### 3.1454 $$\int (d+e x) (a^2+2 a b x+b^2 x^2) \, dx$$

Optimal. Leaf size=38 $\frac{(a+b x)^3 (b d-a e)}{3 b^2}+\frac{e (a+b x)^4}{4 b^2}$

[Out]

((b*d - a*e)*(a + b*x)^3)/(3*b^2) + (e*(a + b*x)^4)/(4*b^2)

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Rubi [A]  time = 0.0297702, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {27, 43} $\frac{(a+b x)^3 (b d-a e)}{3 b^2}+\frac{e (a+b x)^4}{4 b^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

((b*d - a*e)*(a + b*x)^3)/(3*b^2) + (e*(a + b*x)^4)/(4*b^2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

Mathematica [A]  time = 0.0098354, size = 46, normalized size = 1.21 $\frac{1}{12} x \left (6 a^2 (2 d+e x)+4 a b x (3 d+2 e x)+b^2 x^2 (4 d+3 e x)\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(x*(6*a^2*(2*d + e*x) + 4*a*b*x*(3*d + 2*e*x) + b^2*x^2*(4*d + 3*e*x)))/12

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Maple [A]  time = 0.039, size = 49, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}e{x}^{4}}{4}}+{\frac{ \left ( 2\,aeb+{b}^{2}d \right ){x}^{3}}{3}}+{\frac{ \left ({a}^{2}e+2\,abd \right ){x}^{2}}{2}}+{a}^{2}dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(b^2*x^2+2*a*b*x+a^2),x)

[Out]

1/4*b^2*e*x^4+1/3*(2*a*b*e+b^2*d)*x^3+1/2*(a^2*e+2*a*b*d)*x^2+a^2*d*x

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Maxima [A]  time = 1.18585, size = 65, normalized size = 1.71 \begin{align*} \frac{1}{4} \, b^{2} e x^{4} + a^{2} d x + \frac{1}{3} \,{\left (b^{2} d + 2 \, a b e\right )} x^{3} + \frac{1}{2} \,{\left (2 \, a b d + a^{2} e\right )} x^{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*e*x^4 + a^2*d*x + 1/3*(b^2*d + 2*a*b*e)*x^3 + 1/2*(2*a*b*d + a^2*e)*x^2

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Fricas [A]  time = 1.69772, size = 115, normalized size = 3.03 \begin{align*} \frac{1}{4} x^{4} e b^{2} + \frac{1}{3} x^{3} d b^{2} + \frac{2}{3} x^{3} e b a + x^{2} d b a + \frac{1}{2} x^{2} e a^{2} + x d a^{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*e*b^2 + 1/3*x^3*d*b^2 + 2/3*x^3*e*b*a + x^2*d*b*a + 1/2*x^2*e*a^2 + x*d*a^2

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Sympy [A]  time = 0.074636, size = 49, normalized size = 1.29 \begin{align*} a^{2} d x + \frac{b^{2} e x^{4}}{4} + x^{3} \left (\frac{2 a b e}{3} + \frac{b^{2} d}{3}\right ) + x^{2} \left (\frac{a^{2} e}{2} + a b d\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(b**2*x**2+2*a*b*x+a**2),x)

[Out]

a**2*d*x + b**2*e*x**4/4 + x**3*(2*a*b*e/3 + b**2*d/3) + x**2*(a**2*e/2 + a*b*d)

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Giac [A]  time = 1.16536, size = 70, normalized size = 1.84 \begin{align*} \frac{1}{4} \, b^{2} x^{4} e + \frac{1}{3} \, b^{2} d x^{3} + \frac{2}{3} \, a b x^{3} e + a b d x^{2} + \frac{1}{2} \, a^{2} x^{2} e + a^{2} d x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4*e + 1/3*b^2*d*x^3 + 2/3*a*b*x^3*e + a*b*d*x^2 + 1/2*a^2*x^2*e + a^2*d*x