3.8.1 \(\int \frac {1}{x (2+3 x^4)^2} \, dx\) [701]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, 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 = {272, 46} \begin {gather*} \frac {1}{8 \left (3 x^4+2\right )}-\frac {1}{16} \log \left (3 x^4+2\right )+\frac {\log (x)}{4} \end {gather*}

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 46

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] && Lt
Q[m + n + 2, 0])

Rule 272

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]]

Rubi steps

\begin {align*} \int \frac {1}{x \left (2+3 x^4\right )^2} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {1}{x (2+3 x)^2} \, dx,x,x^4\right )\\ &=\frac {1}{4} \text {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*}

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Mathematica [A]
time = 0.01, size = 32, normalized size = 1.00 \begin {gather*} \frac {1}{8 \left (2+3 x^4\right )}+\frac {\log (x)}{4}-\frac {1}{16} \log \left (2+3 x^4\right ) \end {gather*}

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 = 0.14, size = 27, normalized size = 0.84

method	result	size
risch	\(\frac {1}{24 x^{4}+16}+\frac {\ln \left (x \right )}{4}-\frac {\ln \left (3 x^{4}+2\right )}{16}\)	\(25\)
default	\(\frac {1}{24 x^{4}+16}+\frac {\ln \left (x \right )}{4}-\frac {\ln \left (3 x^{4}+2\right )}{16}\)	\(27\)
norman	\(-\frac {3 x^{4}}{16 \left (3 x^{4}+2\right )}+\frac {\ln \left (x \right )}{4}-\frac {\ln \left (3 x^{4}+2\right )}{16}\)	\(30\)
meijerg	\(-\frac {3 x^{4}}{16 \left (3 x^{4}+2\right )}-\frac {\ln \left (1+\frac {3 x^{4}}{2}\right )}{16}+\frac {1}{16}+\frac {\ln \left (x \right )}{4}-\frac {\ln \left (2\right )}{16}+\frac {\ln \left (3\right )}{16}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.29, size = 28, normalized size = 0.88 \begin {gather*} \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 {gather*}

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 = 0.37, size = 40, normalized size = 1.25 \begin {gather*} -\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 {gather*}

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 = 0.05, size = 22, normalized size = 0.69 \begin {gather*} \frac {\log {\left (x \right )}}{4} - \frac {\log {\left (3 x^{4} + 2 \right )}}{16} + \frac {1}{24 x^{4} + 16} \end {gather*}

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 = 0.49, size = 35, normalized size = 1.09 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.07, size = 22, normalized size = 0.69 \begin {gather*} \frac {\ln \left (x\right )}{4}-\frac {\ln \left (x^4+\frac {2}{3}\right )}{16}+\frac {1}{24\,\left (x^4+\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(3*x^4 + 2)^2),x)

[Out]

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

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