\(\int \frac {2+x}{(4-5 x^2+x^4)^2} \, dx\) [97]

Optimal result

Integrand size = 16, antiderivative size = 68 \[ \int \frac {2+x}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{12 (1-x)}+\frac {1}{36 (2-x)}-\frac {1}{36 (1+x)}+\frac {1}{18} \log (1-x)-\frac {35}{432} \log (2-x)+\frac {1}{54} \log (1+x)+\frac {1}{144} \log (2+x) \]

[Out]

1/12/(1-x)+1/36/(2-x)-1/36/(1+x)+1/18*ln(1-x)-35/432*ln(2-x)+1/54*ln(1+x)+1/144*ln(2+x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 68, 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, 2099} \[ \int \frac {2+x}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{12 (1-x)}+\frac {1}{36 (2-x)}-\frac {1}{36 (x+1)}+\frac {1}{18} \log (1-x)-\frac {35}{432} \log (2-x)+\frac {1}{54} \log (x+1)+\frac {1}{144} \log (x+2) \]

[In]

Int[(2 + x)/(4 - 5*x^2 + x^4)^2,x]

[Out]

1/(12*(1 - x)) + 1/(36*(2 - x)) - 1/(36*(1 + x)) + Log[1 - x]/18 - (35*Log[2 - x])/432 + Log[1 + x]/54 + Log[2
+ x]/144

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 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(2+x) \left (2-x-2 x^2+x^3\right )^2} \, dx \\ & = \int \left (\frac {1}{36 (-2+x)^2}-\frac {35}{432 (-2+x)}+\frac {1}{12 (-1+x)^2}+\frac {1}{18 (-1+x)}+\frac {1}{36 (1+x)^2}+\frac {1}{54 (1+x)}+\frac {1}{144 (2+x)}\right ) \, dx \\ & = \frac {1}{12 (1-x)}+\frac {1}{36 (2-x)}-\frac {1}{36 (1+x)}+\frac {1}{18} \log (1-x)-\frac {35}{432} \log (2-x)+\frac {1}{54} \log (1+x)+\frac {1}{144} \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \frac {2+x}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{432} \left (\frac {12 \left (5+6 x-5 x^2\right )}{2-x-2 x^2+x^3}+24 \log (1-x)-35 \log (2-x)+8 \log (1+x)+3 \log (2+x)\right ) \]

[In]

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

[Out]

((12*(5 + 6*x - 5*x^2))/(2 - x - 2*x^2 + x^3) + 24*Log[1 - x] - 35*Log[2 - x] + 8*Log[1 + x] + 3*Log[2 + x])/4
32

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {\ln \left (x +2\right )}{144}-\frac {1}{36 \left (x +1\right )}+\frac {\ln \left (x +1\right )}{54}-\frac {1}{12 \left (x -1\right )}+\frac {\ln \left (x -1\right )}{18}-\frac {1}{36 \left (x -2\right )}-\frac {35 \ln \left (x -2\right )}{432}\)	\(47\)
risch	\(\frac {-\frac {5}{36} x^{2}+\frac {1}{6} x +\frac {5}{36}}{x^{3}-2 x^{2}-x +2}-\frac {35 \ln \left (x -2\right )}{432}+\frac {\ln \left (x -1\right )}{18}+\frac {\ln \left (x +1\right )}{54}+\frac {\ln \left (x +2\right )}{144}\)	\(52\)
norman	\(\frac {-\frac {1}{9} x^{2}+\frac {17}{36} x -\frac {5}{36} x^{3}+\frac {5}{18}}{x^{4}-5 x^{2}+4}-\frac {35 \ln \left (x -2\right )}{432}+\frac {\ln \left (x -1\right )}{18}+\frac {\ln \left (x +1\right )}{54}+\frac {\ln \left (x +2\right )}{144}\)	\(54\)
parallelrisch	\(-\frac {35 \ln \left (x -2\right ) x^{3}-24 \ln \left (x -1\right ) x^{3}-8 \ln \left (x +1\right ) x^{3}-3 \ln \left (x +2\right ) x^{3}-60-70 \ln \left (x -2\right ) x^{2}+48 \ln \left (x -1\right ) x^{2}+16 \ln \left (x +1\right ) x^{2}+6 \ln \left (x +2\right ) x^{2}-35 \ln \left (x -2\right ) x +24 \ln \left (x -1\right ) x +8 \ln \left (x +1\right ) x +3 \ln \left (x +2\right ) x +60 x^{2}+70 \ln \left (x -2\right )-48 \ln \left (x -1\right )-16 \ln \left (x +1\right )-6 \ln \left (x +2\right )-72 x}{432 \left (x^{3}-2 x^{2}-x +2\right )}\)	\(152\)

[In]

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

[Out]

1/144*ln(x+2)-1/36/(x+1)+1/54*ln(x+1)-1/12/(x-1)+1/18*ln(x-1)-1/36/(x-2)-35/432*ln(x-2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (50) = 100\).

Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.51 \[ \int \frac {2+x}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {60 \, x^{2} - 3 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )} \log \left (x + 2\right ) - 8 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )} \log \left (x + 1\right ) - 24 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )} \log \left (x - 1\right ) + 35 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )} \log \left (x - 2\right ) - 72 \, x - 60}{432 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )}} \]

[In]

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

[Out]

-1/432*(60*x^2 - 3*(x^3 - 2*x^2 - x + 2)*log(x + 2) - 8*(x^3 - 2*x^2 - x + 2)*log(x + 1) - 24*(x^3 - 2*x^2 - x
+ 2)*log(x - 1) + 35*(x^3 - 2*x^2 - x + 2)*log(x - 2) - 72*x - 60)/(x^3 - 2*x^2 - x + 2)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {2+x}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {- 5 x^{2} + 6 x + 5}{36 x^{3} - 72 x^{2} - 36 x + 72} - \frac {35 \log {\left (x - 2 \right )}}{432} + \frac {\log {\left (x - 1 \right )}}{18} + \frac {\log {\left (x + 1 \right )}}{54} + \frac {\log {\left (x + 2 \right )}}{144} \]

[In]

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

[Out]

(-5*x**2 + 6*x + 5)/(36*x**3 - 72*x**2 - 36*x + 72) - 35*log(x - 2)/432 + log(x - 1)/18 + log(x + 1)/54 + log(
x + 2)/144

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {2+x}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {5 \, x^{2} - 6 \, x - 5}{36 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )}} + \frac {1}{144} \, \log \left (x + 2\right ) + \frac {1}{54} \, \log \left (x + 1\right ) + \frac {1}{18} \, \log \left (x - 1\right ) - \frac {35}{432} \, \log \left (x - 2\right ) \]

[In]

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

[Out]

-1/36*(5*x^2 - 6*x - 5)/(x^3 - 2*x^2 - x + 2) + 1/144*log(x + 2) + 1/54*log(x + 1) + 1/18*log(x - 1) - 35/432*
log(x - 2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \frac {2+x}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {5 \, x^{2} - 6 \, x - 5}{36 \, {\left (x + 1\right )} {\left (x - 1\right )} {\left (x - 2\right )}} + \frac {1}{144} \, \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{54} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{18} \, \log \left ({\left | x - 1 \right |}\right ) - \frac {35}{432} \, \log \left ({\left | x - 2 \right |}\right ) \]

[In]

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

[Out]

-1/36*(5*x^2 - 6*x - 5)/((x + 1)*(x - 1)*(x - 2)) + 1/144*log(abs(x + 2)) + 1/54*log(abs(x + 1)) + 1/18*log(ab
s(x - 1)) - 35/432*log(abs(x - 2))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {2+x}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {\ln \left (x-1\right )}{18}+\frac {\ln \left (x+1\right )}{54}-\frac {35\,\ln \left (x-2\right )}{432}+\frac {\ln \left (x+2\right )}{144}-\frac {-\frac {5\,x^2}{36}+\frac {x}{6}+\frac {5}{36}}{-x^3+2\,x^2+x-2} \]

[In]

int((x + 2)/(x^4 - 5*x^2 + 4)^2,x)

[Out]

log(x - 1)/18 + log(x + 1)/54 - (35*log(x - 2))/432 + log(x + 2)/144 - (x/6 - (5*x^2)/36 + 5/36)/(x + 2*x^2 -
x^3 - 2)