\(\int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx\) [58]

Optimal result

Integrand size = 22, antiderivative size = 29 \[ \int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx=-\frac {1}{2} \log \left (1-x+2 x^2\right )+\frac {1}{2} \log \left (1+x+2 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1178, 642} \[ \int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {1}{2} \log \left (2 x^2+x+1\right )-\frac {1}{2} \log \left (2 x^2-x+1\right ) \]

[In]

Int[(1 - 2*x^2)/(1 + 3*x^2 + 4*x^4),x]

[Out]

-1/2*Log[1 - x + 2*x^2] + Log[1 + x + 2*x^2]/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 1178

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e) - b/c, 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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx=-\frac {1}{2} \log \left (1-x+2 x^2\right )+\frac {1}{2} \log \left (1+x+2 x^2\right ) \]

[In]

Integrate[(1 - 2*x^2)/(1 + 3*x^2 + 4*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(-\frac {\ln \left (x^{2}-\frac {1}{2} x +\frac {1}{2}\right )}{2}+\frac {\ln \left (x^{2}+\frac {1}{2} x +\frac {1}{2}\right )}{2}\)	\(24\)
default	\(-\frac {\ln \left (2 x^{2}-x +1\right )}{2}+\frac {\ln \left (2 x^{2}+x +1\right )}{2}\)	\(26\)
norman	\(-\frac {\ln \left (2 x^{2}-x +1\right )}{2}+\frac {\ln \left (2 x^{2}+x +1\right )}{2}\)	\(26\)
risch	\(-\frac {\ln \left (2 x^{2}-x +1\right )}{2}+\frac {\ln \left (2 x^{2}+x +1\right )}{2}\)	\(26\)

[In]

int((-2*x^2+1)/(4*x^4+3*x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {1}{2} \, \log \left (2 \, x^{2} + x + 1\right ) - \frac {1}{2} \, \log \left (2 \, x^{2} - x + 1\right ) \]

[In]

integrate((-2*x^2+1)/(4*x^4+3*x^2+1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx=- \frac {\log {\left (x^{2} - \frac {x}{2} + \frac {1}{2} \right )}}{2} + \frac {\log {\left (x^{2} + \frac {x}{2} + \frac {1}{2} \right )}}{2} \]

[In]

integrate((-2*x**2+1)/(4*x**4+3*x**2+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {1}{2} \, \log \left (2 \, x^{2} + x + 1\right ) - \frac {1}{2} \, \log \left (2 \, x^{2} - x + 1\right ) \]

[In]

integrate((-2*x^2+1)/(4*x^4+3*x^2+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {1}{2} \, \log \left (2 \, x^{2} + x + 1\right ) - \frac {1}{2} \, \log \left (2 \, x^{2} - x + 1\right ) \]

[In]

integrate((-2*x^2+1)/(4*x^4+3*x^2+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.41 \[ \int \frac {1-2 x^2}{1+3 x^2+4 x^4} \, dx=\mathrm {atanh}\left (\frac {x}{2\,x^2+1}\right ) \]

[In]

int(-(2*x^2 - 1)/(3*x^2 + 4*x^4 + 1),x)

[Out]

atanh(x/(2*x^2 + 1))