3.438 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^2}{x^{11}} \, dx\)

Optimal. Leaf size=19 \[ -\frac{\left (a+b x^2\right )^5}{10 a x^{10}} \]

[Out]

-(a + b*x^2)^5/(10*a*x^10)

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Rubi [A]  time = 0.0061135, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {28, 264} \[ -\frac{\left (a+b x^2\right )^5}{10 a x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^11,x]

[Out]

-(a + b*x^2)^5/(10*a*x^10)

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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^2}{x^{11}} \, dx &=\frac{\int \frac{\left (a b+b^2 x^2\right )^4}{x^{11}} \, dx}{b^4}\\ &=-\frac{\left (a+b x^2\right )^5}{10 a x^{10}}\\ \end{align*}

Mathematica [B]  time = 0.0041784, size = 52, normalized size = 2.74 \[ -\frac{a^2 b^2}{x^6}-\frac{a^3 b}{2 x^8}-\frac{a^4}{10 x^{10}}-\frac{a b^3}{x^4}-\frac{b^4}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x^11,x]

[Out]

-a^4/(10*x^10) - (a^3*b)/(2*x^8) - (a^2*b^2)/x^6 - (a*b^3)/x^4 - b^4/(2*x^2)

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Maple [B]  time = 0.048, size = 47, normalized size = 2.5 \begin{align*} -{\frac{{a}^{4}}{10\,{x}^{10}}}-{\frac{a{b}^{3}}{{x}^{4}}}-{\frac{{b}^{4}}{2\,{x}^{2}}}-{\frac{{a}^{3}b}{2\,{x}^{8}}}-{\frac{{b}^{2}{a}^{2}}{{x}^{6}}} \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^11,x)

[Out]

-1/10*a^4/x^10-a*b^3/x^4-1/2*b^4/x^2-1/2*a^3*b/x^8-b^2*a^2/x^6

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Maxima [B]  time = 0.993977, size = 62, normalized size = 3.26 \begin{align*} -\frac{5 \, b^{4} x^{8} + 10 \, a b^{3} x^{6} + 10 \, a^{2} b^{2} x^{4} + 5 \, a^{3} b x^{2} + a^{4}}{10 \, x^{10}} \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^11,x, algorithm="maxima")

[Out]

-1/10*(5*b^4*x^8 + 10*a*b^3*x^6 + 10*a^2*b^2*x^4 + 5*a^3*b*x^2 + a^4)/x^10

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Fricas [B]  time = 1.6613, size = 103, normalized size = 5.42 \begin{align*} -\frac{5 \, b^{4} x^{8} + 10 \, a b^{3} x^{6} + 10 \, a^{2} b^{2} x^{4} + 5 \, a^{3} b x^{2} + a^{4}}{10 \, x^{10}} \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^11,x, algorithm="fricas")

[Out]

-1/10*(5*b^4*x^8 + 10*a*b^3*x^6 + 10*a^2*b^2*x^4 + 5*a^3*b*x^2 + a^4)/x^10

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Sympy [B]  time = 0.506714, size = 49, normalized size = 2.58 \begin{align*} - \frac{a^{4} + 5 a^{3} b x^{2} + 10 a^{2} b^{2} x^{4} + 10 a b^{3} x^{6} + 5 b^{4} x^{8}}{10 x^{10}} \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**11,x)

[Out]

-(a**4 + 5*a**3*b*x**2 + 10*a**2*b**2*x**4 + 10*a*b**3*x**6 + 5*b**4*x**8)/(10*x**10)

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Giac [B]  time = 1.20888, size = 62, normalized size = 3.26 \begin{align*} -\frac{5 \, b^{4} x^{8} + 10 \, a b^{3} x^{6} + 10 \, a^{2} b^{2} x^{4} + 5 \, a^{3} b x^{2} + a^{4}}{10 \, x^{10}} \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^11,x, algorithm="giac")

[Out]

-1/10*(5*b^4*x^8 + 10*a*b^3*x^6 + 10*a^2*b^2*x^4 + 5*a^3*b*x^2 + a^4)/x^10