3.433 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^2}{x^6} \, dx\)

Optimal. Leaf size=50 \[ -\frac{6 a^2 b^2}{x}-\frac{4 a^3 b}{3 x^3}-\frac{a^4}{5 x^5}+4 a b^3 x+\frac{b^4 x^3}{3} \]

[Out]

-a^4/(5*x^5) - (4*a^3*b)/(3*x^3) - (6*a^2*b^2)/x + 4*a*b^3*x + (b^4*x^3)/3

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Rubi [A]  time = 0.0263534, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {28, 270} \[ -\frac{6 a^2 b^2}{x}-\frac{4 a^3 b}{3 x^3}-\frac{a^4}{5 x^5}+4 a b^3 x+\frac{b^4 x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^6,x]

[Out]

-a^4/(5*x^5) - (4*a^3*b)/(3*x^3) - (6*a^2*b^2)/x + 4*a*b^3*x + (b^4*x^3)/3

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^6} \, dx &=\frac{\int \frac{\left (a b+b^2 x^2\right )^4}{x^6} \, dx}{b^4}\\ &=\frac{\int \left (4 a b^7+\frac{a^4 b^4}{x^6}+\frac{4 a^3 b^5}{x^4}+\frac{6 a^2 b^6}{x^2}+b^8 x^2\right ) \, dx}{b^4}\\ &=-\frac{a^4}{5 x^5}-\frac{4 a^3 b}{3 x^3}-\frac{6 a^2 b^2}{x}+4 a b^3 x+\frac{b^4 x^3}{3}\\ \end{align*}

Mathematica [A]  time = 0.00808, size = 50, normalized size = 1. \[ -\frac{6 a^2 b^2}{x}-\frac{4 a^3 b}{3 x^3}-\frac{a^4}{5 x^5}+4 a b^3 x+\frac{b^4 x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^6,x]

[Out]

-a^4/(5*x^5) - (4*a^3*b)/(3*x^3) - (6*a^2*b^2)/x + 4*a*b^3*x + (b^4*x^3)/3

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Maple [A]  time = 0.046, size = 45, normalized size = 0.9 \begin{align*} -{\frac{{a}^{4}}{5\,{x}^{5}}}-{\frac{4\,{a}^{3}b}{3\,{x}^{3}}}-6\,{\frac{{b}^{2}{a}^{2}}{x}}+4\,a{b}^{3}x+{\frac{{b}^{4}{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^2/x^6,x)

[Out]

-1/5*a^4/x^5-4/3*a^3*b/x^3-6*a^2*b^2/x+4*a*b^3*x+1/3*b^4*x^3

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Maxima [A]  time = 0.993944, size = 63, normalized size = 1.26 \begin{align*} \frac{1}{3} \, b^{4} x^{3} + 4 \, a b^{3} x - \frac{90 \, a^{2} b^{2} x^{4} + 20 \, a^{3} b x^{2} + 3 \, a^{4}}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x^6,x, algorithm="maxima")

[Out]

1/3*b^4*x^3 + 4*a*b^3*x - 1/15*(90*a^2*b^2*x^4 + 20*a^3*b*x^2 + 3*a^4)/x^5

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Fricas [A]  time = 1.67278, size = 104, normalized size = 2.08 \begin{align*} \frac{5 \, b^{4} x^{8} + 60 \, a b^{3} x^{6} - 90 \, a^{2} b^{2} x^{4} - 20 \, a^{3} b x^{2} - 3 \, a^{4}}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x^6,x, algorithm="fricas")

[Out]

1/15*(5*b^4*x^8 + 60*a*b^3*x^6 - 90*a^2*b^2*x^4 - 20*a^3*b*x^2 - 3*a^4)/x^5

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Sympy [A]  time = 0.367966, size = 48, normalized size = 0.96 \begin{align*} 4 a b^{3} x + \frac{b^{4} x^{3}}{3} - \frac{3 a^{4} + 20 a^{3} b x^{2} + 90 a^{2} b^{2} x^{4}}{15 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**2/x**6,x)

[Out]

4*a*b**3*x + b**4*x**3/3 - (3*a**4 + 20*a**3*b*x**2 + 90*a**2*b**2*x**4)/(15*x**5)

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Giac [A]  time = 1.11867, size = 63, normalized size = 1.26 \begin{align*} \frac{1}{3} \, b^{4} x^{3} + 4 \, a b^{3} x - \frac{90 \, a^{2} b^{2} x^{4} + 20 \, a^{3} b x^{2} + 3 \, a^{4}}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x^6,x, algorithm="giac")

[Out]

1/3*b^4*x^3 + 4*a*b^3*x - 1/15*(90*a^2*b^2*x^4 + 20*a^3*b*x^2 + 3*a^4)/x^5