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

Optimal. Leaf size=31 \[ \log (x+2) (d-2 e+4 f)+x (e-4 f)+\frac {1}{2} f (x+2)^2 \]

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Rubi [A]  time = 0.05, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1586, 698} \begin {gather*} \log (x+2) (d-2 e+4 f)+x (e-4 f)+\frac {1}{2} f (x+2)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

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

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.97 \begin {gather*} \log (x+2) (d-2 e+4 f)+\frac {1}{2} (x+2) (2 e+f (x-6)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d+e x+f x^2\right ) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.22, size = 27, normalized size = 0.87 \begin {gather*} \frac {1}{2} \, f x^{2} + {\left (e - 2 \, f\right )} x + {\left (d - 2 \, e + 4 \, f\right )} \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 30, normalized size = 0.97 \begin {gather*} \frac {1}{2} \, f x^{2} - 2 \, f x + x e + {\left (d + 4 \, f - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 35, normalized size = 1.13 \begin {gather*} \frac {f \,x^{2}}{2}+d \ln \left (x +2\right )+e x -2 e \ln \left (x +2\right )-2 f x +4 f \ln \left (x +2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 27, normalized size = 0.87 \begin {gather*} \frac {1}{2} \, f x^{2} + {\left (e - 2 \, f\right )} x + {\left (d - 2 \, e + 4 \, f\right )} \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 27, normalized size = 0.87 \begin {gather*} x\,\left (e-2\,f\right )+\frac {f\,x^2}{2}+\ln \left (x+2\right )\,\left (d-2\,e+4\,f\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(e - 2*f) + (f*x^2)/2 + log(x + 2)*(d - 2*e + 4*f)

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sympy [A]  time = 0.15, size = 26, normalized size = 0.84 \begin {gather*} \frac {f x^{2}}{2} + x \left (e - 2 f\right ) + \left (d - 2 e + 4 f\right ) \log {\left (x + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f*x**2/2 + x*(e - 2*f) + (d - 2*e + 4*f)*log(x + 2)

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