3.692 \(\int \frac{1}{x (2+3 x^4)} \, dx\)

Optimal. Leaf size=19 \[ \frac{\log (x)}{2}-\frac{1}{8} \log \left (3 x^4+2\right ) \]

[Out]

Log[x]/2 - Log[2 + 3*x^4]/8

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Rubi [A]  time = 0.0081227, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {266, 36, 29, 31} \[ \frac{\log (x)}{2}-\frac{1}{8} \log \left (3 x^4+2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(2 + 3*x^4)),x]

[Out]

Log[x]/2 - Log[2 + 3*x^4]/8

Rule 266

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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{1}{x \left (2+3 x^4\right )} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x (2+3 x)} \, dx,x,x^4\right )\\ &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^4\right )-\frac{3}{8} \operatorname{Subst}\left (\int \frac{1}{2+3 x} \, dx,x,x^4\right )\\ &=\frac{\log (x)}{2}-\frac{1}{8} \log \left (2+3 x^4\right )\\ \end{align*}

Mathematica [A]  time = 0.0034749, size = 19, normalized size = 1. \[ \frac{\log (x)}{2}-\frac{1}{8} \log \left (3 x^4+2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(2 + 3*x^4)),x]

[Out]

Log[x]/2 - Log[2 + 3*x^4]/8

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Maple [A]  time = 0.003, size = 16, normalized size = 0.8 \begin{align*}{\frac{\ln \left ( x \right ) }{2}}-{\frac{\ln \left ( 3\,{x}^{4}+2 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(3*x^4+2),x)

[Out]

1/2*ln(x)-1/8*ln(3*x^4+2)

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Maxima [A]  time = 1.02881, size = 23, normalized size = 1.21 \begin{align*} -\frac{1}{8} \, \log \left (3 \, x^{4} + 2\right ) + \frac{1}{8} \, \log \left (x^{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(3*x^4+2),x, algorithm="maxima")

[Out]

-1/8*log(3*x^4 + 2) + 1/8*log(x^4)

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Fricas [A]  time = 1.62523, size = 46, normalized size = 2.42 \begin{align*} -\frac{1}{8} \, \log \left (3 \, x^{4} + 2\right ) + \frac{1}{2} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(3*x^4+2),x, algorithm="fricas")

[Out]

-1/8*log(3*x^4 + 2) + 1/2*log(x)

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Sympy [A]  time = 0.128638, size = 14, normalized size = 0.74 \begin{align*} \frac{\log{\left (x \right )}}{2} - \frac{\log{\left (3 x^{4} + 2 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(3*x**4+2),x)

[Out]

log(x)/2 - log(3*x**4 + 2)/8

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Giac [A]  time = 1.12402, size = 23, normalized size = 1.21 \begin{align*} -\frac{1}{8} \, \log \left (3 \, x^{4} + 2\right ) + \frac{1}{8} \, \log \left (x^{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(3*x^4+2),x, algorithm="giac")

[Out]

-1/8*log(3*x^4 + 2) + 1/8*log(x^4)