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

Optimal. Leaf size=68 \[ \frac{1}{12 (1-x)}+\frac{1}{36 (2-x)}-\frac{1}{36 (x+1)}+\frac{1}{18} \log (1-x)-\frac{35}{432} \log (2-x)+\frac{1}{54} \log (x+1)+\frac{1}{144} \log (x+2) \]

[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

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Rubi [A]  time = 0.057872, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1586, 2074} \[ \frac{1}{12 (1-x)}+\frac{1}{36 (2-x)}-\frac{1}{36 (x+1)}+\frac{1}{18} \log (1-x)-\frac{35}{432} \log (2-x)+\frac{1}{54} \log (x+1)+\frac{1}{144} \log (x+2) \]

Antiderivative was successfully verified.

[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 1586

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 2074

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*} \int \frac{2+x}{\left (4-5 x^2+x^4\right )^2} \, dx &=\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]  time = 0.0288866, size = 60, normalized size = 0.88 \[ \frac{1}{432} \left (\frac{12 \left (-5 x^2+6 x+5\right )}{x^3-2 x^2-x+2}+24 \log (1-x)-35 \log (2-x)+8 \log (x+1)+3 \log (x+2)\right ) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.013, size = 47, normalized size = 0.7 \begin{align*}{\frac{\ln \left ( 2+x \right ) }{144}}-{\frac{1}{36+36\,x}}+{\frac{\ln \left ( 1+x \right ) }{54}}-{\frac{1}{36\,x-72}}-{\frac{35\,\ln \left ( x-2 \right ) }{432}}-{\frac{1}{12\,x-12}}+{\frac{\ln \left ( x-1 \right ) }{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.95761, size = 70, normalized size = 1.03 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 1.73935, size = 271, normalized size = 3.99 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.279434, size = 53, normalized size = 0.78 \begin{align*} - \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.10012, size = 76, normalized size = 1.12 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))