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

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

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 31

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

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 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]

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*}

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Mathematica [A]  time = 0.00, size = 25, normalized size = 1.00 \[ \frac {1}{24} \log \left (9-x^3\right )-\frac {1}{24} \log \left (1-x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(9 - 10*x^3 + x^6),x]

[Out]

-1/24*Log[1 - x^3] + Log[9 - x^3]/24

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IntegrateAlgebraic [A]  time = 0.01, size = 16, normalized size = 0.64 \[ \frac {1}{12} \tanh ^{-1}\left (\frac {5}{4}-\frac {x^3}{4}\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/(9 - 10*x^3 + x^6),x]

[Out]

ArcTanh[5/4 - x^3/4]/12

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fricas [A]  time = 1.40, size = 17, normalized size = 0.68 \[ -\frac {1}{24} \, \log \left (x^{3} - 1\right ) + \frac {1}{24} \, \log \left (x^{3} - 9\right ) \]

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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giac [A]  time = 0.91, size = 19, normalized size = 0.76 \[ -\frac {1}{24} \, \log \left ({\left | x^{3} - 1 \right |}\right ) + \frac {1}{24} \, \log \left ({\left | x^{3} - 9 \right |}\right ) \]

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))

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maple [A]  time = 0.03, size = 18, normalized size = 0.72

method	result	size
default	\(\frac {\ln \left (x^{3}-9\right )}{24}-\frac {\ln \left (x^{3}-1\right )}{24}\)	\(18\)
risch	\(\frac {\ln \left (x^{3}-9\right )}{24}-\frac {\ln \left (x^{3}-1\right )}{24}\)	\(18\)
norman	\(-\frac {\ln \left (-1+x \right )}{24}+\frac {\ln \left (x^{3}-9\right )}{24}-\frac {\ln \left (x^{2}+x +1\right )}{24}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 17, normalized size = 0.68 \[ -\frac {1}{24} \, \log \left (x^{3} - 1\right ) + \frac {1}{24} \, \log \left (x^{3} - 9\right ) \]

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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mupad [B]  time = 0.58, size = 16, normalized size = 0.64 \[ \frac {\mathrm {atanh}\left (\frac {81}{320\,\left (\frac {5\,x^3}{4}-\frac {9}{8}\right )}-\frac {41}{40}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^6 - 10*x^3 + 9),x)

[Out]

atanh(81/(320*((5*x^3)/4 - 9/8)) - 41/40)/12

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sympy [A]  time = 0.11, size = 15, normalized size = 0.60 \[ \frac {\log {\left (x^{3} - 9 \right )}}{24} - \frac {\log {\left (x^{3} - 1 \right )}}{24} \]

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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