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

Optimal. Leaf size=14 \[ (d-2 e) \log (x+2)+e x \]

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1586, 43} \begin {gather*} (d-2 e) \log (x+2)+e x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

e*x + (d - 2*e)*Log[2 + x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx &=\int \frac {d+e x}{2+x} \, dx\\ &=\int \left (e+\frac {d-2 e}{2+x}\right ) \, dx\\ &=e x+(d-2 e) \log (2+x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.14 \begin {gather*} (d-2 e) \log (x+2)+e (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

e*(2 + x) + (d - 2*e)*Log[2 + x]

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

[Out]

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

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fricas [A]  time = 1.38, size = 14, normalized size = 1.00 \begin {gather*} e x + {\left (d - 2 \, e\right )} \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x + (d - 2*e)*log(x + 2)

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giac [A]  time = 0.33, size = 17, normalized size = 1.21 \begin {gather*} x e + {\left (d - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e + (d - 2*e)*log(abs(x + 2))

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maple [A]  time = 0.00, size = 18, normalized size = 1.29 \begin {gather*} d \ln \left (x +2\right )+e x -2 e \ln \left (x +2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x+d*ln(x+2)-2*e*ln(x+2)

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maxima [A]  time = 0.44, size = 14, normalized size = 1.00 \begin {gather*} e x + {\left (d - 2 \, e\right )} \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x + (d - 2*e)*log(x + 2)

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mupad [B]  time = 0.73, size = 14, normalized size = 1.00 \begin {gather*} \ln \left (x+2\right )\,\left (d-2\,e\right )+e\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 2)*(d - 2*e) + e*x

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sympy [A]  time = 0.12, size = 12, normalized size = 0.86 \begin {gather*} e x + \left (d - 2 e\right ) \log {\left (x + 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x + (d - 2*e)*log(x + 2)

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