3.147 \(\int \frac{(a+b x^2)^2 (c+d x^2)}{x^2} \, dx\)

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

[Out]

-((a^2*c)/x) + a*(2*b*c + a*d)*x + (b*(b*c + 2*a*d)*x^3)/3 + (b^2*d*x^5)/5

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Rubi [A]  time = 0.0256699, 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} \[ -\frac{a^2 c}{x}+\frac{1}{3} b x^3 (2 a d+b c)+a x (a d+2 b c)+\frac{1}{5} b^2 d x^5 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2))/x^2,x]

[Out]

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

Mathematica [A]  time = 0.0161062, size = 48, normalized size = 1. \[ -\frac{a^2 c}{x}+\frac{1}{3} b x^3 (2 a d+b c)+a x (a d+2 b c)+\frac{1}{5} b^2 d x^5 \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2))/x^2,x]

[Out]

-((a^2*c)/x) + a*(2*b*c + a*d)*x + (b*(b*c + 2*a*d)*x^3)/3 + (b^2*d*x^5)/5

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Maple [A]  time = 0.003, size = 49, normalized size = 1. \begin{align*}{\frac{{b}^{2}d{x}^{5}}{5}}+{\frac{2\,{x}^{3}abd}{3}}+{\frac{{x}^{3}{b}^{2}c}{3}}+{a}^{2}dx+2\,abcx-{\frac{{a}^{2}c}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)/x^2,x)

[Out]

1/5*b^2*d*x^5+2/3*x^3*a*b*d+1/3*x^3*b^2*c+a^2*d*x+2*a*b*c*x-a^2*c/x

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Maxima [A]  time = 1.00418, size = 65, normalized size = 1.35 \begin{align*} \frac{1}{5} \, b^{2} d x^{5} + \frac{1}{3} \,{\left (b^{2} c + 2 \, a b d\right )} x^{3} - \frac{a^{2} c}{x} +{\left (2 \, a b c + a^{2} d\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*d*x^5 + 1/3*(b^2*c + 2*a*b*d)*x^3 - a^2*c/x + (2*a*b*c + a^2*d)*x

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Fricas [A]  time = 1.26258, size = 116, normalized size = 2.42 \begin{align*} \frac{3 \, b^{2} d x^{6} + 5 \,{\left (b^{2} c + 2 \, a b d\right )} x^{4} - 15 \, a^{2} c + 15 \,{\left (2 \, a b c + a^{2} d\right )} x^{2}}{15 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.281134, size = 48, normalized size = 1. \begin{align*} - \frac{a^{2} c}{x} + \frac{b^{2} d x^{5}}{5} + x^{3} \left (\frac{2 a b d}{3} + \frac{b^{2} c}{3}\right ) + x \left (a^{2} d + 2 a b c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)/x**2,x)

[Out]

-a**2*c/x + b**2*d*x**5/5 + x**3*(2*a*b*d/3 + b**2*c/3) + x*(a**2*d + 2*a*b*c)

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Giac [A]  time = 1.16103, size = 65, normalized size = 1.35 \begin{align*} \frac{1}{5} \, b^{2} d x^{5} + \frac{1}{3} \, b^{2} c x^{3} + \frac{2}{3} \, a b d x^{3} + 2 \, a b c x + a^{2} d x - \frac{a^{2} c}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*d*x^5 + 1/3*b^2*c*x^3 + 2/3*a*b*d*x^3 + 2*a*b*c*x + a^2*d*x - a^2*c/x