3.1026 \(\int (a+b x)^2 (A+B x) \, dx\)

Optimal. Leaf size=38 \[ \frac{(a+b x)^3 (A b-a B)}{3 b^2}+\frac{B (a+b x)^4}{4 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0279523, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {43} \[ \frac{(a+b x)^3 (A b-a B)}{3 b^2}+\frac{B (a+b x)^4}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2*(A + B*x),x]

[Out]

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

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

Mathematica [A]  time = 0.0094946, size = 46, normalized size = 1.21 \[ \frac{1}{12} x \left (6 a^2 (2 A+B x)+4 a b x (3 A+2 B x)+b^2 x^2 (4 A+3 B x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2*(A + B*x),x]

[Out]

(x*(6*a^2*(2*A + B*x) + 4*a*b*x*(3*A + 2*B*x) + b^2*x^2*(4*A + 3*B*x)))/12

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Maple [A]  time = 0.001, size = 49, normalized size = 1.3 \begin{align*}{\frac{B{b}^{2}{x}^{4}}{4}}+{\frac{ \left ( A{b}^{2}+2\,Bba \right ){x}^{3}}{3}}+{\frac{ \left ( 2\,Aba+B{a}^{2} \right ){x}^{2}}{2}}+{a}^{2}Ax \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06367, size = 65, normalized size = 1.71 \begin{align*} \frac{1}{4} \, B b^{2} x^{4} + A a^{2} x + \frac{1}{3} \,{\left (2 \, B a b + A b^{2}\right )} x^{3} + \frac{1}{2} \,{\left (B a^{2} + 2 \, A a b\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.52715, size = 115, normalized size = 3.03 \begin{align*} \frac{1}{4} x^{4} b^{2} B + \frac{2}{3} x^{3} b a B + \frac{1}{3} x^{3} b^{2} A + \frac{1}{2} x^{2} a^{2} B + x^{2} b a A + x a^{2} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.06705, size = 49, normalized size = 1.29 \begin{align*} A a^{2} x + \frac{B b^{2} x^{4}}{4} + x^{3} \left (\frac{A b^{2}}{3} + \frac{2 B a b}{3}\right ) + x^{2} \left (A a b + \frac{B a^{2}}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.08984, size = 66, normalized size = 1.74 \begin{align*} \frac{1}{4} \, B b^{2} x^{4} + \frac{2}{3} \, B a b x^{3} + \frac{1}{3} \, A b^{2} x^{3} + \frac{1}{2} \, B a^{2} x^{2} + A a b x^{2} + A a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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