3.1.92 \(\int \frac {x^9}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=19 \[ \frac {x^4}{4 b \left (b+c x^2\right )^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1584, 264} \begin {gather*} \frac {x^4}{4 b \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(b*x^2 + c*x^4)^3,x]

[Out]

x^4/(4*b*(b + c*x^2)^2)

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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x^9}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {x^3}{\left (b+c x^2\right )^3} \, dx\\ &=\frac {x^4}{4 b \left (b+c x^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.26 \begin {gather*} -\frac {b+2 c x^2}{4 c^2 \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(b*x^2 + c*x^4)^3,x]

[Out]

-1/4*(b + 2*c*x^2)/(c^2*(b + c*x^2)^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^9}{\left (b x^2+c x^4\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^9/(b*x^2 + c*x^4)^3,x]

[Out]

IntegrateAlgebraic[x^9/(b*x^2 + c*x^4)^3, x]

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fricas [B]  time = 0.81, size = 36, normalized size = 1.89 \begin {gather*} -\frac {2 \, c x^{2} + b}{4 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 22, normalized size = 1.16 \begin {gather*} -\frac {2 \, c x^{2} + b}{4 \, {\left (c x^{2} + b\right )}^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-1/4*(2*c*x^2 + b)/((c*x^2 + b)^2*c^2)

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maple [A]  time = 0.01, size = 31, normalized size = 1.63 \begin {gather*} \frac {b}{4 \left (c \,x^{2}+b \right )^{2} c^{2}}-\frac {1}{2 \left (c \,x^{2}+b \right ) c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(c*x^4+b*x^2)^3,x)

[Out]

1/4*b/c^2/(c*x^2+b)^2-1/2/c^2/(c*x^2+b)

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maxima [B]  time = 1.33, size = 36, normalized size = 1.89 \begin {gather*} -\frac {2 \, c x^{2} + b}{4 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{2} + b^{2} c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.18, size = 37, normalized size = 1.95 \begin {gather*} -\frac {\frac {b}{4\,c^2}+\frac {x^2}{2\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^2 + c*x^4)^3,x)

[Out]

-(b/(4*c^2) + x^2/(2*c))/(b^2 + c^2*x^4 + 2*b*c*x^2)

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sympy [B]  time = 0.31, size = 36, normalized size = 1.89 \begin {gather*} \frac {- b - 2 c x^{2}}{4 b^{2} c^{2} + 8 b c^{3} x^{2} + 4 c^{4} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(c*x**4+b*x**2)**3,x)

[Out]

(-b - 2*c*x**2)/(4*b**2*c**2 + 8*b*c**3*x**2 + 4*c**4*x**4)

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