3.18 \(\int \frac{(a+b x^3)^2 (A+B x^3)}{x^5} \, dx\)

Optimal. Leaf size=53 \[ -\frac{a^2 A}{4 x^4}+\frac{1}{2} b x^2 (2 a B+A b)-\frac{a (a B+2 A b)}{x}+\frac{1}{5} b^2 B x^5 \]

[Out]

-(a^2*A)/(4*x^4) - (a*(2*A*b + a*B))/x + (b*(A*b + 2*a*B)*x^2)/2 + (b^2*B*x^5)/5

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Rubi [A]  time = 0.030959, 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} \[ -\frac{a^2 A}{4 x^4}+\frac{1}{2} b x^2 (2 a B+A b)-\frac{a (a B+2 A b)}{x}+\frac{1}{5} b^2 B x^5 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*(A + B*x^3))/x^5,x]

[Out]

-(a^2*A)/(4*x^4) - (a*(2*A*b + a*B))/x + (b*(A*b + 2*a*B)*x^2)/2 + (b^2*B*x^5)/5

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

Mathematica [A]  time = 0.0156971, size = 51, normalized size = 0.96 \[ \frac{-5 a^2 A+10 b x^6 (2 a B+A b)-20 a x^3 (a B+2 A b)+4 b^2 B x^9}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*(A + B*x^3))/x^5,x]

[Out]

(-5*a^2*A - 20*a*(2*A*b + a*B)*x^3 + 10*b*(A*b + 2*a*B)*x^6 + 4*b^2*B*x^9)/(20*x^4)

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Maple [A]  time = 0.006, size = 50, normalized size = 0.9 \begin{align*}{\frac{{b}^{2}B{x}^{5}}{5}}+{\frac{A{x}^{2}{b}^{2}}{2}}+B{x}^{2}ab-{\frac{A{a}^{2}}{4\,{x}^{4}}}-{\frac{a \left ( 2\,Ab+Ba \right ) }{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^5,x)

[Out]

1/5*b^2*B*x^5+1/2*A*x^2*b^2+B*x^2*a*b-1/4*a^2*A/x^4-a*(2*A*b+B*a)/x

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Maxima [A]  time = 1.17113, size = 72, normalized size = 1.36 \begin{align*} \frac{1}{5} \, B b^{2} x^{5} + \frac{1}{2} \,{\left (2 \, B a b + A b^{2}\right )} x^{2} - \frac{4 \,{\left (B a^{2} + 2 \, A a b\right )} x^{3} + A a^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^5,x, algorithm="maxima")

[Out]

1/5*B*b^2*x^5 + 1/2*(2*B*a*b + A*b^2)*x^2 - 1/4*(4*(B*a^2 + 2*A*a*b)*x^3 + A*a^2)/x^4

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Fricas [A]  time = 1.70558, size = 119, normalized size = 2.25 \begin{align*} \frac{4 \, B b^{2} x^{9} + 10 \,{\left (2 \, B a b + A b^{2}\right )} x^{6} - 20 \,{\left (B a^{2} + 2 \, A a b\right )} x^{3} - 5 \, A a^{2}}{20 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^5,x, algorithm="fricas")

[Out]

1/20*(4*B*b^2*x^9 + 10*(2*B*a*b + A*b^2)*x^6 - 20*(B*a^2 + 2*A*a*b)*x^3 - 5*A*a^2)/x^4

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Sympy [A]  time = 0.442399, size = 51, normalized size = 0.96 \begin{align*} \frac{B b^{2} x^{5}}{5} + x^{2} \left (\frac{A b^{2}}{2} + B a b\right ) - \frac{A a^{2} + x^{3} \left (8 A a b + 4 B a^{2}\right )}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(B*x**3+A)/x**5,x)

[Out]

B*b**2*x**5/5 + x**2*(A*b**2/2 + B*a*b) - (A*a**2 + x**3*(8*A*a*b + 4*B*a**2))/(4*x**4)

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Giac [A]  time = 1.1991, size = 73, normalized size = 1.38 \begin{align*} \frac{1}{5} \, B b^{2} x^{5} + B a b x^{2} + \frac{1}{2} \, A b^{2} x^{2} - \frac{4 \, B a^{2} x^{3} + 8 \, A a b x^{3} + A a^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^5,x, algorithm="giac")

[Out]

1/5*B*b^2*x^5 + B*a*b*x^2 + 1/2*A*b^2*x^2 - 1/4*(4*B*a^2*x^3 + 8*A*a*b*x^3 + A*a^2)/x^4