3.22 \(\int \frac{(a+b x^3)^2 (A+B x^3)}{x^9} \, dx\)

Optimal. Leaf size=50 \[ -\frac{a^2 A}{8 x^8}-\frac{a (a B+2 A b)}{5 x^5}-\frac{b (2 a B+A b)}{2 x^2}+b^2 B x \]

[Out]

-(a^2*A)/(8*x^8) - (a*(2*A*b + a*B))/(5*x^5) - (b*(A*b + 2*a*B))/(2*x^2) + b^2*B*x

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Rubi [A]  time = 0.0287994, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {448} \[ -\frac{a^2 A}{8 x^8}-\frac{a (a B+2 A b)}{5 x^5}-\frac{b (2 a B+A b)}{2 x^2}+b^2 B x \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*(A + B*x^3))/x^9,x]

[Out]

-(a^2*A)/(8*x^8) - (a*(2*A*b + a*B))/(5*x^5) - (b*(A*b + 2*a*B))/(2*x^2) + b^2*B*x

Rule 448

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3\right )^2 \left (A+B x^3\right )}{x^9} \, dx &=\int \left (b^2 B+\frac{a^2 A}{x^9}+\frac{a (2 A b+a B)}{x^6}+\frac{b (A b+2 a B)}{x^3}\right ) \, dx\\ &=-\frac{a^2 A}{8 x^8}-\frac{a (2 A b+a B)}{5 x^5}-\frac{b (A b+2 a B)}{2 x^2}+b^2 B x\\ \end{align*}

Mathematica [A]  time = 0.021259, size = 50, normalized size = 1. \[ -\frac{a^2 A}{8 x^8}-\frac{a (a B+2 A b)}{5 x^5}-\frac{b (2 a B+A b)}{2 x^2}+b^2 B x \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*(A + B*x^3))/x^9,x]

[Out]

-(a^2*A)/(8*x^8) - (a*(2*A*b + a*B))/(5*x^5) - (b*(A*b + 2*a*B))/(2*x^2) + b^2*B*x

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Maple [A]  time = 0.006, size = 45, normalized size = 0.9 \begin{align*} -{\frac{A{a}^{2}}{8\,{x}^{8}}}-{\frac{a \left ( 2\,Ab+Ba \right ) }{5\,{x}^{5}}}-{\frac{b \left ( Ab+2\,Ba \right ) }{2\,{x}^{2}}}+{b}^{2}Bx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(B*x^3+A)/x^9,x)

[Out]

-1/8*a^2*A/x^8-1/5*a*(2*A*b+B*a)/x^5-1/2*b*(A*b+2*B*a)/x^2+b^2*B*x

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Maxima [A]  time = 1.17527, size = 69, normalized size = 1.38 \begin{align*} B b^{2} x - \frac{20 \,{\left (2 \, B a b + A b^{2}\right )} x^{6} + 8 \,{\left (B a^{2} + 2 \, A a b\right )} x^{3} + 5 \, A a^{2}}{40 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^9,x, algorithm="maxima")

[Out]

B*b^2*x - 1/40*(20*(2*B*a*b + A*b^2)*x^6 + 8*(B*a^2 + 2*A*a*b)*x^3 + 5*A*a^2)/x^8

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Fricas [A]  time = 1.33918, size = 119, normalized size = 2.38 \begin{align*} \frac{40 \, B b^{2} x^{9} - 20 \,{\left (2 \, B a b + A b^{2}\right )} x^{6} - 8 \,{\left (B a^{2} + 2 \, A a b\right )} x^{3} - 5 \, A a^{2}}{40 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^9,x, algorithm="fricas")

[Out]

1/40*(40*B*b^2*x^9 - 20*(2*B*a*b + A*b^2)*x^6 - 8*(B*a^2 + 2*A*a*b)*x^3 - 5*A*a^2)/x^8

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Sympy [A]  time = 1.10184, size = 51, normalized size = 1.02 \begin{align*} B b^{2} x - \frac{5 A a^{2} + x^{6} \left (20 A b^{2} + 40 B a b\right ) + x^{3} \left (16 A a b + 8 B a^{2}\right )}{40 x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(B*x**3+A)/x**9,x)

[Out]

B*b**2*x - (5*A*a**2 + x**6*(20*A*b**2 + 40*B*a*b) + x**3*(16*A*a*b + 8*B*a**2))/(40*x**8)

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Giac [A]  time = 1.19952, size = 72, normalized size = 1.44 \begin{align*} B b^{2} x - \frac{40 \, B a b x^{6} + 20 \, A b^{2} x^{6} + 8 \, B a^{2} x^{3} + 16 \, A a b x^{3} + 5 \, A a^{2}}{40 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^9,x, algorithm="giac")

[Out]

B*b^2*x - 1/40*(40*B*a*b*x^6 + 20*A*b^2*x^6 + 8*B*a^2*x^3 + 16*A*a*b*x^3 + 5*A*a^2)/x^8