\(\int \frac {2+x}{4-5 x^2+x^4} \, dx\) [79]

Optimal result

Integrand size = 16, antiderivative size = 29 \[ \int \frac {2+x}{4-5 x^2+x^4} \, dx=-\frac {1}{2} \log (1-x)+\frac {1}{3} \log (2-x)+\frac {1}{6} \log (1+x) \]

[Out]

-1/2*ln(1-x)+1/3*ln(2-x)+1/6*ln(1+x)

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1600, 2083} \[ \int \frac {2+x}{4-5 x^2+x^4} \, dx=-\frac {1}{2} \log (1-x)+\frac {1}{3} \log (2-x)+\frac {1}{6} \log (x+1) \]

[In]

Int[(2 + x)/(4 - 5*x^2 + x^4),x]

[Out]

-1/2*Log[1 - x] + Log[2 - x]/3 + Log[1 + x]/6

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 2083

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P]}, Int[ExpandIntegrand[u^p, x], x] /;  !SumQ[NonfreeFactors[u,
x]]] /; PolyQ[P, x] && ILtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{2-x-2 x^2+x^3} \, dx \\ & = \int \left (\frac {1}{3 (-2+x)}-\frac {1}{2 (-1+x)}+\frac {1}{6 (1+x)}\right ) \, dx \\ & = -\frac {1}{2} \log (1-x)+\frac {1}{3} \log (2-x)+\frac {1}{6} \log (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {2+x}{4-5 x^2+x^4} \, dx=-\frac {1}{2} \log (1-x)+\frac {1}{3} \log (2-x)+\frac {1}{6} \log (1+x) \]

[In]

Integrate[(2 + x)/(4 - 5*x^2 + x^4),x]

[Out]

-1/2*Log[1 - x] + Log[2 - x]/3 + Log[1 + x]/6

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x -2\right )}{3}\)	\(20\)
norman	\(\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x -2\right )}{3}\)	\(20\)
risch	\(\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x -2\right )}{3}\)	\(20\)
parallelrisch	\(\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x -2\right )}{3}\)	\(20\)

[In]

int((x+2)/(x^4-5*x^2+4),x,method=_RETURNVERBOSE)

[Out]

1/6*ln(x+1)-1/2*ln(x-1)+1/3*ln(x-2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {2+x}{4-5 x^2+x^4} \, dx=\frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{2} \, \log \left (x - 1\right ) + \frac {1}{3} \, \log \left (x - 2\right ) \]

[In]

integrate((2+x)/(x^4-5*x^2+4),x, algorithm="fricas")

[Out]

1/6*log(x + 1) - 1/2*log(x - 1) + 1/3*log(x - 2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {2+x}{4-5 x^2+x^4} \, dx=\frac {\log {\left (x - 2 \right )}}{3} - \frac {\log {\left (x - 1 \right )}}{2} + \frac {\log {\left (x + 1 \right )}}{6} \]

[In]

integrate((2+x)/(x**4-5*x**2+4),x)

[Out]

log(x - 2)/3 - log(x - 1)/2 + log(x + 1)/6

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {2+x}{4-5 x^2+x^4} \, dx=\frac {1}{6} \, \log \left (x + 1\right ) - \frac {1}{2} \, \log \left (x - 1\right ) + \frac {1}{3} \, \log \left (x - 2\right ) \]

[In]

integrate((2+x)/(x^4-5*x^2+4),x, algorithm="maxima")

[Out]

1/6*log(x + 1) - 1/2*log(x - 1) + 1/3*log(x - 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {2+x}{4-5 x^2+x^4} \, dx=\frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{3} \, \log \left ({\left | x - 2 \right |}\right ) \]

[In]

integrate((2+x)/(x^4-5*x^2+4),x, algorithm="giac")

[Out]

1/6*log(abs(x + 1)) - 1/2*log(abs(x - 1)) + 1/3*log(abs(x - 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {2+x}{4-5 x^2+x^4} \, dx=\frac {\ln \left (x+1\right )}{6}-\frac {\ln \left (x-1\right )}{2}+\frac {\ln \left (x-2\right )}{3} \]

[In]

int((x + 2)/(x^4 - 5*x^2 + 4),x)

[Out]

log(x + 1)/6 - log(x - 1)/2 + log(x - 2)/3