3.154 \(\int \frac {x^2}{3+4 x^3+x^6} \, dx\)

Optimal. Leaf size=10 \[ -\frac {1}{3} \tanh ^{-1}\left (x^3+2\right ) \]

[Out]

-1/3*arctanh(x^3+2)

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Rubi [B]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 2.10, 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} \[ \frac {1}{6} \log \left (x^3+1\right )-\frac {1}{6} \log \left (x^3+3\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(3 + 4*x^3 + x^6),x]

[Out]

Log[1 + x^3]/6 - Log[3 + x^3]/6

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}{3+4 x^3+x^6} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{3+4 x+x^2} \, dx,x,x^3\right )\\ &=\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^3\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{3+x} \, dx,x,x^3\right )\\ &=\frac {1}{6} \log \left (1+x^3\right )-\frac {1}{6} \log \left (3+x^3\right )\\ \end {align*}

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Mathematica [B]  time = 0.00, size = 21, normalized size = 2.10 \[ \frac {1}{6} \log \left (x^3+1\right )-\frac {1}{6} \log \left (x^3+3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(3 + 4*x^3 + x^6),x]

[Out]

Log[1 + x^3]/6 - Log[3 + x^3]/6

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fricas [B]  time = 1.13, size = 17, normalized size = 1.70 \[ -\frac {1}{6} \, \log \left (x^{3} + 3\right ) + \frac {1}{6} \, \log \left (x^{3} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6+4*x^3+3),x, algorithm="fricas")

[Out]

-1/6*log(x^3 + 3) + 1/6*log(x^3 + 1)

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giac [B]  time = 0.33, size = 19, normalized size = 1.90 \[ -\frac {1}{6} \, \log \left ({\left | x^{3} + 3 \right |}\right ) + \frac {1}{6} \, \log \left ({\left | x^{3} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6+4*x^3+3),x, algorithm="giac")

[Out]

-1/6*log(abs(x^3 + 3)) + 1/6*log(abs(x^3 + 1))

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maple [B]  time = 0.00, size = 18, normalized size = 1.80 \[ \frac {\ln \left (x^{3}+1\right )}{6}-\frac {\ln \left (x^{3}+3\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^6+4*x^3+3),x)

[Out]

1/6*ln(x^3+1)-1/6*ln(x^3+3)

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maxima [B]  time = 0.47, size = 17, normalized size = 1.70 \[ -\frac {1}{6} \, \log \left (x^{3} + 3\right ) + \frac {1}{6} \, \log \left (x^{3} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^6+4*x^3+3),x, algorithm="maxima")

[Out]

-1/6*log(x^3 + 3) + 1/6*log(x^3 + 1)

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mupad [B]  time = 0.38, size = 16, normalized size = 1.60 \[ \frac {\mathrm {atanh}\left (\frac {9}{2\,\left (8\,x^3+6\right )}+\frac {5}{4}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(4*x^3 + x^6 + 3),x)

[Out]

atanh(9/(2*(8*x^3 + 6)) + 5/4)/3

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sympy [A]  time = 0.11, size = 15, normalized size = 1.50 \[ \frac {\log {\left (x^{3} + 1 \right )}}{6} - \frac {\log {\left (x^{3} + 3 \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**6+4*x**3+3),x)

[Out]

log(x**3 + 1)/6 - log(x**3 + 3)/6

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