### 3.1830 $$\int (d+e x)^2 (a d e+(c d^2+a e^2) x+c d e x^2) \, dx$$

Optimal. Leaf size=39 $\frac{1}{4} (d+e x)^4 \left (a-\frac{c d^2}{e^2}\right )+\frac{c d (d+e x)^5}{5 e^2}$

[Out]

((a - (c*d^2)/e^2)*(d + e*x)^4)/4 + (c*d*(d + e*x)^5)/(5*e^2)

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Rubi [A]  time = 0.0171029, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.061, Rules used = {626, 43} $\frac{1}{4} (d+e x)^4 \left (a-\frac{c d^2}{e^2}\right )+\frac{c d (d+e x)^5}{5 e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((a - (c*d^2)/e^2)*(d + e*x)^4)/4 + (c*d*(d + e*x)^5)/(5*e^2)

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 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)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx &=\int (a e+c d x) (d+e x)^3 \, dx\\ &=\int \left (\frac{\left (-c d^2+a e^2\right ) (d+e x)^3}{e}+\frac{c d (d+e x)^4}{e}\right ) \, dx\\ &=\frac{1}{4} \left (a-\frac{c d^2}{e^2}\right ) (d+e x)^4+\frac{c d (d+e x)^5}{5 e^2}\\ \end{align*}

Mathematica [A]  time = 0.0195682, size = 73, normalized size = 1.87 $\frac{1}{20} x \left (5 a e \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+c d x \left (20 d^2 e x+10 d^3+15 d e^2 x^2+4 e^3 x^3\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x*(5*a*e*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + c*d*x*(10*d^3 + 20*d^2*e*x + 15*d*e^2*x^2 + 4*e^3*x^3)
))/20

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Maple [B]  time = 0.04, size = 112, normalized size = 2.9 \begin{align*}{\frac{{e}^{3}dc{x}^{5}}{5}}+{\frac{ \left ( 2\,{d}^{2}{e}^{2}c+{e}^{2} \left ( a{e}^{2}+c{d}^{2} \right ) \right ){x}^{4}}{4}}+{\frac{ \left ( c{d}^{3}e+2\,de \left ( a{e}^{2}+c{d}^{2} \right ) +ad{e}^{3} \right ){x}^{3}}{3}}+{\frac{ \left ({d}^{2} \left ( a{e}^{2}+c{d}^{2} \right ) +2\,a{d}^{2}{e}^{2} \right ){x}^{2}}{2}}+{d}^{3}aex \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.1614, size = 101, normalized size = 2.59 \begin{align*} \frac{1}{5} \, c d e^{3} x^{5} + a d^{3} e x + \frac{1}{4} \,{\left (3 \, c d^{2} e^{2} + a e^{4}\right )} x^{4} +{\left (c d^{3} e + a d e^{3}\right )} x^{3} + \frac{1}{2} \,{\left (c d^{4} + 3 \, a d^{2} e^{2}\right )} x^{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.32734, size = 174, normalized size = 4.46 \begin{align*} \frac{1}{5} x^{5} e^{3} d c + \frac{3}{4} x^{4} e^{2} d^{2} c + \frac{1}{4} x^{4} e^{4} a + x^{3} e d^{3} c + x^{3} e^{3} d a + \frac{1}{2} x^{2} d^{4} c + \frac{3}{2} x^{2} e^{2} d^{2} a + x e d^{3} a \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.115237, size = 80, normalized size = 2.05 \begin{align*} a d^{3} e x + \frac{c d e^{3} x^{5}}{5} + x^{4} \left (\frac{a e^{4}}{4} + \frac{3 c d^{2} e^{2}}{4}\right ) + x^{3} \left (a d e^{3} + c d^{3} e\right ) + x^{2} \left (\frac{3 a d^{2} e^{2}}{2} + \frac{c d^{4}}{2}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.24991, size = 101, normalized size = 2.59 \begin{align*} \frac{1}{5} \, c d x^{5} e^{3} + \frac{3}{4} \, c d^{2} x^{4} e^{2} + c d^{3} x^{3} e + \frac{1}{2} \, c d^{4} x^{2} + \frac{1}{4} \, a x^{4} e^{4} + a d x^{3} e^{3} + \frac{3}{2} \, a d^{2} x^{2} e^{2} + a d^{3} x e \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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