\(\int \frac {-x+2 x^3}{1-x^2+x^4} \, dx\) [196]

Optimal result

Integrand size = 22, antiderivative size = 15 \[ \int \frac {-x+2 x^3}{1-x^2+x^4} \, dx=\frac {1}{2} \log \left (1-x^2+x^4\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1601} \[ \int \frac {-x+2 x^3}{1-x^2+x^4} \, dx=\frac {1}{2} \log \left (x^4-x^2+1\right ) \]

[In]

Int[(-x + 2*x^3)/(1 - x^2 + x^4),x]

[Out]

Log[1 - x^2 + x^4]/2

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \log \left (1-x^2+x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-x+2 x^3}{1-x^2+x^4} \, dx=\frac {1}{2} \log \left (1-x^2+x^4\right ) \]

[In]

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

[Out]

Log[1 - x^2 + x^4]/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\ln \left (x^{4}-x^{2}+1\right )}{2}\)	\(14\)
norman	\(\frac {\ln \left (x^{4}-x^{2}+1\right )}{2}\)	\(14\)
risch	\(\frac {\ln \left (x^{4}-x^{2}+1\right )}{2}\)	\(14\)
parallelrisch	\(\frac {\ln \left (x^{4}-x^{2}+1\right )}{2}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {-x+2 x^3}{1-x^2+x^4} \, dx=\frac {1}{2} \, \log \left (x^{4} - x^{2} + 1\right ) \]

[In]

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

[Out]

1/2*log(x^4 - x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {-x+2 x^3}{1-x^2+x^4} \, dx=\frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{2} \]

[In]

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

[Out]

log(x**4 - x**2 + 1)/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {-x+2 x^3}{1-x^2+x^4} \, dx=\frac {1}{2} \, \log \left (x^{4} - x^{2} + 1\right ) \]

[In]

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

[Out]

1/2*log(x^4 - x^2 + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {-x+2 x^3}{1-x^2+x^4} \, dx=\frac {1}{2} \, \log \left (x^{4} - x^{2} + 1\right ) \]

[In]

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

[Out]

1/2*log(x^4 - x^2 + 1)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {-x+2 x^3}{1-x^2+x^4} \, dx=\frac {\ln \left (x^4-x^2+1\right )}{2} \]

[In]

int(-(x - 2*x^3)/(x^4 - x^2 + 1),x)

[Out]

log(x^4 - x^2 + 1)/2