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

Optimal. Leaf size=50 \[ a^2 A x+\frac{1}{7} b x^7 (2 a B+A b)+\frac{1}{4} a x^4 (a B+2 A b)+\frac{1}{10} b^2 B x^{10} \]

[Out]

a^2*A*x + (a*(2*A*b + a*B)*x^4)/4 + (b*(A*b + 2*a*B)*x^7)/7 + (b^2*B*x^10)/10

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Rubi [A]  time = 0.0254353, antiderivative size = 50, 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 = {373} \[ a^2 A x+\frac{1}{7} b x^7 (2 a B+A b)+\frac{1}{4} a x^4 (a B+2 A b)+\frac{1}{10} b^2 B x^{10} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^2*(A + B*x^3),x]

[Out]

a^2*A*x + (a*(2*A*b + a*B)*x^4)/4 + (b*(A*b + 2*a*B)*x^7)/7 + (b^2*B*x^10)/10

Rule 373

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

Rubi steps

\begin{align*} \int \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx &=\int \left (a^2 A+a (2 A b+a B) x^3+b (A b+2 a B) x^6+b^2 B x^9\right ) \, dx\\ &=a^2 A x+\frac{1}{4} a (2 A b+a B) x^4+\frac{1}{7} b (A b+2 a B) x^7+\frac{1}{10} b^2 B x^{10}\\ \end{align*}

Mathematica [A]  time = 0.0071112, size = 50, normalized size = 1. \[ a^2 A x+\frac{1}{7} b x^7 (2 a B+A b)+\frac{1}{4} a x^4 (a B+2 A b)+\frac{1}{10} b^2 B x^{10} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*A*x + (a*(2*A*b + a*B)*x^4)/4 + (b*(A*b + 2*a*B)*x^7)/7 + (b^2*B*x^10)/10

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Maple [A]  time = 0.002, size = 49, normalized size = 1. \begin{align*}{\frac{{b}^{2}B{x}^{10}}{10}}+{\frac{ \left ({b}^{2}A+2\,abB \right ){x}^{7}}{7}}+{\frac{ \left ( 2\,abA+{a}^{2}B \right ){x}^{4}}{4}}+{a}^{2}Ax \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^2*B*x^10+1/7*(A*b^2+2*B*a*b)*x^7+1/4*(2*A*a*b+B*a^2)*x^4+a^2*A*x

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Maxima [A]  time = 1.28197, size = 65, normalized size = 1.3 \begin{align*} \frac{1}{10} \, B b^{2} x^{10} + \frac{1}{7} \,{\left (2 \, B a b + A b^{2}\right )} x^{7} + \frac{1}{4} \,{\left (B a^{2} + 2 \, A a b\right )} x^{4} + A a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*b^2*x^10 + 1/7*(2*B*a*b + A*b^2)*x^7 + 1/4*(B*a^2 + 2*A*a*b)*x^4 + A*a^2*x

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Fricas [A]  time = 1.22204, size = 123, normalized size = 2.46 \begin{align*} \frac{1}{10} x^{10} b^{2} B + \frac{2}{7} x^{7} b a B + \frac{1}{7} x^{7} b^{2} A + \frac{1}{4} x^{4} a^{2} B + \frac{1}{2} x^{4} b a A + x a^{2} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*x^10*b^2*B + 2/7*x^7*b*a*B + 1/7*x^7*b^2*A + 1/4*x^4*a^2*B + 1/2*x^4*b*a*A + x*a^2*A

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Sympy [A]  time = 0.066957, size = 51, normalized size = 1.02 \begin{align*} A a^{2} x + \frac{B b^{2} x^{10}}{10} + x^{7} \left (\frac{A b^{2}}{7} + \frac{2 B a b}{7}\right ) + x^{4} \left (\frac{A a b}{2} + \frac{B a^{2}}{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**2*x + B*b**2*x**10/10 + x**7*(A*b**2/7 + 2*B*a*b/7) + x**4*(A*a*b/2 + B*a**2/4)

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Giac [A]  time = 1.16846, size = 68, normalized size = 1.36 \begin{align*} \frac{1}{10} \, B b^{2} x^{10} + \frac{2}{7} \, B a b x^{7} + \frac{1}{7} \, A b^{2} x^{7} + \frac{1}{4} \, B a^{2} x^{4} + \frac{1}{2} \, A a b x^{4} + A a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*b^2*x^10 + 2/7*B*a*b*x^7 + 1/7*A*b^2*x^7 + 1/4*B*a^2*x^4 + 1/2*A*a*b*x^4 + A*a^2*x