\(\int \frac {1-x^2}{1+x^4} \, dx\) [86]

Optimal result

Integrand size = 15, antiderivative size = 46 \[ \int \frac {1-x^2}{1+x^4} \, dx=-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{2 \sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1179, 642} \[ \int \frac {1-x^2}{1+x^4} \, dx=\frac {\log \left (x^2+\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{2 \sqrt {2}} \]

[In]

Int[(1 - x^2)/(1 + x^4),x]

[Out]

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

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 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = -\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{2 \sqrt {2}} \\ & = -\frac {\log \left (1-\sqrt {2} x+x^2\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{2 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \frac {1-x^2}{1+x^4} \, dx=\frac {-\log \left (-1+\sqrt {2} x-x^2\right )+\log \left (1+\sqrt {2} x+x^2\right )}{2 \sqrt {2}} \]

[In]

Integrate[(1 - x^2)/(1 + x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\ln \left (1+x^{2}-x \sqrt {2}\right ) \sqrt {2}}{4}+\frac {\ln \left (1+x^{2}+x \sqrt {2}\right ) \sqrt {2}}{4}\)	\(35\)
default	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+x \sqrt {2}}{1+x^{2}-x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{8}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-x \sqrt {2}}{1+x^{2}+x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{8}\)	\(104\)
meijerg	\(-\frac {x^{3} \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {x \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}\)	\(268\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int \frac {1-x^2}{1+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (\frac {x^{4} + 4 \, x^{2} + 2 \, \sqrt {2} {\left (x^{3} + x\right )} + 1}{x^{4} + 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1-x^2}{1+x^4} \, dx=- \frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{4} + \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{4} \]

[In]

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

[Out]

-sqrt(2)*log(x**2 - sqrt(2)*x + 1)/4 + sqrt(2)*log(x**2 + sqrt(2)*x + 1)/4

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int \frac {1-x^2}{1+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.74 \[ \int \frac {1-x^2}{1+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.39 \[ \int \frac {1-x^2}{1+x^4} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,x}{x^2+1}\right )}{2} \]

[In]

int(-(x^2 - 1)/(x^4 + 1),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*x)/(x^2 + 1)))/2