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

Optimal. Leaf size=55 \[ \frac{1}{3} A b^2 x^3+\frac{1}{5} c x^5 (A c+2 b B)+\frac{1}{4} b x^4 (2 A c+b B)+\frac{1}{6} B c^2 x^6 \]

[Out]

(A*b^2*x^3)/3 + (b*(b*B + 2*A*c)*x^4)/4 + (c*(2*b*B + A*c)*x^5)/5 + (B*c^2*x^6)/6

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Rubi [A]  time = 0.0340721, antiderivative size = 55, 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 = {631} \[ \frac{1}{3} A b^2 x^3+\frac{1}{5} c x^5 (A c+2 b B)+\frac{1}{4} b x^4 (2 A c+b B)+\frac{1}{6} B c^2 x^6 \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(b*x + c*x^2)^2,x]

[Out]

(A*b^2*x^3)/3 + (b*(b*B + 2*A*c)*x^4)/4 + (c*(2*b*B + A*c)*x^5)/5 + (B*c^2*x^6)/6

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 (b x+c x^2\right )^2 \, dx &=\int \left (A b^2 x^2+b (b B+2 A c) x^3+c (2 b B+A c) x^4+B c^2 x^5\right ) \, dx\\ &=\frac{1}{3} A b^2 x^3+\frac{1}{4} b (b B+2 A c) x^4+\frac{1}{5} c (2 b B+A c) x^5+\frac{1}{6} B c^2 x^6\\ \end{align*}

Mathematica [A]  time = 0.0112767, size = 49, normalized size = 0.89 \[ \frac{1}{60} x^3 \left (20 A b^2+12 c x^2 (A c+2 b B)+15 b x (2 A c+b B)+10 B c^2 x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(20*A*b^2 + 15*b*(b*B + 2*A*c)*x + 12*c*(2*b*B + A*c)*x^2 + 10*B*c^2*x^3))/60

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Maple [A]  time = 0., size = 52, normalized size = 1. \begin{align*}{\frac{B{c}^{2}{x}^{6}}{6}}+{\frac{ \left ( A{c}^{2}+2\,Bbc \right ){x}^{5}}{5}}+{\frac{ \left ( 2\,Abc+{b}^{2}B \right ){x}^{4}}{4}}+{\frac{A{b}^{2}{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*c^2*x^6+1/5*(A*c^2+2*B*b*c)*x^5+1/4*(2*A*b*c+B*b^2)*x^4+1/3*A*b^2*x^3

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Maxima [A]  time = 1.14642, size = 69, normalized size = 1.25 \begin{align*} \frac{1}{6} \, B c^{2} x^{6} + \frac{1}{3} \, A b^{2} x^{3} + \frac{1}{5} \,{\left (2 \, B b c + A c^{2}\right )} x^{5} + \frac{1}{4} \,{\left (B b^{2} + 2 \, A b c\right )} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*c^2*x^6 + 1/3*A*b^2*x^3 + 1/5*(2*B*b*c + A*c^2)*x^5 + 1/4*(B*b^2 + 2*A*b*c)*x^4

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Fricas [A]  time = 1.62258, size = 128, normalized size = 2.33 \begin{align*} \frac{1}{6} x^{6} c^{2} B + \frac{2}{5} x^{5} c b B + \frac{1}{5} x^{5} c^{2} A + \frac{1}{4} x^{4} b^{2} B + \frac{1}{2} x^{4} c b A + \frac{1}{3} x^{3} b^{2} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*c^2*B + 2/5*x^5*c*b*B + 1/5*x^5*c^2*A + 1/4*x^4*b^2*B + 1/2*x^4*c*b*A + 1/3*x^3*b^2*A

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Sympy [A]  time = 0.079924, size = 54, normalized size = 0.98 \begin{align*} \frac{A b^{2} x^{3}}{3} + \frac{B c^{2} x^{6}}{6} + x^{5} \left (\frac{A c^{2}}{5} + \frac{2 B b c}{5}\right ) + x^{4} \left (\frac{A b c}{2} + \frac{B b^{2}}{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*b**2*x**3/3 + B*c**2*x**6/6 + x**5*(A*c**2/5 + 2*B*b*c/5) + x**4*(A*b*c/2 + B*b**2/4)

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Giac [A]  time = 1.11641, size = 72, normalized size = 1.31 \begin{align*} \frac{1}{6} \, B c^{2} x^{6} + \frac{2}{5} \, B b c x^{5} + \frac{1}{5} \, A c^{2} x^{5} + \frac{1}{4} \, B b^{2} x^{4} + \frac{1}{2} \, A b c x^{4} + \frac{1}{3} \, A b^{2} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*c^2*x^6 + 2/5*B*b*c*x^5 + 1/5*A*c^2*x^5 + 1/4*B*b^2*x^4 + 1/2*A*b*c*x^4 + 1/3*A*b^2*x^3