\(\int \frac {1}{2-x^2+x^4} \, dx\) [45]

Optimal result

Integrand size = 12, antiderivative size = 196 \[ \int \frac {1}{2-x^2+x^4} \, dx=-\frac {1}{2} \sqrt {\frac {1}{14} \left (1+2 \sqrt {2}\right )} \arctan \left (\frac {\sqrt {1+2 \sqrt {2}}-2 x}{\sqrt {-1+2 \sqrt {2}}}\right )+\frac {1}{2} \sqrt {\frac {1}{14} \left (1+2 \sqrt {2}\right )} \arctan \left (\frac {\sqrt {1+2 \sqrt {2}}+2 x}{\sqrt {-1+2 \sqrt {2}}}\right )-\frac {\log \left (\sqrt {2}-\sqrt {1+2 \sqrt {2}} x+x^2\right )}{4 \sqrt {2 \left (1+2 \sqrt {2}\right )}}+\frac {\log \left (\sqrt {2}+\sqrt {1+2 \sqrt {2}} x+x^2\right )}{4 \sqrt {2 \left (1+2 \sqrt {2}\right )}} \]

[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)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {1108, 648, 632, 210, 642} \[ \int \frac {1}{2-x^2+x^4} \, dx=-\frac {1}{2} \sqrt {\frac {1}{14} \left (1+2 \sqrt {2}\right )} \arctan \left (\frac {\sqrt {1+2 \sqrt {2}}-2 x}{\sqrt {2 \sqrt {2}-1}}\right )+\frac {1}{2} \sqrt {\frac {1}{14} \left (1+2 \sqrt {2}\right )} \arctan \left (\frac {2 x+\sqrt {1+2 \sqrt {2}}}{\sqrt {2 \sqrt {2}-1}}\right )-\frac {\log \left (x^2-\sqrt {1+2 \sqrt {2}} x+\sqrt {2}\right )}{4 \sqrt {2 \left (1+2 \sqrt {2}\right )}}+\frac {\log \left (x^2+\sqrt {1+2 \sqrt {2}} x+\sqrt {2}\right )}{4 \sqrt {2 \left (1+2 \sqrt {2}\right )}} \]

[In]

Int[(2 - x^2 + x^4)^(-1),x]

[Out]

-1/2*(Sqrt[(1 + 2*Sqrt[2])/14]*ArcTan[(Sqrt[1 + 2*Sqrt[2]] - 2*x)/Sqrt[-1 + 2*Sqrt[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 210

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

Rule 632

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 642

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 648

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 1108

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*} \text {integral}& = \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 {\text {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 {\text {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 {\arctan \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 {\arctan \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] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.46 \[ \int \frac {1}{2-x^2+x^4} \, dx=-\frac {i \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \left (-1-i \sqrt {7}\right )}}\right )}{\sqrt {\frac {7}{2} \left (-1-i \sqrt {7}\right )}}+\frac {i \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \left (-1+i \sqrt {7}\right )}}\right )}{\sqrt {\frac {7}{2} \left (-1+i \sqrt {7}\right )}} \]

[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]

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.13 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.18

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+2\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3}-\textit {\_R}}\right )}{2}\)	\(35\)
default	\(\frac {\left (\sqrt {1+2 \sqrt {2}}\, \sqrt {2}-4 \sqrt {1+2 \sqrt {2}}\right ) \ln \left (x^{2}+\sqrt {2}-x \sqrt {1+2 \sqrt {2}}\right )}{56}+\frac {\left (7 \sqrt {2}+\frac {\left (\sqrt {1+2 \sqrt {2}}\, \sqrt {2}-4 \sqrt {1+2 \sqrt {2}}\right ) \sqrt {1+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 x -\sqrt {1+2 \sqrt {2}}}{\sqrt {-1+2 \sqrt {2}}}\right )}{14 \sqrt {-1+2 \sqrt {2}}}+\frac {\left (-\sqrt {1+2 \sqrt {2}}\, \sqrt {2}+4 \sqrt {1+2 \sqrt {2}}\right ) \ln \left (x^{2}+\sqrt {2}+x \sqrt {1+2 \sqrt {2}}\right )}{56}+\frac {\left (7 \sqrt {2}-\frac {\left (-\sqrt {1+2 \sqrt {2}}\, \sqrt {2}+4 \sqrt {1+2 \sqrt {2}}\right ) \sqrt {1+2 \sqrt {2}}}{2}\right ) \arctan \left (\frac {2 x +\sqrt {1+2 \sqrt {2}}}{\sqrt {-1+2 \sqrt {2}}}\right )}{14 \sqrt {-1+2 \sqrt {2}}}\)	\(253\)

[In]

int(1/(x^4-x^2+2),x,method=_RETURNVERBOSE)

[Out]

1/2*sum(1/(2*_R^3-_R)*ln(x-_R),_R=RootOf(_Z^4-_Z^2+2))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.71 \[ \int \frac {1}{2-x^2+x^4} \, dx=\frac {1}{28} \, \sqrt {7} \sqrt {i \, \sqrt {7} - 1} \log \left ({\left (\sqrt {7} - i\right )} \sqrt {i \, \sqrt {7} - 1} + 4 \, x\right ) - \frac {1}{28} \, \sqrt {7} \sqrt {i \, \sqrt {7} - 1} \log \left (-{\left (\sqrt {7} - i\right )} \sqrt {i \, \sqrt {7} - 1} + 4 \, x\right ) + \frac {1}{28} \, \sqrt {7} \sqrt {-i \, \sqrt {7} - 1} \log \left ({\left (\sqrt {7} + i\right )} \sqrt {-i \, \sqrt {7} - 1} + 4 \, x\right ) - \frac {1}{28} \, \sqrt {7} \sqrt {-i \, \sqrt {7} - 1} \log \left (-{\left (\sqrt {7} + i\right )} \sqrt {-i \, \sqrt {7} - 1} + 4 \, x\right ) \]

[In]

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

[Out]

1/28*sqrt(7)*sqrt(I*sqrt(7) - 1)*log((sqrt(7) - I)*sqrt(I*sqrt(7) - 1) + 4*x) - 1/28*sqrt(7)*sqrt(I*sqrt(7) -
1)*log(-(sqrt(7) - I)*sqrt(I*sqrt(7) - 1) + 4*x) + 1/28*sqrt(7)*sqrt(-I*sqrt(7) - 1)*log((sqrt(7) + I)*sqrt(-I
*sqrt(7) - 1) + 4*x) - 1/28*sqrt(7)*sqrt(-I*sqrt(7) - 1)*log(-(sqrt(7) + I)*sqrt(-I*sqrt(7) - 1) + 4*x)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.12 \[ \int \frac {1}{2-x^2+x^4} \, dx=\operatorname {RootSum} {\left (1568 t^{4} + 28 t^{2} + 1, \left ( t \mapsto t \log {\left (- 112 t^{3} + 6 t + x \right )} \right )\right )} \]

[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)))

Maxima [F]

\[ \int \frac {1}{2-x^2+x^4} \, dx=\int { \frac {1}{x^{4} - x^{2} + 2} \,d x } \]

[In]

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

[Out]

integrate(1/(x^4 - x^2 + 2), x)

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.31 \[ \int \frac {1}{2-x^2+x^4} \, dx=\frac {1}{112} \, \sqrt {7} {\left (\sqrt {7} 2^{\frac {1}{4}} \sqrt {-2 \, \sqrt {2} + 8} + 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 8}\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\frac {1}{2}} \sqrt {\sqrt {2} + 4} + 2 \, x\right )}}{4 \, \sqrt {-\frac {1}{8} \, \sqrt {2} + \frac {1}{2}}}\right ) + \frac {1}{112} \, \sqrt {7} {\left (\sqrt {7} 2^{\frac {1}{4}} \sqrt {-2 \, \sqrt {2} + 8} + 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 8}\right )} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\frac {1}{2}} \sqrt {\sqrt {2} + 4} - 2 \, x\right )}}{4 \, \sqrt {-\frac {1}{8} \, \sqrt {2} + \frac {1}{2}}}\right ) + \frac {1}{224} \, \sqrt {7} {\left (\sqrt {7} 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 8} - 2^{\frac {1}{4}} \sqrt {-2 \, \sqrt {2} + 8}\right )} \log \left (2^{\frac {1}{4}} \sqrt {\frac {1}{2}} x \sqrt {\sqrt {2} + 4} + x^{2} + \sqrt {2}\right ) - \frac {1}{224} \, \sqrt {7} {\left (\sqrt {7} 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 8} - 2^{\frac {1}{4}} \sqrt {-2 \, \sqrt {2} + 8}\right )} \log \left (-2^{\frac {1}{4}} \sqrt {\frac {1}{2}} x \sqrt {\sqrt {2} + 4} + x^{2} + \sqrt {2}\right ) \]

[In]

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

[Out]

1/112*sqrt(7)*(sqrt(7)*2^(1/4)*sqrt(-2*sqrt(2) + 8) + 2^(1/4)*sqrt(2*sqrt(2) + 8))*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)) + 1/112*sqrt(7)*(sqrt(7)*2^(1/4)*sqrt(-2*sqrt(2)
+ 8) + 2^(1/4)*sqrt(2*sqrt(2) + 8))*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)*(sqrt(7)*2^(1/4)*sqrt(2*sqrt(2) + 8) - 2^(1/4)*sqrt(-2*sqrt(2) + 8))*log(2^(1
/4)*sqrt(1/2)*x*sqrt(sqrt(2) + 4) + x^2 + sqrt(2)) - 1/224*sqrt(7)*(sqrt(7)*2^(1/4)*sqrt(2*sqrt(2) + 8) - 2^(1
/4)*sqrt(-2*sqrt(2) + 8))*log(-2^(1/4)*sqrt(1/2)*x*sqrt(sqrt(2) + 4) + x^2 + sqrt(2))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.67 \[ \int \frac {1}{2-x^2+x^4} \, dx=\frac {\mathrm {atan}\left (\frac {x\,\sqrt {-7-\sqrt {7}\,7{}\mathrm {i}}\,1{}\mathrm {i}}{4\,\left (\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )}+\frac {\sqrt {7}\,x\,\sqrt {-7-\sqrt {7}\,7{}\mathrm {i}}}{28\,\left (\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )}\right )\,\sqrt {-7-\sqrt {7}\,7{}\mathrm {i}}\,1{}\mathrm {i}}{14}+\frac {\sqrt {7}\,\mathrm {atan}\left (\frac {x\,\sqrt {-1+\sqrt {7}\,1{}\mathrm {i}}}{4\,\left (-\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )}-\frac {\sqrt {7}\,x\,\sqrt {-1+\sqrt {7}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{4\,\left (-\frac {1}{2}+\frac {\sqrt {7}\,1{}\mathrm {i}}{2}\right )}\right )\,\sqrt {-1+\sqrt {7}\,1{}\mathrm {i}}\,1{}\mathrm {i}}{14} \]

[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