3.333 \(\int \frac{1+x+2 x^2+2 x^3}{x^2+x^3+x^4} \, dx\)

Optimal. Leaf size=13 \[ \log \left (x^2+x+1\right )-\frac{1}{x} \]

[Out]

-x^(-1) + Log[1 + x + x^2]

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Rubi [A]  time = 0.0490023, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1594, 1628, 628} \[ \log \left (x^2+x+1\right )-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x + 2*x^2 + 2*x^3)/(x^2 + x^3 + x^4),x]

[Out]

-x^(-1) + Log[1 + x + x^2]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.005564, size = 13, normalized size = 1. \[ \log \left (x^2+x+1\right )-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + 2*x^2 + 2*x^3)/(x^2 + x^3 + x^4),x]

[Out]

-x^(-1) + Log[1 + x + x^2]

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Maple [A]  time = 0.003, size = 14, normalized size = 1.1 \begin{align*} -{x}^{-1}+\ln \left ({x}^{2}+x+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3+2*x^2+x+1)/(x^4+x^3+x^2),x)

[Out]

-1/x+ln(x^2+x+1)

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Maxima [A]  time = 1.03172, size = 18, normalized size = 1.38 \begin{align*} -\frac{1}{x} + \log \left (x^{2} + x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+2*x^2+x+1)/(x^4+x^3+x^2),x, algorithm="maxima")

[Out]

-1/x + log(x^2 + x + 1)

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Fricas [A]  time = 1.43135, size = 38, normalized size = 2.92 \begin{align*} \frac{x \log \left (x^{2} + x + 1\right ) - 1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+2*x^2+x+1)/(x^4+x^3+x^2),x, algorithm="fricas")

[Out]

(x*log(x^2 + x + 1) - 1)/x

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Sympy [A]  time = 0.092617, size = 10, normalized size = 0.77 \begin{align*} \log{\left (x^{2} + x + 1 \right )} - \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3+2*x**2+x+1)/(x**4+x**3+x**2),x)

[Out]

log(x**2 + x + 1) - 1/x

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Giac [A]  time = 1.13992, size = 18, normalized size = 1.38 \begin{align*} -\frac{1}{x} + \log \left (x^{2} + x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+2*x^2+x+1)/(x^4+x^3+x^2),x, algorithm="giac")

[Out]

-1/x + log(x^2 + x + 1)