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

Optimal. Leaf size=47 \[ \frac{1}{3} x^3 (a B+A b)+\frac{1}{2} a A x^2+\frac{1}{4} x^4 (A c+b B)+\frac{1}{5} B c x^5 \]

[Out]

(a*A*x^2)/2 + ((A*b + a*B)*x^3)/3 + ((b*B + A*c)*x^4)/4 + (B*c*x^5)/5

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Rubi [A]  time = 0.0369795, antiderivative size = 47, 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 = {765} \[ \frac{1}{3} x^3 (a B+A b)+\frac{1}{2} a A x^2+\frac{1}{4} x^4 (A c+b B)+\frac{1}{5} B c x^5 \]

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*(a + b*x + c*x^2),x]

[Out]

(a*A*x^2)/2 + ((A*b + a*B)*x^3)/3 + ((b*B + A*c)*x^4)/4 + (B*c*x^5)/5

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.013442, size = 41, normalized size = 0.87 \[ \frac{1}{60} x^2 \left (20 x (a B+A b)+30 a A+15 x^2 (A c+b B)+12 B c x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(30*a*A + 20*(A*b + a*B)*x + 15*(b*B + A*c)*x^2 + 12*B*c*x^3))/60

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Maple [A]  time = 0.002, size = 40, normalized size = 0.9 \begin{align*}{\frac{aA{x}^{2}}{2}}+{\frac{ \left ( Ab+aB \right ){x}^{3}}{3}}+{\frac{ \left ( Ac+bB \right ){x}^{4}}{4}}+{\frac{Bc{x}^{5}}{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*A*x^2+1/3*(A*b+B*a)*x^3+1/4*(A*c+B*b)*x^4+1/5*B*c*x^5

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Maxima [A]  time = 1.18031, size = 53, normalized size = 1.13 \begin{align*} \frac{1}{5} \, B c x^{5} + \frac{1}{4} \,{\left (B b + A c\right )} x^{4} + \frac{1}{2} \, A a x^{2} + \frac{1}{3} \,{\left (B a + A b\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*c*x^5 + 1/4*(B*b + A*c)*x^4 + 1/2*A*a*x^2 + 1/3*(B*a + A*b)*x^3

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Fricas [A]  time = 1.16329, size = 112, normalized size = 2.38 \begin{align*} \frac{1}{5} x^{5} c B + \frac{1}{4} x^{4} b B + \frac{1}{4} x^{4} c A + \frac{1}{3} x^{3} a B + \frac{1}{3} x^{3} b A + \frac{1}{2} x^{2} a A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*c*B + 1/4*x^4*b*B + 1/4*x^4*c*A + 1/3*x^3*a*B + 1/3*x^3*b*A + 1/2*x^2*a*A

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Sympy [A]  time = 0.203798, size = 42, normalized size = 0.89 \begin{align*} \frac{A a x^{2}}{2} + \frac{B c x^{5}}{5} + x^{4} \left (\frac{A c}{4} + \frac{B b}{4}\right ) + x^{3} \left (\frac{A b}{3} + \frac{B a}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*x**2/2 + B*c*x**5/5 + x**4*(A*c/4 + B*b/4) + x**3*(A*b/3 + B*a/3)

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Giac [A]  time = 1.26496, size = 58, normalized size = 1.23 \begin{align*} \frac{1}{5} \, B c x^{5} + \frac{1}{4} \, B b x^{4} + \frac{1}{4} \, A c x^{4} + \frac{1}{3} \, B a x^{3} + \frac{1}{3} \, A b x^{3} + \frac{1}{2} \, A a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*c*x^5 + 1/4*B*b*x^4 + 1/4*A*c*x^4 + 1/3*B*a*x^3 + 1/3*A*b*x^3 + 1/2*A*a*x^2