3.153 \(\int \frac {x^5}{3+4 x^3+x^6} \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{2} \log \left (x^3+3\right )-\frac {1}{6} \log \left (x^3+1\right ) \]

[Out]

-1/6*ln(x^3+1)+1/2*ln(x^3+3)

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Rubi [A]  time = 0.01, antiderivative size = 21, 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 = {1357, 632, 31} \[ \frac {1}{2} \log \left (x^3+3\right )-\frac {1}{6} \log \left (x^3+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^5/(3 + 4*x^3 + x^6),x]

[Out]

-Log[1 + x^3]/6 + Log[3 + x^3]/2

Rule 31

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

Rule 632

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

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.00 \[ \frac {1}{2} \log \left (x^3+3\right )-\frac {1}{6} \log \left (x^3+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(3 + 4*x^3 + x^6),x]

[Out]

-1/6*Log[1 + x^3] + Log[3 + x^3]/2

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fricas [A]  time = 0.92, size = 17, normalized size = 0.81 \[ \frac {1}{2} \, \log \left (x^{3} + 3\right ) - \frac {1}{6} \, \log \left (x^{3} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^6+4*x^3+3),x, algorithm="fricas")

[Out]

1/2*log(x^3 + 3) - 1/6*log(x^3 + 1)

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giac [A]  time = 0.36, size = 19, normalized size = 0.90 \[ \frac {1}{2} \, \log \left ({\left | x^{3} + 3 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x^{3} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^6+4*x^3+3),x, algorithm="giac")

[Out]

1/2*log(abs(x^3 + 3)) - 1/6*log(abs(x^3 + 1))

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maple [A]  time = 0.00, size = 18, normalized size = 0.86 \[ -\frac {\ln \left (x^{3}+1\right )}{6}+\frac {\ln \left (x^{3}+3\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(x^6+4*x^3+3),x)

[Out]

-1/6*ln(x^3+1)+1/2*ln(x^3+3)

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maxima [A]  time = 0.52, size = 17, normalized size = 0.81 \[ \frac {1}{2} \, \log \left (x^{3} + 3\right ) - \frac {1}{6} \, \log \left (x^{3} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^6+4*x^3+3),x, algorithm="maxima")

[Out]

1/2*log(x^3 + 3) - 1/6*log(x^3 + 1)

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mupad [B]  time = 0.05, size = 17, normalized size = 0.81 \[ \frac {\ln \left (x^3+3\right )}{2}-\frac {\ln \left (x^3+1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(4*x^3 + x^6 + 3),x)

[Out]

log(x^3 + 3)/2 - log(x^3 + 1)/6

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sympy [A]  time = 0.12, size = 15, normalized size = 0.71 \[ - \frac {\log {\left (x^{3} + 1 \right )}}{6} + \frac {\log {\left (x^{3} + 3 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(x**6+4*x**3+3),x)

[Out]

-log(x**3 + 1)/6 + log(x**3 + 3)/2

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