3.149 \(\int \frac{(a+b x^2)^2 (c+d x^2)}{x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0278166, 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} \[ -\frac{a^2 c}{3 x^3}+b x (2 a d+b c)-\frac{a (a d+2 b c)}{x}+\frac{1}{3} b^2 d x^3 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2))/x^4,x]

[Out]

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

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

Mathematica [A]  time = 0.0193895, size = 50, normalized size = 1.04 \[ \frac{a^2 (-d)-2 a b c}{x}-\frac{a^2 c}{3 x^3}+b x (2 a d+b c)+\frac{1}{3} b^2 d x^3 \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2))/x^4,x]

[Out]

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

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Maple [A]  time = 0.004, size = 46, normalized size = 1. \begin{align*}{\frac{{b}^{2}d{x}^{3}}{3}}+2\,abdx+{b}^{2}cx-{\frac{{a}^{2}c}{3\,{x}^{3}}}-{\frac{a \left ( ad+2\,bc \right ) }{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^4,x)

[Out]

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

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Maxima [A]  time = 1.07129, size = 68, normalized size = 1.42 \begin{align*} \frac{1}{3} \, b^{2} d x^{3} +{\left (b^{2} c + 2 \, a b d\right )} x - \frac{a^{2} c + 3 \,{\left (2 \, a b c + a^{2} d\right )} x^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.11661, size = 109, normalized size = 2.27 \begin{align*} \frac{b^{2} d x^{6} + 3 \,{\left (b^{2} c + 2 \, a b d\right )} x^{4} - a^{2} c - 3 \,{\left (2 \, a b c + a^{2} d\right )} x^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.381036, size = 49, normalized size = 1.02 \begin{align*} \frac{b^{2} d x^{3}}{3} + x \left (2 a b d + b^{2} c\right ) - \frac{a^{2} c + x^{2} \left (3 a^{2} d + 6 a b c\right )}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)/x**4,x)

[Out]

b**2*d*x**3/3 + x*(2*a*b*d + b**2*c) - (a**2*c + x**2*(3*a**2*d + 6*a*b*c))/(3*x**3)

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Giac [A]  time = 1.15255, size = 68, normalized size = 1.42 \begin{align*} \frac{1}{3} \, b^{2} d x^{3} + b^{2} c x + 2 \, a b d x - \frac{6 \, a b c x^{2} + 3 \, a^{2} d x^{2} + a^{2} c}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)/x^4,x, algorithm="giac")

[Out]

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