\(\int \frac {1-2 x^2}{1+4 x^4} \, dx\) [61]

Optimal result

Integrand size = 17, antiderivative size = 31 \[ \int \frac {1-2 x^2}{1+4 x^4} \, dx=-\frac {1}{4} \log \left (1-2 x+2 x^2\right )+\frac {1}{4} \log \left (1+2 x+2 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1179, 642} \[ \int \frac {1-2 x^2}{1+4 x^4} \, dx=\frac {1}{4} \log \left (2 x^2+2 x+1\right )-\frac {1}{4} \log \left (2 x^2-2 x+1\right ) \]

[In]

Int[(1 - 2*x^2)/(1 + 4*x^4),x]

[Out]

-1/4*Log[1 - 2*x + 2*x^2] + Log[1 + 2*x + 2*x^2]/4

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x^2}{1+4 x^4} \, dx=-\frac {1}{4} \log \left (1-2 x+2 x^2\right )+\frac {1}{4} \log \left (1+2 x+2 x^2\right ) \]

[In]

Integrate[(1 - 2*x^2)/(1 + 4*x^4),x]

[Out]

-1/4*Log[1 - 2*x + 2*x^2] + Log[1 + 2*x + 2*x^2]/4

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(-\frac {\ln \left (x^{2}-x +\frac {1}{2}\right )}{4}+\frac {\ln \left (x^{2}+x +\frac {1}{2}\right )}{4}\)	\(22\)
default	\(-\frac {\ln \left (2 x^{2}-2 x +1\right )}{4}+\frac {\ln \left (2 x^{2}+2 x +1\right )}{4}\)	\(28\)
norman	\(-\frac {\ln \left (2 x^{2}-2 x +1\right )}{4}+\frac {\ln \left (2 x^{2}+2 x +1\right )}{4}\)	\(28\)
risch	\(-\frac {\ln \left (2 x^{2}-2 x +1\right )}{4}+\frac {\ln \left (2 x^{2}+2 x +1\right )}{4}\)	\(28\)
meijerg	\(-\frac {\sqrt {2}\, \left (\frac {x^{3} \sqrt {2}\, \ln \left (1-2 \left (x^{4}\right )^{\frac {1}{4}}+2 \sqrt {x^{4}}\right )}{2 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\left (x^{4}\right )^{\frac {1}{4}}}{1-\left (x^{4}\right )^{\frac {1}{4}}}\right )}{\left (x^{4}\right )^{\frac {3}{4}}}-\frac {x^{3} \sqrt {2}\, \ln \left (1+2 \left (x^{4}\right )^{\frac {1}{4}}+2 \sqrt {x^{4}}\right )}{2 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\left (x^{4}\right )^{\frac {1}{4}}}{1+\left (x^{4}\right )^{\frac {1}{4}}}\right )}{\left (x^{4}\right )^{\frac {3}{4}}}\right )}{8}+\frac {\sqrt {2}\, \left (-\frac {x \sqrt {2}\, \ln \left (1-2 \left (x^{4}\right )^{\frac {1}{4}}+2 \sqrt {x^{4}}\right )}{2 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\left (x^{4}\right )^{\frac {1}{4}}}{1-\left (x^{4}\right )^{\frac {1}{4}}}\right )}{\left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \ln \left (1+2 \left (x^{4}\right )^{\frac {1}{4}}+2 \sqrt {x^{4}}\right )}{2 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\left (x^{4}\right )^{\frac {1}{4}}}{1+\left (x^{4}\right )^{\frac {1}{4}}}\right )}{\left (x^{4}\right )^{\frac {1}{4}}}\right )}{8}\)	\(242\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1-2 x^2}{1+4 x^4} \, dx=\frac {1}{4} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - 2 \, x + 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {1-2 x^2}{1+4 x^4} \, dx=- \frac {\log {\left (x^{2} - x + \frac {1}{2} \right )}}{4} + \frac {\log {\left (x^{2} + x + \frac {1}{2} \right )}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1-2 x^2}{1+4 x^4} \, dx=\frac {1}{4} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - 2 \, x + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {1-2 x^2}{1+4 x^4} \, dx=\frac {1}{4} \, \log \left (x^{2} + \sqrt {2} \left (\frac {1}{4}\right )^{\frac {1}{4}} x + \frac {1}{2}\right ) - \frac {1}{4} \, \log \left (x^{2} - \sqrt {2} \left (\frac {1}{4}\right )^{\frac {1}{4}} x + \frac {1}{2}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.48 \[ \int \frac {1-2 x^2}{1+4 x^4} \, dx=\frac {\mathrm {atanh}\left (\frac {2\,x}{2\,x^2+1}\right )}{2} \]

[In]

int(-(2*x^2 - 1)/(4*x^4 + 1),x)

[Out]

atanh((2*x)/(2*x^2 + 1))/2