### 3.234 $$\int (d+e x) (b x+c x^2)^2 \, dx$$

Optimal. Leaf size=55 $\frac{1}{3} b^2 d x^3+\frac{1}{5} c x^5 (2 b e+c d)+\frac{1}{4} b x^4 (b e+2 c d)+\frac{1}{6} c^2 e x^6$

[Out]

(b^2*d*x^3)/3 + (b*(2*c*d + b*e)*x^4)/4 + (c*(c*d + 2*b*e)*x^5)/5 + (c^2*e*x^6)/6

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Rubi [A]  time = 0.0400556, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.059, Rules used = {631} $\frac{1}{3} b^2 d x^3+\frac{1}{5} c x^5 (2 b e+c d)+\frac{1}{4} b x^4 (b e+2 c d)+\frac{1}{6} c^2 e x^6$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)*(b*x + c*x^2)^2,x]

[Out]

(b^2*d*x^3)/3 + (b*(2*c*d + b*e)*x^4)/4 + (c*(c*d + 2*b*e)*x^5)/5 + (c^2*e*x^6)/6

Rule 631

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps

\begin{align*} \int (d+e x) \left (b x+c x^2\right )^2 \, dx &=\int \left (b^2 d x^2+b (2 c d+b e) x^3+c (c d+2 b e) x^4+c^2 e x^5\right ) \, dx\\ &=\frac{1}{3} b^2 d x^3+\frac{1}{4} b (2 c d+b e) x^4+\frac{1}{5} c (c d+2 b e) x^5+\frac{1}{6} c^2 e x^6\\ \end{align*}

Mathematica [A]  time = 0.0101526, size = 50, normalized size = 0.91 $\frac{1}{60} x^3 \left (5 b^2 (4 d+3 e x)+6 b c x (5 d+4 e x)+2 c^2 x^2 (6 d+5 e x)\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x^3*(5*b^2*(4*d + 3*e*x) + 6*b*c*x*(5*d + 4*e*x) + 2*c^2*x^2*(6*d + 5*e*x)))/60

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Maple [A]  time = 0.043, size = 52, normalized size = 1. \begin{align*}{\frac{{c}^{2}e{x}^{6}}{6}}+{\frac{ \left ( 2\,bce+d{c}^{2} \right ){x}^{5}}{5}}+{\frac{ \left ({b}^{2}e+2\,bcd \right ){x}^{4}}{4}}+{\frac{{b}^{2}d{x}^{3}}{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c^2*e*x^6+1/5*(2*b*c*e+c^2*d)*x^5+1/4*(b^2*e+2*b*c*d)*x^4+1/3*b^2*d*x^3

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Maxima [A]  time = 1.15484, size = 69, normalized size = 1.25 \begin{align*} \frac{1}{6} \, c^{2} e x^{6} + \frac{1}{3} \, b^{2} d x^{3} + \frac{1}{5} \,{\left (c^{2} d + 2 \, b c e\right )} x^{5} + \frac{1}{4} \,{\left (2 \, b c d + b^{2} e\right )} x^{4} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c^2*e*x^6 + 1/3*b^2*d*x^3 + 1/5*(c^2*d + 2*b*c*e)*x^5 + 1/4*(2*b*c*d + b^2*e)*x^4

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Fricas [A]  time = 1.45162, size = 128, normalized size = 2.33 \begin{align*} \frac{1}{6} x^{6} e c^{2} + \frac{1}{5} x^{5} d c^{2} + \frac{2}{5} x^{5} e c b + \frac{1}{2} x^{4} d c b + \frac{1}{4} x^{4} e b^{2} + \frac{1}{3} x^{3} d b^{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*e*c^2 + 1/5*x^5*d*c^2 + 2/5*x^5*e*c*b + 1/2*x^4*d*c*b + 1/4*x^4*e*b^2 + 1/3*x^3*d*b^2

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Sympy [A]  time = 0.278477, size = 54, normalized size = 0.98 \begin{align*} \frac{b^{2} d x^{3}}{3} + \frac{c^{2} e x^{6}}{6} + x^{5} \left (\frac{2 b c e}{5} + \frac{c^{2} d}{5}\right ) + x^{4} \left (\frac{b^{2} e}{4} + \frac{b c d}{2}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**2*d*x**3/3 + c**2*e*x**6/6 + x**5*(2*b*c*e/5 + c**2*d/5) + x**4*(b**2*e/4 + b*c*d/2)

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Giac [A]  time = 1.4691, size = 76, normalized size = 1.38 \begin{align*} \frac{1}{6} \, c^{2} x^{6} e + \frac{1}{5} \, c^{2} d x^{5} + \frac{2}{5} \, b c x^{5} e + \frac{1}{2} \, b c d x^{4} + \frac{1}{4} \, b^{2} x^{4} e + \frac{1}{3} \, b^{2} d x^{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c^2*x^6*e + 1/5*c^2*d*x^5 + 2/5*b*c*x^5*e + 1/2*b*c*d*x^4 + 1/4*b^2*x^4*e + 1/3*b^2*d*x^3