3.11 \(\int x^2 (a+b x^3)^2 (A+B x^3) \, dx\)

Optimal. Leaf size=42 \[ \frac{\left (a+b x^3\right )^3 (A b-a B)}{9 b^2}+\frac{B \left (a+b x^3\right )^4}{12 b^2} \]

[Out]

((A*b - a*B)*(a + b*x^3)^3)/(9*b^2) + (B*(a + b*x^3)^4)/(12*b^2)

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Rubi [A]  time = 0.0721012, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {444, 43} \[ \frac{\left (a+b x^3\right )^3 (A b-a B)}{9 b^2}+\frac{B \left (a+b x^3\right )^4}{12 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^3)^2*(A + B*x^3),x]

[Out]

((A*b - a*B)*(a + b*x^3)^3)/(9*b^2) + (B*(a + b*x^3)^4)/(12*b^2)

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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 x^2 \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int (a+b x)^2 (A+B x) \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{(A b-a B) (a+b x)^2}{b}+\frac{B (a+b x)^3}{b}\right ) \, dx,x,x^3\right )\\ &=\frac{(A b-a B) \left (a+b x^3\right )^3}{9 b^2}+\frac{B \left (a+b x^3\right )^4}{12 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0145113, size = 51, normalized size = 1.21 \[ \frac{1}{36} x^3 \left (12 a^2 A+4 b x^6 (2 a B+A b)+6 a x^3 (a B+2 A b)+3 b^2 B x^9\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^3)^2*(A + B*x^3),x]

[Out]

(x^3*(12*a^2*A + 6*a*(2*A*b + a*B)*x^3 + 4*b*(A*b + 2*a*B)*x^6 + 3*b^2*B*x^9))/36

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Maple [A]  time = 0.001, size = 52, normalized size = 1.2 \begin{align*}{\frac{{b}^{2}B{x}^{12}}{12}}+{\frac{ \left ({b}^{2}A+2\,abB \right ){x}^{9}}{9}}+{\frac{ \left ( 2\,abA+{a}^{2}B \right ){x}^{6}}{6}}+{\frac{{a}^{2}A{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^3+a)^2*(B*x^3+A),x)

[Out]

1/12*b^2*B*x^12+1/9*(A*b^2+2*B*a*b)*x^9+1/6*(2*A*a*b+B*a^2)*x^6+1/3*a^2*A*x^3

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Maxima [A]  time = 1.12715, size = 69, normalized size = 1.64 \begin{align*} \frac{1}{12} \, B b^{2} x^{12} + \frac{1}{9} \,{\left (2 \, B a b + A b^{2}\right )} x^{9} + \frac{1}{6} \,{\left (B a^{2} + 2 \, A a b\right )} x^{6} + \frac{1}{3} \, A a^{2} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a)^2*(B*x^3+A),x, algorithm="maxima")

[Out]

1/12*B*b^2*x^12 + 1/9*(2*B*a*b + A*b^2)*x^9 + 1/6*(B*a^2 + 2*A*a*b)*x^6 + 1/3*A*a^2*x^3

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Fricas [A]  time = 1.26638, size = 131, normalized size = 3.12 \begin{align*} \frac{1}{12} x^{12} b^{2} B + \frac{2}{9} x^{9} b a B + \frac{1}{9} x^{9} b^{2} A + \frac{1}{6} x^{6} a^{2} B + \frac{1}{3} x^{6} b a A + \frac{1}{3} x^{3} a^{2} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a)^2*(B*x^3+A),x, algorithm="fricas")

[Out]

1/12*x^12*b^2*B + 2/9*x^9*b*a*B + 1/9*x^9*b^2*A + 1/6*x^6*a^2*B + 1/3*x^6*b*a*A + 1/3*x^3*a^2*A

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Sympy [A]  time = 0.067793, size = 54, normalized size = 1.29 \begin{align*} \frac{A a^{2} x^{3}}{3} + \frac{B b^{2} x^{12}}{12} + x^{9} \left (\frac{A b^{2}}{9} + \frac{2 B a b}{9}\right ) + x^{6} \left (\frac{A a b}{3} + \frac{B a^{2}}{6}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**3+a)**2*(B*x**3+A),x)

[Out]

A*a**2*x**3/3 + B*b**2*x**12/12 + x**9*(A*b**2/9 + 2*B*a*b/9) + x**6*(A*a*b/3 + B*a**2/6)

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Giac [A]  time = 1.18901, size = 72, normalized size = 1.71 \begin{align*} \frac{1}{12} \, B b^{2} x^{12} + \frac{2}{9} \, B a b x^{9} + \frac{1}{9} \, A b^{2} x^{9} + \frac{1}{6} \, B a^{2} x^{6} + \frac{1}{3} \, A a b x^{6} + \frac{1}{3} \, A a^{2} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^3+a)^2*(B*x^3+A),x, algorithm="giac")

[Out]

1/12*B*b^2*x^12 + 2/9*B*a*b*x^9 + 1/9*A*b^2*x^9 + 1/6*B*a^2*x^6 + 1/3*A*a*b*x^6 + 1/3*A*a^2*x^3