3.846 \(\int (A+B x) (a+b x+c x^2) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0277259, 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} \[ \frac{1}{2} x^2 (a B+A b)+a A x+\frac{1}{3} x^3 (A c+b B)+\frac{1}{4} B c x^4 \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(a + b*x + c*x^2),x]

[Out]

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

Mathematica [A]  time = 0.013361, size = 42, normalized size = 1. \[ \frac{1}{2} x^2 (a B+A b)+a A x+\frac{1}{3} x^3 (A c+b B)+\frac{1}{4} B c x^4 \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*(a + b*x + c*x^2),x]

[Out]

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

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Maple [A]  time = 0.001, size = 37, normalized size = 0.9 \begin{align*} aAx+{\frac{ \left ( Ab+aB \right ){x}^{2}}{2}}+{\frac{ \left ( Ac+bB \right ){x}^{3}}{3}}+{\frac{Bc{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03019, size = 49, normalized size = 1.17 \begin{align*} \frac{1}{4} \, B c x^{4} + \frac{1}{3} \,{\left (B b + A c\right )} x^{3} + A a x + \frac{1}{2} \,{\left (B a + A b\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.08533, size = 104, normalized size = 2.48 \begin{align*} \frac{1}{4} x^{4} c B + \frac{1}{3} x^{3} b B + \frac{1}{3} x^{3} c A + \frac{1}{2} x^{2} a B + \frac{1}{2} x^{2} b A + x a A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.143956, size = 39, normalized size = 0.93 \begin{align*} A a x + \frac{B c x^{4}}{4} + x^{3} \left (\frac{A c}{3} + \frac{B b}{3}\right ) + x^{2} \left (\frac{A b}{2} + \frac{B a}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21008, size = 54, normalized size = 1.29 \begin{align*} \frac{1}{4} \, B c x^{4} + \frac{1}{3} \, B b x^{3} + \frac{1}{3} \, A c x^{3} + \frac{1}{2} \, B a x^{2} + \frac{1}{2} \, A b x^{2} + A a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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