\(\int \frac {1}{1-x^2+x^4} \, dx\) [14]

Optimal result

Integrand size = 12, antiderivative size = 74 \[ \int \frac {1}{1-x^2+x^4} \, dx=-\frac {1}{2} \arctan \left (\sqrt {3}-2 x\right )+\frac {1}{2} \arctan \left (\sqrt {3}+2 x\right )-\frac {\log \left (1-\sqrt {3} x+x^2\right )}{4 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}} \]

[Out]

1/2*arctan(2*x-3^(1/2))+1/2*arctan(2*x+3^(1/2))-1/12*ln(1+x^2-x*3^(1/2))*3^(1/2)+1/12*ln(1+x^2+x*3^(1/2))*3^(1
/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, 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} \[ \int \frac {1}{1-x^2+x^4} \, dx=-\frac {1}{2} \arctan \left (\sqrt {3}-2 x\right )+\frac {1}{2} \arctan \left (2 x+\sqrt {3}\right )-\frac {\log \left (x^2-\sqrt {3} x+1\right )}{4 \sqrt {3}}+\frac {\log \left (x^2+\sqrt {3} x+1\right )}{4 \sqrt {3}} \]

[In]

Int[(1 - x^2 + x^4)^(-1),x]

[Out]

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04 \[ \int \frac {1}{1-x^2+x^4} \, dx=\frac {i \left (\sqrt {-1-i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )-\sqrt {-1+i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )\right )}{\sqrt {6}} \]

[In]

Integrate[(1 - x^2 + x^4)^(-1),x]

[Out]

(I*(Sqrt[-1 - I*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x)/2] - Sqrt[-1 + I*Sqrt[3]]*ArcTan[((1 + I*Sqrt[3])*x)/2]))/
Sqrt[6]

Maple [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.47

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3}-\textit {\_R}}\right )}{2}\)	\(35\)
default	\(\frac {\arctan \left (2 x -\sqrt {3}\right )}{2}+\frac {\arctan \left (2 x +\sqrt {3}\right )}{2}-\frac {\ln \left (1+x^{2}-x \sqrt {3}\right ) \sqrt {3}}{12}+\frac {\ln \left (1+x^{2}+x \sqrt {3}\right ) \sqrt {3}}{12}\)	\(57\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.12 \[ \int \frac {1}{1-x^2+x^4} \, dx=-\frac {1}{12} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (i \, \sqrt {3} - 3\right )} + 12 \, x\right ) + \frac {1}{12} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} + 3\right )} + 12 \, x\right ) + \frac {1}{12} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (\sqrt {6} {\left (i \, \sqrt {3} + 3\right )} \sqrt {-i \, \sqrt {3} - 1} + 12 \, x\right ) - \frac {1}{12} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {-i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} - 3\right )} + 12 \, x\right ) \]

[In]

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

[Out]

-1/12*sqrt(6)*sqrt(I*sqrt(3) - 1)*log(sqrt(6)*sqrt(I*sqrt(3) - 1)*(I*sqrt(3) - 3) + 12*x) + 1/12*sqrt(6)*sqrt(
I*sqrt(3) - 1)*log(sqrt(6)*sqrt(I*sqrt(3) - 1)*(-I*sqrt(3) + 3) + 12*x) + 1/12*sqrt(6)*sqrt(-I*sqrt(3) - 1)*lo
g(sqrt(6)*(I*sqrt(3) + 3)*sqrt(-I*sqrt(3) - 1) + 12*x) - 1/12*sqrt(6)*sqrt(-I*sqrt(3) - 1)*log(sqrt(6)*sqrt(-I
*sqrt(3) - 1)*(-I*sqrt(3) - 3) + 12*x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85 \[ \int \frac {1}{1-x^2+x^4} \, dx=- \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} + \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} + \frac {\operatorname {atan}{\left (2 x - \sqrt {3} \right )}}{2} + \frac {\operatorname {atan}{\left (2 x + \sqrt {3} \right )}}{2} \]

[In]

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

[Out]

-sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 + sqrt(3)*log(x**2 + sqrt(3)*x + 1)/12 + atan(2*x - sqrt(3))/2 + atan(2*
x + sqrt(3))/2

Maxima [F]

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

[In]

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

[Out]

integrate(1/(x^4 - x^2 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76 \[ \int \frac {1}{1-x^2+x^4} \, dx=\frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{2} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{2} \, \arctan \left (2 \, x - \sqrt {3}\right ) \]

[In]

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

[Out]

1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) - 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + 1/2*arctan(2*x + sqrt(3)) + 1/
2*arctan(2*x - sqrt(3))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64 \[ \int \frac {1}{1-x^2+x^4} \, dx=\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \]

[In]

int(1/(x^4 - x^2 + 1),x)

[Out]

atan((2*x)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/6 - 1/2) + atan((2*x)/(3^(1/2)*1i + 1))*((3^(1/2)*1i)/6 + 1/2)