### 3.2167 $$\int (3-4 x+x^2)^2 \, dx$$

Optimal. Leaf size=28 $-\frac{1}{5} (3-x)^5-\frac{4}{3} (3-x)^3+(x-3)^4$

[Out]

(-4*(3 - x)^3)/3 - (3 - x)^5/5 + (-3 + x)^4

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Rubi [A]  time = 0.0106319, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {610, 43} $-\frac{1}{5} (3-x)^5-\frac{4}{3} (3-x)^3+(x-3)^4$

Antiderivative was successfully veriﬁed.

[In]

Int[(3 - 4*x + x^2)^2,x]

[Out]

(-4*(3 - x)^3)/3 - (3 - x)^5/5 + (-3 + x)^4

Rule 610

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[1/c^p, Int[Simp[
b/2 - q/2 + c*x, x]^p*Simp[b/2 + q/2 + c*x, x]^p, x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && IGt
Q[p, 0] && PerfectSquareQ[b^2 - 4*a*c]

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])

Rubi steps

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

Mathematica [A]  time = 0.0006362, size = 28, normalized size = 1. $\frac{x^5}{5}-2 x^4+\frac{22 x^3}{3}-12 x^2+9 x$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(3 - 4*x + x^2)^2,x]

[Out]

9*x - 12*x^2 + (22*x^3)/3 - 2*x^4 + x^5/5

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Maple [A]  time = 0.039, size = 25, normalized size = 0.9 \begin{align*}{\frac{{x}^{5}}{5}}-2\,{x}^{4}+{\frac{22\,{x}^{3}}{3}}-12\,{x}^{2}+9\,x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5-2*x^4+22/3*x^3-12*x^2+9*x

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Maxima [A]  time = 0.978717, size = 32, normalized size = 1.14 \begin{align*} \frac{1}{5} \, x^{5} - 2 \, x^{4} + \frac{22}{3} \, x^{3} - 12 \, x^{2} + 9 \, x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5 - 2*x^4 + 22/3*x^3 - 12*x^2 + 9*x

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Fricas [A]  time = 1.4252, size = 58, normalized size = 2.07 \begin{align*} \frac{1}{5} x^{5} - 2 x^{4} + \frac{22}{3} x^{3} - 12 x^{2} + 9 x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5 - 2*x^4 + 22/3*x^3 - 12*x^2 + 9*x

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Sympy [A]  time = 0.058233, size = 24, normalized size = 0.86 \begin{align*} \frac{x^{5}}{5} - 2 x^{4} + \frac{22 x^{3}}{3} - 12 x^{2} + 9 x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**5/5 - 2*x**4 + 22*x**3/3 - 12*x**2 + 9*x

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Giac [A]  time = 1.12871, size = 32, normalized size = 1.14 \begin{align*} \frac{1}{5} \, x^{5} - 2 \, x^{4} + \frac{22}{3} \, x^{3} - 12 \, x^{2} + 9 \, x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5 - 2*x^4 + 22/3*x^3 - 12*x^2 + 9*x