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

Optimal. Leaf size=32 \[ \frac{1}{8 \left (3 x^4+2\right )}-\frac{1}{16} \log \left (3 x^4+2\right )+\frac{\log (x)}{4} \]

[Out]

1/(8*(2 + 3*x^4)) + Log[x]/4 - Log[2 + 3*x^4]/16

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Rubi [A]  time = 0.0199567, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 44} \[ \frac{1}{8 \left (3 x^4+2\right )}-\frac{1}{16} \log \left (3 x^4+2\right )+\frac{\log (x)}{4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(2 + 3*x^4)^2),x]

[Out]

1/(8*(2 + 3*x^4)) + Log[x]/4 - Log[2 + 3*x^4]/16

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{x \left (2+3 x^4\right )^2} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x (2+3 x)^2} \, dx,x,x^4\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (\frac{1}{4 x}-\frac{3}{2 (2+3 x)^2}-\frac{3}{4 (2+3 x)}\right ) \, dx,x,x^4\right )\\ &=\frac{1}{8 \left (2+3 x^4\right )}+\frac{\log (x)}{4}-\frac{1}{16} \log \left (2+3 x^4\right )\\ \end{align*}

Mathematica [A]  time = 0.0081473, size = 32, normalized size = 1. \[ \frac{1}{8 \left (3 x^4+2\right )}-\frac{1}{16} \log \left (3 x^4+2\right )+\frac{\log (x)}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(2 + 3*x^4)^2),x]

[Out]

1/(8*(2 + 3*x^4)) + Log[x]/4 - Log[2 + 3*x^4]/16

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Maple [A]  time = 0.012, size = 27, normalized size = 0.8 \begin{align*}{\frac{1}{24\,{x}^{4}+16}}+{\frac{\ln \left ( x \right ) }{4}}-{\frac{\ln \left ( 3\,{x}^{4}+2 \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/(3*x^4+2)+1/4*ln(x)-1/16*ln(3*x^4+2)

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Maxima [A]  time = 0.982029, size = 38, normalized size = 1.19 \begin{align*} \frac{1}{8 \,{\left (3 \, x^{4} + 2\right )}} - \frac{1}{16} \, \log \left (3 \, x^{4} + 2\right ) + \frac{1}{16} \, \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)^2,x, algorithm="maxima")

[Out]

1/8/(3*x^4 + 2) - 1/16*log(3*x^4 + 2) + 1/16*log(x^4)

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Fricas [A]  time = 1.66831, size = 101, normalized size = 3.16 \begin{align*} -\frac{{\left (3 \, x^{4} + 2\right )} \log \left (3 \, x^{4} + 2\right ) - 4 \,{\left (3 \, x^{4} + 2\right )} \log \left (x\right ) - 2}{16 \,{\left (3 \, x^{4} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((3*x^4 + 2)*log(3*x^4 + 2) - 4*(3*x^4 + 2)*log(x) - 2)/(3*x^4 + 2)

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Sympy [A]  time = 0.160597, size = 22, normalized size = 0.69 \begin{align*} \frac{\log{\left (x \right )}}{4} - \frac{\log{\left (3 x^{4} + 2 \right )}}{16} + \frac{1}{24 x^{4} + 16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/4 - log(3*x**4 + 2)/16 + 1/(24*x**4 + 16)

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Giac [A]  time = 1.11793, size = 47, normalized size = 1.47 \begin{align*} \frac{3 \, x^{4} + 4}{16 \,{\left (3 \, x^{4} + 2\right )}} - \frac{1}{16} \, \log \left (3 \, x^{4} + 2\right ) + \frac{1}{16} \, \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)^2,x, algorithm="giac")

[Out]

1/16*(3*x^4 + 4)/(3*x^4 + 2) - 1/16*log(3*x^4 + 2) + 1/16*log(x^4)