### 3.2122 $$\int (a+b x+c x^2)^2 \, dx$$

Optimal. Leaf size=46 $a^2 x+\frac{1}{3} x^3 \left (2 a c+b^2\right )+a b x^2+\frac{1}{2} b c x^4+\frac{c^2 x^5}{5}$

[Out]

a^2*x + a*b*x^2 + ((b^2 + 2*a*c)*x^3)/3 + (b*c*x^4)/2 + (c^2*x^5)/5

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Rubi [A]  time = 0.0229114, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {611} $a^2 x+\frac{1}{3} x^3 \left (2 a c+b^2\right )+a b x^2+\frac{1}{2} b c x^4+\frac{c^2 x^5}{5}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^2,x]

[Out]

a^2*x + a*b*x^2 + ((b^2 + 2*a*c)*x^3)/3 + (b*c*x^4)/2 + (c^2*x^5)/5

Rule 611

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && (EqQ[a, 0] ||  !PerfectSquareQ[b^2 - 4*a*c])

Rubi steps

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

Mathematica [A]  time = 0.0062207, size = 46, normalized size = 1. $a^2 x+\frac{1}{3} x^3 \left (2 a c+b^2\right )+a b x^2+\frac{1}{2} b c x^4+\frac{c^2 x^5}{5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^2,x]

[Out]

a^2*x + a*b*x^2 + ((b^2 + 2*a*c)*x^3)/3 + (b*c*x^4)/2 + (c^2*x^5)/5

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Maple [A]  time = 0.039, size = 41, normalized size = 0.9 \begin{align*}{a}^{2}x+ab{x}^{2}+{\frac{ \left ( 2\,ac+{b}^{2} \right ){x}^{3}}{3}}+{\frac{bc{x}^{4}}{2}}+{\frac{{c}^{2}{x}^{5}}{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^2,x)

[Out]

a^2*x+a*b*x^2+1/3*(2*a*c+b^2)*x^3+1/2*b*c*x^4+1/5*c^2*x^5

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Maxima [A]  time = 0.979495, size = 61, normalized size = 1.33 \begin{align*} \frac{1}{5} \, c^{2} x^{5} + \frac{1}{2} \, b c x^{4} + \frac{1}{3} \, b^{2} x^{3} + a^{2} x + \frac{1}{3} \,{\left (2 \, c x^{3} + 3 \, b x^{2}\right )} a \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

1/5*c^2*x^5 + 1/2*b*c*x^4 + 1/3*b^2*x^3 + a^2*x + 1/3*(2*c*x^3 + 3*b*x^2)*a

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Fricas [A]  time = 1.70207, size = 99, normalized size = 2.15 \begin{align*} \frac{1}{5} x^{5} c^{2} + \frac{1}{2} x^{4} c b + \frac{1}{3} x^{3} b^{2} + \frac{2}{3} x^{3} c a + x^{2} b a + x a^{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

1/5*x^5*c^2 + 1/2*x^4*c*b + 1/3*x^3*b^2 + 2/3*x^3*c*a + x^2*b*a + x*a^2

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Sympy [A]  time = 0.065928, size = 42, normalized size = 0.91 \begin{align*} a^{2} x + a b x^{2} + \frac{b c x^{4}}{2} + \frac{c^{2} x^{5}}{5} + x^{3} \left (\frac{2 a c}{3} + \frac{b^{2}}{3}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**2,x)

[Out]

a**2*x + a*b*x**2 + b*c*x**4/2 + c**2*x**5/5 + x**3*(2*a*c/3 + b**2/3)

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Giac [A]  time = 1.09587, size = 57, normalized size = 1.24 \begin{align*} \frac{1}{5} \, c^{2} x^{5} + \frac{1}{2} \, b c x^{4} + \frac{1}{3} \, b^{2} x^{3} + \frac{2}{3} \, a c x^{3} + a b x^{2} + a^{2} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

1/5*c^2*x^5 + 1/2*b*c*x^4 + 1/3*b^2*x^3 + 2/3*a*c*x^3 + a*b*x^2 + a^2*x