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

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

[Out]

-1/28*arctan((-2*x+(1+2*2^(1/2))^(1/2))/(-1+2*2^(1/2))^(1/2))*(14+28*2^(1/2))^(1/2)+1/28*arctan((2*x+(1+2*2^(1
/2))^(1/2))/(-1+2*2^(1/2))^(1/2))*(14+28*2^(1/2))^(1/2)-1/4*ln(x^2+2^(1/2)-x*(1+2*2^(1/2))^(1/2))/(2+4*2^(1/2)
)^(1/2)+1/4*ln(x^2+2^(1/2)+x*(1+2*2^(1/2))^(1/2))/(2+4*2^(1/2))^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 196, normalized size of antiderivative = 1.00, 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} \[ -\frac {\log \left (x^2-\sqrt {1+2 \sqrt {2}} x+\sqrt {2}\right )}{4 \sqrt {2 \left (1+2 \sqrt {2}\right )}}+\frac {\log \left (x^2+\sqrt {1+2 \sqrt {2}} x+\sqrt {2}\right )}{4 \sqrt {2 \left (1+2 \sqrt {2}\right )}}-\frac {1}{2} \sqrt {\frac {1}{14} \left (1+2 \sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {1+2 \sqrt {2}}-2 x}{\sqrt {2 \sqrt {2}-1}}\right )+\frac {1}{2} \sqrt {\frac {1}{14} \left (1+2 \sqrt {2}\right )} \tan ^{-1}\left (\frac {2 x+\sqrt {1+2 \sqrt {2}}}{\sqrt {2 \sqrt {2}-1}}\right ) \]

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 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 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 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]

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 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]

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*}

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Mathematica [C]  time = 0.08, 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]]
)/Sqrt[(7*(-1 + I*Sqrt[7]))/2]

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fricas [B]  time = 0.44, size = 275, normalized size = 1.40 \[ -\frac {1}{28} \cdot 8^{\frac {1}{4}} \sqrt {7} \sqrt {2} \sqrt {\sqrt {2} + 4} \arctan \left (-\frac {1}{28} \cdot 8^{\frac {3}{4}} \sqrt {7} \sqrt {2} x \sqrt {\sqrt {2} + 4} + \frac {1}{56} \cdot 8^{\frac {3}{4}} \sqrt {7} \sqrt {2} \sqrt {4 \, x^{2} + 2 \cdot 8^{\frac {1}{4}} x \sqrt {\sqrt {2} + 4} + 4 \, \sqrt {2}} \sqrt {\sqrt {2} + 4} - \frac {1}{7} \, \sqrt {7} {\left (2 \, \sqrt {2} + 1\right )}\right ) - \frac {1}{28} \cdot 8^{\frac {1}{4}} \sqrt {7} \sqrt {2} \sqrt {\sqrt {2} + 4} \arctan \left (-\frac {1}{28} \cdot 8^{\frac {3}{4}} \sqrt {7} \sqrt {2} x \sqrt {\sqrt {2} + 4} + \frac {1}{56} \cdot 8^{\frac {3}{4}} \sqrt {7} \sqrt {2} \sqrt {4 \, x^{2} - 2 \cdot 8^{\frac {1}{4}} x \sqrt {\sqrt {2} + 4} + 4 \, \sqrt {2}} \sqrt {\sqrt {2} + 4} + \frac {1}{7} \, \sqrt {7} {\left (2 \, \sqrt {2} + 1\right )}\right ) - \frac {1}{112} \cdot 8^{\frac {1}{4}} \sqrt {\sqrt {2} + 4} {\left (\sqrt {2} - 4\right )} \log \left (4 \, x^{2} + 2 \cdot 8^{\frac {1}{4}} x \sqrt {\sqrt {2} + 4} + 4 \, \sqrt {2}\right ) + \frac {1}{112} \cdot 8^{\frac {1}{4}} \sqrt {\sqrt {2} + 4} {\left (\sqrt {2} - 4\right )} \log \left (4 \, x^{2} - 2 \cdot 8^{\frac {1}{4}} x \sqrt {\sqrt {2} + 4} + 4 \, \sqrt {2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/28*8^(1/4)*sqrt(7)*sqrt(2)*sqrt(sqrt(2) + 4)*arctan(-1/28*8^(3/4)*sqrt(7)*sqrt(2)*x*sqrt(sqrt(2) + 4) + 1/5
6*8^(3/4)*sqrt(7)*sqrt(2)*sqrt(4*x^2 + 2*8^(1/4)*x*sqrt(sqrt(2) + 4) + 4*sqrt(2))*sqrt(sqrt(2) + 4) - 1/7*sqrt
(7)*(2*sqrt(2) + 1)) - 1/28*8^(1/4)*sqrt(7)*sqrt(2)*sqrt(sqrt(2) + 4)*arctan(-1/28*8^(3/4)*sqrt(7)*sqrt(2)*x*s
qrt(sqrt(2) + 4) + 1/56*8^(3/4)*sqrt(7)*sqrt(2)*sqrt(4*x^2 - 2*8^(1/4)*x*sqrt(sqrt(2) + 4) + 4*sqrt(2))*sqrt(s
qrt(2) + 4) + 1/7*sqrt(7)*(2*sqrt(2) + 1)) - 1/112*8^(1/4)*sqrt(sqrt(2) + 4)*(sqrt(2) - 4)*log(4*x^2 + 2*8^(1/
4)*x*sqrt(sqrt(2) + 4) + 4*sqrt(2)) + 1/112*8^(1/4)*sqrt(sqrt(2) + 4)*(sqrt(2) - 4)*log(4*x^2 - 2*8^(1/4)*x*sq
rt(sqrt(2) + 4) + 4*sqrt(2))

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giac [A]  time = 2.00, size = 252, normalized size = 1.29 \[ \frac {1}{224} \, \sqrt {7} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {\sqrt {2} + 4} + \sqrt {7} 2^{\frac {1}{4}} \sqrt {-8 \, \sqrt {2} + 32}\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\frac {1}{2}} \sqrt {\sqrt {2} + 4} + 2 \, x\right )}}{4 \, \sqrt {-\frac {1}{8} \, \sqrt {2} + \frac {1}{2}}}\right ) + \frac {1}{224} \, \sqrt {7} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {\sqrt {2} + 4} + \sqrt {7} 2^{\frac {1}{4}} \sqrt {-8 \, \sqrt {2} + 32}\right )} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\frac {1}{2}} \sqrt {\sqrt {2} + 4} - 2 \, x\right )}}{4 \, \sqrt {-\frac {1}{8} \, \sqrt {2} + \frac {1}{2}}}\right ) + \frac {1}{448} \, \sqrt {7} {\left (2 \, \sqrt {7} 2^{\frac {3}{4}} \sqrt {\sqrt {2} + 4} - 2^{\frac {1}{4}} \sqrt {-8 \, \sqrt {2} + 32}\right )} \log \left (2^{\frac {1}{4}} \sqrt {\frac {1}{2}} x \sqrt {\sqrt {2} + 4} + x^{2} + \sqrt {2}\right ) - \frac {1}{448} \, \sqrt {7} {\left (2 \, \sqrt {7} 2^{\frac {3}{4}} \sqrt {\sqrt {2} + 4} - 2^{\frac {1}{4}} \sqrt {-8 \, \sqrt {2} + 32}\right )} \log \left (-2^{\frac {1}{4}} \sqrt {\frac {1}{2}} x \sqrt {\sqrt {2} + 4} + x^{2} + \sqrt {2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/224*sqrt(7)*(2*2^(3/4)*sqrt(sqrt(2) + 4) + sqrt(7)*2^(1/4)*sqrt(-8*sqrt(2) + 32))*arctan(1/4*2^(3/4)*(2^(1/4
)*sqrt(1/2)*sqrt(sqrt(2) + 4) + 2*x)/sqrt(-1/8*sqrt(2) + 1/2)) + 1/224*sqrt(7)*(2*2^(3/4)*sqrt(sqrt(2) + 4) +
sqrt(7)*2^(1/4)*sqrt(-8*sqrt(2) + 32))*arctan(-1/4*2^(3/4)*(2^(1/4)*sqrt(1/2)*sqrt(sqrt(2) + 4) - 2*x)/sqrt(-1
/8*sqrt(2) + 1/2)) + 1/448*sqrt(7)*(2*sqrt(7)*2^(3/4)*sqrt(sqrt(2) + 4) - 2^(1/4)*sqrt(-8*sqrt(2) + 32))*log(2
^(1/4)*sqrt(1/2)*x*sqrt(sqrt(2) + 4) + x^2 + sqrt(2)) - 1/448*sqrt(7)*(2*sqrt(7)*2^(3/4)*sqrt(sqrt(2) + 4) - 2
^(1/4)*sqrt(-8*sqrt(2) + 32))*log(-2^(1/4)*sqrt(1/2)*x*sqrt(sqrt(2) + 4) + x^2 + sqrt(2))

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maple [B]  time = 0.11, size = 386, normalized size = 1.97 \[ \frac {\left (1+2 \sqrt {2}\right ) \sqrt {2}\, \arctan \left (\frac {2 x -\sqrt {1+2 \sqrt {2}}}{\sqrt {-1+2 \sqrt {2}}}\right )}{28 \sqrt {-1+2 \sqrt {2}}}-\frac {\left (1+2 \sqrt {2}\right ) \arctan \left (\frac {2 x -\sqrt {1+2 \sqrt {2}}}{\sqrt {-1+2 \sqrt {2}}}\right )}{7 \sqrt {-1+2 \sqrt {2}}}+\frac {\sqrt {2}\, \arctan \left (\frac {2 x -\sqrt {1+2 \sqrt {2}}}{\sqrt {-1+2 \sqrt {2}}}\right )}{2 \sqrt {-1+2 \sqrt {2}}}+\frac {\left (1+2 \sqrt {2}\right ) \sqrt {2}\, \arctan \left (\frac {2 x +\sqrt {1+2 \sqrt {2}}}{\sqrt {-1+2 \sqrt {2}}}\right )}{28 \sqrt {-1+2 \sqrt {2}}}-\frac {\left (1+2 \sqrt {2}\right ) \arctan \left (\frac {2 x +\sqrt {1+2 \sqrt {2}}}{\sqrt {-1+2 \sqrt {2}}}\right )}{7 \sqrt {-1+2 \sqrt {2}}}+\frac {\sqrt {2}\, \arctan \left (\frac {2 x +\sqrt {1+2 \sqrt {2}}}{\sqrt {-1+2 \sqrt {2}}}\right )}{2 \sqrt {-1+2 \sqrt {2}}}+\frac {\sqrt {1+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (x^{2}-\sqrt {1+2 \sqrt {2}}\, x +\sqrt {2}\right )}{56}-\frac {\sqrt {1+2 \sqrt {2}}\, \ln \left (x^{2}-\sqrt {1+2 \sqrt {2}}\, x +\sqrt {2}\right )}{14}-\frac {\sqrt {1+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (x^{2}+\sqrt {1+2 \sqrt {2}}\, x +\sqrt {2}\right )}{56}+\frac {\sqrt {1+2 \sqrt {2}}\, \ln \left (x^{2}+\sqrt {1+2 \sqrt {2}}\, x +\sqrt {2}\right )}{14} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} - x^{2} + 2}\,{d x} \]

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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mupad [B]  time = 0.11, size = 132, normalized size = 0.67 \[ \frac {\mathrm {atan}\left (\frac {x\,\sqrt {-7-\sqrt {7}\,7{}\mathrm {i}}\,1{}\mathrm {i}}{4\,\left (\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )}+\frac {\sqrt {7}\,x\,\sqrt {-7-\sqrt {7}\,7{}\mathrm {i}}}{28\,\left (\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )}\right )\,\sqrt {-7-\sqrt {7}\,7{}\mathrm {i}}\,1{}\mathrm {i}}{14}+\frac {\sqrt {7}\,\mathrm {atan}\left (\frac {x\,\sqrt {-1+\sqrt {7}\,1{}\mathrm {i}}}{4\,\left (-\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )}-\frac {\sqrt {7}\,x\,\sqrt {-1+\sqrt {7}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{4\,\left (-\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )}\right )\,\sqrt {-1+\sqrt {7}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{14} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4 - x^2 + 2),x)

[Out]

(atan((x*(- 7^(1/2)*7i - 7)^(1/2)*1i)/(4*((7^(1/2)*1i)/2 + 1/2)) + (7^(1/2)*x*(- 7^(1/2)*7i - 7)^(1/2))/(28*((
7^(1/2)*1i)/2 + 1/2)))*(- 7^(1/2)*7i - 7)^(1/2)*1i)/14 + (7^(1/2)*atan((x*(7^(1/2)*1i - 1)^(1/2))/(4*((7^(1/2)
*1i)/2 - 1/2)) - (7^(1/2)*x*(7^(1/2)*1i - 1)^(1/2)*1i)/(4*((7^(1/2)*1i)/2 - 1/2)))*(7^(1/2)*1i - 1)^(1/2)*1i)/
14

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sympy [A]  time = 0.54, size = 24, normalized size = 0.12 \[ \operatorname {RootSum} {\left (1568 t^{4} + 28 t^{2} + 1, \left (t \mapsto t \log {\left (- 112 t^{3} + 6 t + x \right )} \right )\right )} \]

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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