3.2110 \(\int (d+e x) (a+b x+c x^2) \, dx\)

Optimal. Leaf size=42 \[ \frac{1}{2} x^2 (a e+b d)+a d x+\frac{1}{3} x^3 (b e+c d)+\frac{1}{4} c e x^4 \]

[Out]

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

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Rubi [A]  time = 0.028922, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {631} \[ \frac{1}{2} x^2 (a e+b d)+a d x+\frac{1}{3} x^3 (b e+c d)+\frac{1}{4} c e x^4 \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*x + c*x^2),x]

[Out]

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

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

Mathematica [A]  time = 0.0129499, size = 42, normalized size = 1. \[ \frac{1}{2} x^2 (a e+b d)+a d x+\frac{1}{3} x^3 (b e+c d)+\frac{1}{4} c e x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 37, normalized size = 0.9 \begin{align*} adx+{\frac{ \left ( ae+bd \right ){x}^{2}}{2}}+{\frac{ \left ( be+cd \right ){x}^{3}}{3}}+{\frac{ce{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.967219, size = 49, normalized size = 1.17 \begin{align*} \frac{1}{4} \, c e x^{4} + \frac{1}{3} \,{\left (c d + b e\right )} x^{3} + a d x + \frac{1}{2} \,{\left (b d + a e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.81696, size = 104, normalized size = 2.48 \begin{align*} \frac{1}{4} x^{4} e c + \frac{1}{3} x^{3} d c + \frac{1}{3} x^{3} e b + \frac{1}{2} x^{2} d b + \frac{1}{2} x^{2} e a + x d a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.063283, size = 39, normalized size = 0.93 \begin{align*} a d x + \frac{c e x^{4}}{4} + x^{3} \left (\frac{b e}{3} + \frac{c d}{3}\right ) + x^{2} \left (\frac{a e}{2} + \frac{b 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+a),x)

[Out]

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

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Giac [A]  time = 1.08335, size = 58, normalized size = 1.38 \begin{align*} \frac{1}{4} \, c x^{4} e + \frac{1}{3} \, c d x^{3} + \frac{1}{3} \, b x^{3} e + \frac{1}{2} \, b d x^{2} + \frac{1}{2} \, a x^{2} e + a d x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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