3.76 \(\int \frac{(2-3 x+x^2) (d+e x+f x^2+g x^3)}{4-5 x^2+x^4} \, dx\)

Optimal. Leaf size=47 \[ \log (x+1) (d-e+f-g)-\log (x+2) (d-2 e+4 f-8 g)+x (f-3 g)+\frac{g x^2}{2} \]

[Out]

(f - 3*g)*x + (g*x^2)/2 + (d - e + f - g)*Log[1 + x] - (d - 2*e + 4*f - 8*g)*Log[2 + x]

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Rubi [A]  time = 0.0677106, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1586, 1657, 632, 31} \[ \log (x+1) (d-e+f-g)-\log (x+2) (d-2 e+4 f-8 g)+x (f-3 g)+\frac{g x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f - 3*g)*x + (g*x^2)/2 + (d - e + f - g)*Log[1 + x] - (d - 2*e + 4*f - 8*g)*Log[2 + x]

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 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 31

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

Rubi steps

\begin{align*} \int \frac{\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx &=\int \frac{d+e x+f x^2+g x^3}{2+3 x+x^2} \, dx\\ &=\int \left (f-3 g+g x+\frac{d-2 f+6 g+(e-3 f+7 g) x}{2+3 x+x^2}\right ) \, dx\\ &=(f-3 g) x+\frac{g x^2}{2}+\int \frac{d-2 f+6 g+(e-3 f+7 g) x}{2+3 x+x^2} \, dx\\ &=(f-3 g) x+\frac{g x^2}{2}-(d-2 e+4 f-8 g) \int \frac{1}{2+x} \, dx+(d-e+f-g) \int \frac{1}{1+x} \, dx\\ &=(f-3 g) x+\frac{g x^2}{2}+(d-e+f-g) \log (1+x)-(d-2 e+4 f-8 g) \log (2+x)\\ \end{align*}

Mathematica [A]  time = 0.020238, size = 44, normalized size = 0.94 \[ \log (x+1) (d-e+f-g)-\log (x+2) (d-2 e+4 f-8 g)+f x+\frac{1}{2} g (x-6) x \]

Antiderivative was successfully verified.

[In]

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

[Out]

f*x + (g*(-6 + x)*x)/2 + (d - e + f - g)*Log[1 + x] - (d - 2*e + 4*f - 8*g)*Log[2 + x]

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Maple [A]  time = 0.006, size = 69, normalized size = 1.5 \begin{align*}{\frac{g{x}^{2}}{2}}+fx-3\,gx-\ln \left ( 2+x \right ) d+2\,\ln \left ( 2+x \right ) e-4\,\ln \left ( 2+x \right ) f+8\,\ln \left ( 2+x \right ) g+\ln \left ( 1+x \right ) d-\ln \left ( 1+x \right ) e+\ln \left ( 1+x \right ) f-\ln \left ( 1+x \right ) g \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-3*x+2)*(g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x)

[Out]

1/2*g*x^2+f*x-3*g*x-ln(2+x)*d+2*ln(2+x)*e-4*ln(2+x)*f+8*ln(2+x)*g+ln(1+x)*d-ln(1+x)*e+ln(1+x)*f-ln(1+x)*g

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Maxima [A]  time = 0.962906, size = 61, normalized size = 1.3 \begin{align*} \frac{1}{2} \, g x^{2} +{\left (f - 3 \, g\right )} x -{\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left (x + 2\right ) +{\left (d - e + f - g\right )} \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*g*x^2 + (f - 3*g)*x - (d - 2*e + 4*f - 8*g)*log(x + 2) + (d - e + f - g)*log(x + 1)

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Fricas [A]  time = 1.50892, size = 120, normalized size = 2.55 \begin{align*} \frac{1}{2} \, g x^{2} +{\left (f - 3 \, g\right )} x -{\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left (x + 2\right ) +{\left (d - e + f - g\right )} \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*g*x^2 + (f - 3*g)*x - (d - 2*e + 4*f - 8*g)*log(x + 2) + (d - e + f - g)*log(x + 1)

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Sympy [A]  time = 0.885272, size = 66, normalized size = 1.4 \begin{align*} \frac{g x^{2}}{2} + x \left (f - 3 g\right ) + \left (- d + 2 e - 4 f + 8 g\right ) \log{\left (x + \frac{4 d - 6 e + 10 f - 18 g}{2 d - 3 e + 5 f - 9 g} \right )} + \left (d - e + f - g\right ) \log{\left (x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-3*x+2)*(g*x**3+f*x**2+e*x+d)/(x**4-5*x**2+4),x)

[Out]

g*x**2/2 + x*(f - 3*g) + (-d + 2*e - 4*f + 8*g)*log(x + (4*d - 6*e + 10*f - 18*g)/(2*d - 3*e + 5*f - 9*g)) + (
d - e + f - g)*log(x + 1)

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Giac [A]  time = 1.11015, size = 66, normalized size = 1.4 \begin{align*} \frac{1}{2} \, g x^{2} + f x - 3 \, g x -{\left (d + 4 \, f - 8 \, g - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) +{\left (d + f - g - e\right )} \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*g*x^2 + f*x - 3*g*x - (d + 4*f - 8*g - 2*e)*log(abs(x + 2)) + (d + f - g - e)*log(abs(x + 1))