3.157 \(\int \frac{x^2}{9-10 x^3+x^6} \, dx\)

Optimal. Leaf size=25 \[ \frac{1}{24} \log \left (9-x^3\right )-\frac{1}{24} \log \left (1-x^3\right ) \]

[Out]

-Log[1 - x^3]/24 + Log[9 - x^3]/24

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Rubi [A]  time = 0.0171892, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1352, 616, 31} \[ \frac{1}{24} \log \left (9-x^3\right )-\frac{1}{24} \log \left (1-x^3\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(9 - 10*x^3 + x^6),x]

[Out]

-Log[1 - x^3]/24 + Log[9 - x^3]/24

Rule 1352

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

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{x^2}{9-10 x^3+x^6} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{9-10 x+x^2} \, dx,x,x^3\right )\\ &=\frac{1}{24} \operatorname{Subst}\left (\int \frac{1}{-9+x} \, dx,x,x^3\right )-\frac{1}{24} \operatorname{Subst}\left (\int \frac{1}{-1+x} \, dx,x,x^3\right )\\ &=-\frac{1}{24} \log \left (1-x^3\right )+\frac{1}{24} \log \left (9-x^3\right )\\ \end{align*}

Mathematica [A]  time = 0.0038913, size = 25, normalized size = 1. \[ \frac{1}{24} \log \left (9-x^3\right )-\frac{1}{24} \log \left (1-x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(9 - 10*x^3 + x^6),x]

[Out]

-Log[1 - x^3]/24 + Log[9 - x^3]/24

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Maple [A]  time = 0.005, size = 18, normalized size = 0.7 \begin{align*} -{\frac{\ln \left ({x}^{3}-1 \right ) }{24}}+{\frac{\ln \left ({x}^{3}-9 \right ) }{24}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^6-10*x^3+9),x)

[Out]

-1/24*ln(x^3-1)+1/24*ln(x^3-9)

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Maxima [A]  time = 0.928193, size = 23, normalized size = 0.92 \begin{align*} -\frac{1}{24} \, \log \left (x^{3} - 1\right ) + \frac{1}{24} \, \log \left (x^{3} - 9\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6-10*x^3+9),x, algorithm="maxima")

[Out]

-1/24*log(x^3 - 1) + 1/24*log(x^3 - 9)

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Fricas [A]  time = 2.04915, size = 54, normalized size = 2.16 \begin{align*} -\frac{1}{24} \, \log \left (x^{3} - 1\right ) + \frac{1}{24} \, \log \left (x^{3} - 9\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6-10*x^3+9),x, algorithm="fricas")

[Out]

-1/24*log(x^3 - 1) + 1/24*log(x^3 - 9)

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Sympy [A]  time = 0.10225, size = 15, normalized size = 0.6 \begin{align*} \frac{\log{\left (x^{3} - 9 \right )}}{24} - \frac{\log{\left (x^{3} - 1 \right )}}{24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**6-10*x**3+9),x)

[Out]

log(x**3 - 9)/24 - log(x**3 - 1)/24

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Giac [A]  time = 1.05928, size = 26, normalized size = 1.04 \begin{align*} -\frac{1}{24} \, \log \left ({\left | x^{3} - 1 \right |}\right ) + \frac{1}{24} \, \log \left ({\left | x^{3} - 9 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6-10*x^3+9),x, algorithm="giac")

[Out]

-1/24*log(abs(x^3 - 1)) + 1/24*log(abs(x^3 - 9))