\(\int \frac {1}{x (3+4 x^3+x^6)} \, dx\) [155]

Optimal result

Integrand size = 16, antiderivative size = 27 \[ \int \frac {1}{x \left (3+4 x^3+x^6\right )} \, dx=\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^3\right )+\frac {1}{18} \log \left (3+x^3\right ) \]

[Out]

1/3*ln(x)-1/6*ln(x^3+1)+1/18*ln(x^3+3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1371, 719, 29, 646, 31} \[ \int \frac {1}{x \left (3+4 x^3+x^6\right )} \, dx=-\frac {1}{6} \log \left (x^3+1\right )+\frac {1}{18} \log \left (x^3+3\right )+\frac {\log (x)}{3} \]

[In]

Int[1/(x*(3 + 4*x^3 + x^6)),x]

[Out]

Log[x]/3 - Log[1 + x^3]/6 + Log[3 + x^3]/18

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]

Rule 646

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 719

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

Rule 1371

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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{x \left (3+4 x+x^2\right )} \, dx,x,x^3\right ) \\ & = \frac {1}{9} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^3\right )+\frac {1}{9} \text {Subst}\left (\int \frac {-4-x}{3+4 x+x^2} \, dx,x,x^3\right ) \\ & = \frac {\log (x)}{3}+\frac {1}{18} \text {Subst}\left (\int \frac {1}{3+x} \, dx,x,x^3\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^3\right ) \\ & = \frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^3\right )+\frac {1}{18} \log \left (3+x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (3+4 x^3+x^6\right )} \, dx=\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^3\right )+\frac {1}{18} \log \left (3+x^3\right ) \]

[In]

Integrate[1/(x*(3 + 4*x^3 + x^6)),x]

[Out]

Log[x]/3 - Log[1 + x^3]/6 + Log[3 + x^3]/18

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x^{3}+1\right )}{6}+\frac {\ln \left (x^{3}+3\right )}{18}\)	\(22\)
default	\(\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{3}+3\right )}{18}-\frac {\ln \left (x^{2}-x +1\right )}{6}\)	\(31\)
norman	\(\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{3}+3\right )}{18}-\frac {\ln \left (x^{2}-x +1\right )}{6}\)	\(31\)
parallelrisch	\(\frac {\ln \left (x \right )}{3}-\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{3}+3\right )}{18}-\frac {\ln \left (x^{2}-x +1\right )}{6}\)	\(31\)

[In]

int(1/x/(x^6+4*x^3+3),x,method=_RETURNVERBOSE)

[Out]

1/3*ln(x)-1/6*ln(x^3+1)+1/18*ln(x^3+3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (3+4 x^3+x^6\right )} \, dx=\frac {1}{18} \, \log \left (x^{3} + 3\right ) - \frac {1}{6} \, \log \left (x^{3} + 1\right ) + \frac {1}{3} \, \log \left (x\right ) \]

[In]

integrate(1/x/(x^6+4*x^3+3),x, algorithm="fricas")

[Out]

1/18*log(x^3 + 3) - 1/6*log(x^3 + 1) + 1/3*log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x \left (3+4 x^3+x^6\right )} \, dx=\frac {\log {\left (x \right )}}{3} - \frac {\log {\left (x^{3} + 1 \right )}}{6} + \frac {\log {\left (x^{3} + 3 \right )}}{18} \]

[In]

integrate(1/x/(x**6+4*x**3+3),x)

[Out]

log(x)/3 - log(x**3 + 1)/6 + log(x**3 + 3)/18

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x \left (3+4 x^3+x^6\right )} \, dx=\frac {1}{18} \, \log \left (x^{3} + 3\right ) - \frac {1}{6} \, \log \left (x^{3} + 1\right ) + \frac {1}{9} \, \log \left (x^{3}\right ) \]

[In]

integrate(1/x/(x^6+4*x^3+3),x, algorithm="maxima")

[Out]

1/18*log(x^3 + 3) - 1/6*log(x^3 + 1) + 1/9*log(x^3)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (3+4 x^3+x^6\right )} \, dx=\frac {1}{18} \, \log \left ({\left | x^{3} + 3 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x^{3} + 1 \right |}\right ) + \frac {1}{3} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/x/(x^6+4*x^3+3),x, algorithm="giac")

[Out]

1/18*log(abs(x^3 + 3)) - 1/6*log(abs(x^3 + 1)) + 1/3*log(abs(x))

Mupad [B] (verification not implemented)

Time = 8.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (3+4 x^3+x^6\right )} \, dx=\frac {\ln \left (x^3+3\right )}{18}-\frac {\ln \left (x^3+1\right )}{6}+\frac {\ln \left (x\right )}{3} \]

[In]

int(1/(x*(4*x^3 + x^6 + 3)),x)

[Out]

log(x^3 + 3)/18 - log(x^3 + 1)/6 + log(x)/3