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

Optimal. Leaf size=48 \[ -\frac{a^2 A}{x}+\frac{1}{3} b x^3 (2 a B+A b)+a x (a B+2 A b)+\frac{1}{5} b^2 B x^5 \]

[Out]

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

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Rubi [A]  time = 0.026731, antiderivative size = 48, 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} \[ -\frac{a^2 A}{x}+\frac{1}{3} b x^3 (2 a B+A b)+a x (a B+2 A b)+\frac{1}{5} b^2 B x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*A)/x) + a*(2*A*b + a*B)*x + (b*(A*b + 2*a*B)*x^3)/3 + (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^2\right )^2 \left (A+B x^2\right )}{x^2} \, dx &=\int \left (a (2 A b+a B)+\frac{a^2 A}{x^2}+b (A b+2 a B) x^2+b^2 B x^4\right ) \, dx\\ &=-\frac{a^2 A}{x}+a (2 A b+a B) x+\frac{1}{3} b (A b+2 a B) x^3+\frac{1}{5} b^2 B x^5\\ \end{align*}

Mathematica [A]  time = 0.0154579, size = 48, normalized size = 1. \[ -\frac{a^2 A}{x}+\frac{1}{3} b x^3 (2 a B+A b)+a x (a B+2 A b)+\frac{1}{5} b^2 B x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 49, normalized size = 1. \begin{align*}{\frac{{b}^{2}B{x}^{5}}{5}}+{\frac{A{x}^{3}{b}^{2}}{3}}+{\frac{2\,B{x}^{3}ab}{3}}+2\,abAx+{a}^{2}Bx-{\frac{A{a}^{2}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.975647, size = 65, normalized size = 1.35 \begin{align*} \frac{1}{5} \, B b^{2} x^{5} + \frac{1}{3} \,{\left (2 \, B a b + A b^{2}\right )} x^{3} - \frac{A a^{2}}{x} +{\left (B a^{2} + 2 \, A a b\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.42518, size = 116, normalized size = 2.42 \begin{align*} \frac{3 \, B b^{2} x^{6} + 5 \,{\left (2 \, B a b + A b^{2}\right )} x^{4} - 15 \, A a^{2} + 15 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2}}{15 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*B*b^2*x^6 + 5*(2*B*a*b + A*b^2)*x^4 - 15*A*a^2 + 15*(B*a^2 + 2*A*a*b)*x^2)/x

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Sympy [A]  time = 0.275702, size = 48, normalized size = 1. \begin{align*} - \frac{A a^{2}}{x} + \frac{B b^{2} x^{5}}{5} + x^{3} \left (\frac{A b^{2}}{3} + \frac{2 B a b}{3}\right ) + x \left (2 A a b + B a^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13251, size = 65, normalized size = 1.35 \begin{align*} \frac{1}{5} \, B b^{2} x^{5} + \frac{2}{3} \, B a b x^{3} + \frac{1}{3} \, A b^{2} x^{3} + B a^{2} x + 2 \, A a b x - \frac{A a^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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