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

Optimal. Leaf size=53 \[ -\frac{a^2 A}{x}+\frac{1}{5} b x^5 (2 a B+A b)+\frac{1}{2} a x^2 (a B+2 A b)+\frac{1}{8} b^2 B x^8 \]

[Out]

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

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Rubi [A]  time = 0.0282968, antiderivative size = 53, 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} \[ -\frac{a^2 A}{x}+\frac{1}{5} b x^5 (2 a B+A b)+\frac{1}{2} a x^2 (a B+2 A b)+\frac{1}{8} b^2 B x^8 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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^2} \, dx &=\int \left (\frac{a^2 A}{x^2}+a (2 A b+a B) x+b (A b+2 a B) x^4+b^2 B x^7\right ) \, dx\\ &=-\frac{a^2 A}{x}+\frac{1}{2} a (2 A b+a B) x^2+\frac{1}{5} b (A b+2 a B) x^5+\frac{1}{8} b^2 B x^8\\ \end{align*}

Mathematica [A]  time = 0.0164872, size = 53, normalized size = 1. \[ -\frac{a^2 A}{x}+\frac{1}{5} b x^5 (2 a B+A b)+\frac{1}{2} a x^2 (a B+2 A b)+\frac{1}{8} b^2 B x^8 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 53, normalized size = 1. \begin{align*}{\frac{{b}^{2}B{x}^{8}}{8}}+{\frac{A{x}^{5}{b}^{2}}{5}}+{\frac{2\,B{x}^{5}ab}{5}}+A{x}^{2}ab+{\frac{B{x}^{2}{a}^{2}}{2}}-{\frac{A{a}^{2}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20415, size = 69, normalized size = 1.3 \begin{align*} \frac{1}{8} \, B b^{2} x^{8} + \frac{1}{5} \,{\left (2 \, B a b + A b^{2}\right )} x^{5} + \frac{1}{2} \,{\left (B a^{2} + 2 \, A a b\right )} x^{2} - \frac{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^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.42236, size = 116, normalized size = 2.19 \begin{align*} \frac{5 \, B b^{2} x^{9} + 8 \,{\left (2 \, B a b + A b^{2}\right )} x^{6} + 20 \,{\left (B a^{2} + 2 \, A a b\right )} x^{3} - 40 \, A a^{2}}{40 \, 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^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.280175, size = 49, normalized size = 0.92 \begin{align*} - \frac{A a^{2}}{x} + \frac{B b^{2} x^{8}}{8} + x^{5} \left (\frac{A b^{2}}{5} + \frac{2 B a b}{5}\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**3+a)**2*(B*x**3+A)/x**2,x)

[Out]

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

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Giac [A]  time = 1.10672, size = 70, normalized size = 1.32 \begin{align*} \frac{1}{8} \, B b^{2} x^{8} + \frac{2}{5} \, B a b x^{5} + \frac{1}{5} \, A b^{2} x^{5} + \frac{1}{2} \, B a^{2} x^{2} + A a b x^{2} - \frac{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^2,x, algorithm="giac")

[Out]

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