\sqrt{2}} \right ) \sqrt{-1+2\,\sqrt{2}}}{14}}-{\frac{ \left ( -1+2\,\sqrt{2} \right ) \sqrt{2}}{28\,\sqrt{1+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x+\sqrt{-1+2\,\sqrt{2}}}{\sqrt{1+2\,\sqrt{2}}}} \right ) }-{\frac{-1+2\,\sqrt{2}}{7\,\sqrt{1+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x+\sqrt{-1+2\,\sqrt{2}}}{\sqrt{1+2\,\sqrt{2}}}} \right ) }+{\frac{\sqrt{2}}{2\,\sqrt{1+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x+\sqrt{-1+2\,\sqrt{2}}}{\sqrt{1+2\,\sqrt{2}}}} \right ) }-{\frac{\ln \left ({x}^{2}+\sqrt{2}-x\sqrt{-1+2\,\sqrt{2}} \right ) \sqrt{-1+2\,\sqrt{2}}\sqrt{2}}{56}}-{\frac{\ln \left ({x}^{2}+\sqrt{2}-x\sqrt{-1+2\,\sqrt{2}} \right ) \sqrt{-1+2\,\sqrt{2}}}{14}}-{\frac{ \left ( -1+2\,\sqrt{2} \right ) \sqrt{2}}{28\,\sqrt{1+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x-\sqrt{-1+2\,\sqrt{2}}}{\sqrt{1+2\,\sqrt{2}}}} \right ) }-{\frac{-1+2\,\sqrt{2}}{7\,\sqrt{1+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x-\sqrt{-1+2\,\sqrt{2}}}{\sqrt{1+2\,\sqrt{2}}}} \right ) }+{\frac{\sqrt{2}}{2\,\sqrt{1+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x-\sqrt{-1+2\,\sqrt{2}}}{\sqrt{1+2\,\sqrt{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/56*ln(x^2+2^(1/2)+x*(-1+2*2^(1/2))^(1/2))*(-1+2*2^(1/2))^(1/2)*2^(1/2)+1/14*ln(x^2+2^(1/2)+x*(-1+2*2^(1/2))^
(1/2))*(-1+2*2^(1/2))^(1/2)-1/28/(1+2*2^(1/2))^(1/2)*arctan((2*x+(-1+2*2^(1/2))^(1/2))/(1+2*2^(1/2))^(1/2))*(-
1+2*2^(1/2))*2^(1/2)-1/7/(1+2*2^(1/2))^(1/2)*arctan((2*x+(-1+2*2^(1/2))^(1/2))/(1+2*2^(1/2))^(1/2))*(-1+2*2^(1
/2))+1/2/(1+2*2^(1/2))^(1/2)*arctan((2*x+(-1+2*2^(1/2))^(1/2))/(1+2*2^(1/2))^(1/2))*2^(1/2)-1/56*ln(x^2+2^(1/2
)-x*(-1+2*2^(1/2))^(1/2))*(-1+2*2^(1/2))^(1/2)*2^(1/2)-1/14*ln(x^2+2^(1/2)-x*(-1+2*2^(1/2))^(1/2))*(-1+2*2^(1/
2))^(1/2)-1/28/(1+2*2^(1/2))^(1/2)*arctan((2*x-(-1+2*2^(1/2))^(1/2))/(1+2*2^(1/2))^(1/2))*(-1+2*2^(1/2))*2^(1/
2)-1/7/(1+2*2^(1/2))^(1/2)*arctan((2*x-(-1+2*2^(1/2))^(1/2))/(1+2*2^(1/2))^(1/2))*(-1+2*2^(1/2))+1/2/(1+2*2^(1
/2))^(1/2)*arctan((2*x-(-1+2*2^(1/2))^(1/2))/(1+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} + x^{2} + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^4 + x^2 + 2), x)

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Fricas [B]  time = 2.08491, size = 988, normalized size = 5.04 \begin{align*} -\frac{1}{56} \cdot 8^{\frac{1}{4}} \sqrt{7} \sqrt{2} \sqrt{-4 \, \sqrt{2} + 16} \arctan \left (-\frac{1}{56} \cdot 8^{\frac{3}{4}} \sqrt{7} \sqrt{2} x \sqrt{-4 \, \sqrt{2} + 16} + \frac{1}{112} \cdot 8^{\frac{3}{4}} \sqrt{7} \sqrt{2} \sqrt{4 \, x^{2} + 8^{\frac{1}{4}} x \sqrt{-4 \, \sqrt{2} + 16} + 4 \, \sqrt{2}} \sqrt{-4 \, \sqrt{2} + 16} - \frac{1}{7} \, \sqrt{7}{\left (2 \, \sqrt{2} - 1\right )}\right ) - \frac{1}{56} \cdot 8^{\frac{1}{4}} \sqrt{7} \sqrt{2} \sqrt{-4 \, \sqrt{2} + 16} \arctan \left (-\frac{1}{56} \cdot 8^{\frac{3}{4}} \sqrt{7} \sqrt{2} x \sqrt{-4 \, \sqrt{2} + 16} + \frac{1}{112} \cdot 8^{\frac{3}{4}} \sqrt{7} \sqrt{2} \sqrt{4 \, x^{2} - 8^{\frac{1}{4}} x \sqrt{-4 \, \sqrt{2} + 16} + 4 \, \sqrt{2}} \sqrt{-4 \, \sqrt{2} + 16} + \frac{1}{7} \, \sqrt{7}{\left (2 \, \sqrt{2} - 1\right )}\right ) + \frac{1}{224} \cdot 8^{\frac{1}{4}}{\left (\sqrt{2} + 4\right )} \sqrt{-4 \, \sqrt{2} + 16} \log \left (4 \, x^{2} + 8^{\frac{1}{4}} x \sqrt{-4 \, \sqrt{2} + 16} + 4 \, \sqrt{2}\right ) - \frac{1}{224} \cdot 8^{\frac{1}{4}}{\left (\sqrt{2} + 4\right )} \sqrt{-4 \, \sqrt{2} + 16} \log \left (4 \, x^{2} - 8^{\frac{1}{4}} x \sqrt{-4 \, \sqrt{2} + 16} + 4 \, \sqrt{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/56*8^(1/4)*sqrt(7)*sqrt(2)*sqrt(-4*sqrt(2) + 16)*arctan(-1/56*8^(3/4)*sqrt(7)*sqrt(2)*x*sqrt(-4*sqrt(2) + 1
6) + 1/112*8^(3/4)*sqrt(7)*sqrt(2)*sqrt(4*x^2 + 8^(1/4)*x*sqrt(-4*sqrt(2) + 16) + 4*sqrt(2))*sqrt(-4*sqrt(2) +
16) - 1/7*sqrt(7)*(2*sqrt(2) - 1)) - 1/56*8^(1/4)*sqrt(7)*sqrt(2)*sqrt(-4*sqrt(2) + 16)*arctan(-1/56*8^(3/4)*
sqrt(7)*sqrt(2)*x*sqrt(-4*sqrt(2) + 16) + 1/112*8^(3/4)*sqrt(7)*sqrt(2)*sqrt(4*x^2 - 8^(1/4)*x*sqrt(-4*sqrt(2)
+ 16) + 4*sqrt(2))*sqrt(-4*sqrt(2) + 16) + 1/7*sqrt(7)*(2*sqrt(2) - 1)) + 1/224*8^(1/4)*(sqrt(2) + 4)*sqrt(-4
*sqrt(2) + 16)*log(4*x^2 + 8^(1/4)*x*sqrt(-4*sqrt(2) + 16) + 4*sqrt(2)) - 1/224*8^(1/4)*(sqrt(2) + 4)*sqrt(-4*
sqrt(2) + 16)*log(4*x^2 - 8^(1/4)*x*sqrt(-4*sqrt(2) + 16) + 4*sqrt(2))

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Sympy [A]  time = 0.469528, size = 24, normalized size = 0.12 \begin{align*} \operatorname{RootSum}{\left (1568 t^{4} - 28 t^{2} + 1, \left ( t \mapsto t \log{\left (112 t^{3} + 6 t + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(1568*_t**4 - 28*_t**2 + 1, Lambda(_t, _t*log(112*_t**3 + 6*_t + x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^4 + x^2 + 2), x)