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

Optimal. Leaf size=54 \[ -\frac{a^2}{10 x^{10}}-\frac{2 a c+b^2}{6 x^6}-\frac{a b}{4 x^8}-\frac{b c}{2 x^4}-\frac{c^2}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0374664, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1114, 698} \[ -\frac{a^2}{10 x^{10}}-\frac{2 a c+b^2}{6 x^6}-\frac{a b}{4 x^8}-\frac{b c}{2 x^4}-\frac{c^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^2/x^11,x]

[Out]

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

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2+c x^4\right )^2}{x^{11}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a+b x+c x^2\right )^2}{x^6} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{a^2}{x^6}+\frac{2 a b}{x^5}+\frac{b^2+2 a c}{x^4}+\frac{2 b c}{x^3}+\frac{c^2}{x^2}\right ) \, dx,x,x^2\right )\\ &=-\frac{a^2}{10 x^{10}}-\frac{a b}{4 x^8}-\frac{b^2+2 a c}{6 x^6}-\frac{b c}{2 x^4}-\frac{c^2}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.015071, size = 53, normalized size = 0.98 \[ -\frac{6 a^2+5 a \left (3 b x^2+4 c x^4\right )+10 x^4 \left (b^2+3 b c x^2+3 c^2 x^4\right )}{60 x^{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(6*a^2 + 5*a*(3*b*x^2 + 4*c*x^4) + 10*x^4*(b^2 + 3*b*c*x^2 + 3*c^2*x^4))/(60*x^10)

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Maple [A]  time = 0.048, size = 45, normalized size = 0.8 \begin{align*} -{\frac{bc}{2\,{x}^{4}}}-{\frac{{c}^{2}}{2\,{x}^{2}}}-{\frac{ab}{4\,{x}^{8}}}-{\frac{2\,ac+{b}^{2}}{6\,{x}^{6}}}-{\frac{{a}^{2}}{10\,{x}^{10}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^2/x^11,x)

[Out]

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

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Maxima [A]  time = 0.957288, size = 62, normalized size = 1.15 \begin{align*} -\frac{30 \, c^{2} x^{8} + 30 \, b c x^{6} + 10 \,{\left (b^{2} + 2 \, a c\right )} x^{4} + 15 \, a b x^{2} + 6 \, a^{2}}{60 \, x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^2/x^11,x, algorithm="maxima")

[Out]

-1/60*(30*c^2*x^8 + 30*b*c*x^6 + 10*(b^2 + 2*a*c)*x^4 + 15*a*b*x^2 + 6*a^2)/x^10

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Fricas [A]  time = 1.4316, size = 111, normalized size = 2.06 \begin{align*} -\frac{30 \, c^{2} x^{8} + 30 \, b c x^{6} + 10 \,{\left (b^{2} + 2 \, a c\right )} x^{4} + 15 \, a b x^{2} + 6 \, a^{2}}{60 \, x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^2/x^11,x, algorithm="fricas")

[Out]

-1/60*(30*c^2*x^8 + 30*b*c*x^6 + 10*(b^2 + 2*a*c)*x^4 + 15*a*b*x^2 + 6*a^2)/x^10

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Sympy [A]  time = 2.1357, size = 49, normalized size = 0.91 \begin{align*} - \frac{6 a^{2} + 15 a b x^{2} + 30 b c x^{6} + 30 c^{2} x^{8} + x^{4} \left (20 a c + 10 b^{2}\right )}{60 x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**2/x**11,x)

[Out]

-(6*a**2 + 15*a*b*x**2 + 30*b*c*x**6 + 30*c**2*x**8 + x**4*(20*a*c + 10*b**2))/(60*x**10)

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Giac [A]  time = 1.15229, size = 65, normalized size = 1.2 \begin{align*} -\frac{30 \, c^{2} x^{8} + 30 \, b c x^{6} + 10 \, b^{2} x^{4} + 20 \, a c x^{4} + 15 \, a b x^{2} + 6 \, a^{2}}{60 \, x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^2/x^11,x, algorithm="giac")

[Out]

-1/60*(30*c^2*x^8 + 30*b*c*x^6 + 10*b^2*x^4 + 20*a*c*x^4 + 15*a*b*x^2 + 6*a^2)/x^10