3.91 \(\int \frac {2-3 x+x^2}{(4-5 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=56 \[ -\frac {3 x+5}{12 \left (x^2+3 x+2\right )}-\frac {1}{36} \log (1-x)+\frac {1}{144} \log (2-x)-\frac {7}{36} \log (x+1)+\frac {31}{144} \log (x+2) \]

[Out]

1/12*(-5-3*x)/(x^2+3*x+2)-1/36*ln(1-x)+1/144*ln(2-x)-7/36*ln(1+x)+31/144*ln(2+x)

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Rubi [A]  time = 0.06, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1586, 974, 1072, 632, 31} \[ -\frac {3 x+5}{12 \left (x^2+3 x+2\right )}-\frac {1}{36} \log (1-x)+\frac {1}{144} \log (2-x)-\frac {7}{36} \log (x+1)+\frac {31}{144} \log (x+2) \]

Antiderivative was successfully verified.

[In]

Int[(2 - 3*x + x^2)/(4 - 5*x^2 + x^4)^2,x]

[Out]

-(5 + 3*x)/(12*(2 + 3*x + x^2)) - Log[1 - x]/36 + Log[2 - x]/144 - (7*Log[1 + x])/36 + (31*Log[2 + x])/144

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 974

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((2*a
*c^2*e - b^2*c*e + b^3*f + b*c*(c*d - 3*a*f) + c*(2*c^2*d + b^2*f - c*(b*e + 2*a*f))*x)*(a + b*x + c*x^2)^(p +
1)*(d + e*x + f*x^2)^(q + 1))/((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), x] - Dist[1/
((b^2 - 4*a*c)*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + e*x + f*
x^2)^q*Simp[2*c*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*(p + 1) - (2*c^2*d + b^2*f - c*(b*e + 2*a*f))*(a*f*(
p + 1) - c*d*(p + 2)) - e*(b^2*c*e - 2*a*c^2*e - b^3*f - b*c*(c*d - 3*a*f))*(p + q + 2) + (2*f*(2*a*c^2*e - b^
2*c*e + b^3*f + b*c*(c*d - 3*a*f))*(p + q + 2) - (2*c^2*d + b^2*f - c*(b*e + 2*a*f))*(b*f*(p + 1) - c*e*(2*p +
q + 4)))*x + c*f*(2*c^2*d + b^2*f - c*(b*e + 2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e,
f, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && LtQ[p, -1] && NeQ[(c*d - a*f)^2 - (b*d - a*e)*(c*e
- b*f), 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]

Rule 1072

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)
), x_Symbol] :> With[{q = c^2*d^2 - b*c*d*e + a*c*e^2 + b^2*d*f - 2*a*c*d*f - a*b*e*f + a^2*f^2}, Dist[1/q, In
t[(A*c^2*d - a*c*C*d - A*b*c*e + a*B*c*e + A*b^2*f - a*b*B*f - a*A*c*f + a^2*C*f + c*(B*c*d - b*C*d - A*c*e +
a*C*e + A*b*f - a*B*f)*x)/(a + b*x + c*x^2), x], x] + Dist[1/q, Int[(c*C*d^2 - B*c*d*e + A*c*e^2 + b*B*d*f - A
*c*d*f - a*C*d*f - A*b*e*f + a*A*f^2 - f*(B*c*d - b*C*d - A*c*e + a*C*e + A*b*f - a*B*f)*x)/(d + e*x + f*x^2),
x], x] /; NeQ[q, 0]] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0]

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]

Rubi steps

\begin {align*} \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx &=\int \frac {1}{\left (2-3 x+x^2\right ) \left (2+3 x+x^2\right )^2} \, dx\\ &=-\frac {5+3 x}{12 \left (2+3 x+x^2\right )}+\frac {1}{72} \int \frac {-18+48 x-18 x^2}{\left (2-3 x+x^2\right ) \left (2+3 x+x^2\right )} \, dx\\ &=-\frac {5+3 x}{12 \left (2+3 x+x^2\right )}+\frac {\int \frac {252-108 x}{2-3 x+x^2} \, dx}{5184}+\frac {\int \frac {-900+108 x}{2+3 x+x^2} \, dx}{5184}\\ &=-\frac {5+3 x}{12 \left (2+3 x+x^2\right )}+\frac {1}{144} \int \frac {1}{-2+x} \, dx-\frac {1}{36} \int \frac {1}{-1+x} \, dx-\frac {7}{36} \int \frac {1}{1+x} \, dx+\frac {31}{144} \int \frac {1}{2+x} \, dx\\ &=-\frac {5+3 x}{12 \left (2+3 x+x^2\right )}-\frac {1}{36} \log (1-x)+\frac {1}{144} \log (2-x)-\frac {7}{36} \log (1+x)+\frac {31}{144} \log (2+x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 48, normalized size = 0.86 \[ \frac {1}{144} \left (-\frac {12 (3 x+5)}{x^2+3 x+2}-4 \log (1-x)+\log (2-x)-28 \log (x+1)+31 \log (x+2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - 3*x + x^2)/(4 - 5*x^2 + x^4)^2,x]

[Out]

((-12*(5 + 3*x))/(2 + 3*x + x^2) - 4*Log[1 - x] + Log[2 - x] - 28*Log[1 + x] + 31*Log[2 + x])/144

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fricas [A]  time = 0.74, size = 72, normalized size = 1.29 \[ \frac {31 \, {\left (x^{2} + 3 \, x + 2\right )} \log \left (x + 2\right ) - 28 \, {\left (x^{2} + 3 \, x + 2\right )} \log \left (x + 1\right ) - 4 \, {\left (x^{2} + 3 \, x + 2\right )} \log \left (x - 1\right ) + {\left (x^{2} + 3 \, x + 2\right )} \log \left (x - 2\right ) - 36 \, x - 60}{144 \, {\left (x^{2} + 3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x+2)/(x^4-5*x^2+4)^2,x, algorithm="fricas")

[Out]

1/144*(31*(x^2 + 3*x + 2)*log(x + 2) - 28*(x^2 + 3*x + 2)*log(x + 1) - 4*(x^2 + 3*x + 2)*log(x - 1) + (x^2 + 3
*x + 2)*log(x - 2) - 36*x - 60)/(x^2 + 3*x + 2)

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giac [A]  time = 0.35, size = 46, normalized size = 0.82 \[ -\frac {3 \, x + 5}{12 \, {\left (x + 2\right )} {\left (x + 1\right )}} + \frac {31}{144} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {7}{36} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{36} \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{144} \, \log \left ({\left | x - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x+2)/(x^4-5*x^2+4)^2,x, algorithm="giac")

[Out]

-1/12*(3*x + 5)/((x + 2)*(x + 1)) + 31/144*log(abs(x + 2)) - 7/36*log(abs(x + 1)) - 1/36*log(abs(x - 1)) + 1/1
44*log(abs(x - 2))

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maple [A]  time = 0.01, size = 40, normalized size = 0.71 \[ \frac {31 \ln \left (x +2\right )}{144}+\frac {\ln \left (x -2\right )}{144}-\frac {\ln \left (x -1\right )}{36}-\frac {7 \ln \left (x +1\right )}{36}-\frac {1}{6 \left (x +1\right )}-\frac {1}{12 \left (x +2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-3*x+2)/(x^4-5*x^2+4)^2,x)

[Out]

1/144*ln(x-2)-1/6/(x+1)-7/36*ln(x+1)-1/36*ln(x-1)-1/12/(x+2)+31/144*ln(x+2)

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maxima [A]  time = 0.43, size = 42, normalized size = 0.75 \[ -\frac {3 \, x + 5}{12 \, {\left (x^{2} + 3 \, x + 2\right )}} + \frac {31}{144} \, \log \left (x + 2\right ) - \frac {7}{36} \, \log \left (x + 1\right ) - \frac {1}{36} \, \log \left (x - 1\right ) + \frac {1}{144} \, \log \left (x - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x+2)/(x^4-5*x^2+4)^2,x, algorithm="maxima")

[Out]

-1/12*(3*x + 5)/(x^2 + 3*x + 2) + 31/144*log(x + 2) - 7/36*log(x + 1) - 1/36*log(x - 1) + 1/144*log(x - 2)

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mupad [B]  time = 0.05, size = 42, normalized size = 0.75 \[ \frac {\ln \left (x-2\right )}{144}-\frac {7\,\ln \left (x+1\right )}{36}-\frac {\ln \left (x-1\right )}{36}+\frac {31\,\ln \left (x+2\right )}{144}-\frac {\frac {x}{4}+\frac {5}{12}}{x^2+3\,x+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 3*x + 2)/(x^4 - 5*x^2 + 4)^2,x)

[Out]

log(x - 2)/144 - (7*log(x + 1))/36 - log(x - 1)/36 + (31*log(x + 2))/144 - (x/4 + 5/12)/(3*x + x^2 + 2)

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sympy [A]  time = 0.29, size = 46, normalized size = 0.82 \[ \frac {- 3 x - 5}{12 x^{2} + 36 x + 24} + \frac {\log {\left (x - 2 \right )}}{144} - \frac {\log {\left (x - 1 \right )}}{36} - \frac {7 \log {\left (x + 1 \right )}}{36} + \frac {31 \log {\left (x + 2 \right )}}{144} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-3*x+2)/(x**4-5*x**2+4)**2,x)

[Out]

(-3*x - 5)/(12*x**2 + 36*x + 24) + log(x - 2)/144 - log(x - 1)/36 - 7*log(x + 1)/36 + 31*log(x + 2)/144

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