\(\int \frac {1}{x (1+3 x^4+x^8)} \, dx\) [374]

Optimal result

Integrand size = 16, antiderivative size = 57 \[ \int \frac {1}{x \left (1+3 x^4+x^8\right )} \, dx=\log (x)-\frac {1}{40} \left (5+3 \sqrt {5}\right ) \log \left (3-\sqrt {5}+2 x^4\right )-\frac {1}{40} \left (5-3 \sqrt {5}\right ) \log \left (3+\sqrt {5}+2 x^4\right ) \]

[Out]

ln(x)-1/40*ln(2*x^4+5^(1/2)+3)*(5-3*5^(1/2))-1/40*ln(2*x^4-5^(1/2)+3)*(5+3*5^(1/2))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, 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} \[ \int \frac {1}{x \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{40} \left (5+3 \sqrt {5}\right ) \log \left (2 x^4-\sqrt {5}+3\right )-\frac {1}{40} \left (5-3 \sqrt {5}\right ) \log \left (2 x^4+\sqrt {5}+3\right )+\log (x) \]

[In]

Int[1/(x*(1 + 3*x^4 + x^8)),x]

[Out]

Log[x] - ((5 + 3*Sqrt[5])*Log[3 - Sqrt[5] + 2*x^4])/40 - ((5 - 3*Sqrt[5])*Log[3 + Sqrt[5] + 2*x^4])/40

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}{4} \text {Subst}\left (\int \frac {1}{x \left (1+3 x+x^2\right )} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^4\right )+\frac {1}{4} \text {Subst}\left (\int \frac {-3-x}{1+3 x+x^2} \, dx,x,x^4\right ) \\ & = \log (x)+\frac {1}{40} \left (-5+3 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x} \, dx,x,x^4\right )-\frac {1}{40} \left (5+3 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x} \, dx,x,x^4\right ) \\ & = \log (x)-\frac {1}{40} \left (5+3 \sqrt {5}\right ) \log \left (3-\sqrt {5}+2 x^4\right )-\frac {1}{40} \left (5-3 \sqrt {5}\right ) \log \left (3+\sqrt {5}+2 x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x \left (1+3 x^4+x^8\right )} \, dx=\log (x)+\frac {1}{40} \left (-5-3 \sqrt {5}\right ) \log \left (-3+\sqrt {5}-2 x^4\right )+\frac {1}{40} \left (-5+3 \sqrt {5}\right ) \log \left (3+\sqrt {5}+2 x^4\right ) \]

[In]

Integrate[1/(x*(1 + 3*x^4 + x^8)),x]

[Out]

Log[x] + ((-5 - 3*Sqrt[5])*Log[-3 + Sqrt[5] - 2*x^4])/40 + ((-5 + 3*Sqrt[5])*Log[3 + Sqrt[5] + 2*x^4])/40

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.61

method	result	size
default	\(\ln \left (x \right )-\frac {\ln \left (x^{8}+3 x^{4}+1\right )}{8}+\frac {3 \,\operatorname {arctanh}\left (\frac {\left (2 x^{4}+3\right ) \sqrt {5}}{5}\right ) \sqrt {5}}{20}\)	\(35\)
risch	\(\ln \left (x \right )-\frac {\ln \left (3 x^{4}+\frac {9}{2}+\frac {3 \sqrt {5}}{2}\right )}{8}+\frac {3 \ln \left (3 x^{4}+\frac {9}{2}+\frac {3 \sqrt {5}}{2}\right ) \sqrt {5}}{40}-\frac {3 \ln \left (3 x^{4}-\frac {3 \sqrt {5}}{2}+\frac {9}{2}\right ) \sqrt {5}}{40}-\frac {\ln \left (3 x^{4}-\frac {3 \sqrt {5}}{2}+\frac {9}{2}\right )}{8}\)	\(70\)

[In]

int(1/x/(x^8+3*x^4+1),x,method=_RETURNVERBOSE)

[Out]

ln(x)-1/8*ln(x^8+3*x^4+1)+3/20*arctanh(1/5*(2*x^4+3)*5^(1/2))*5^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (1+3 x^4+x^8\right )} \, dx=\frac {3}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{8} + 6 \, x^{4} + \sqrt {5} {\left (2 \, x^{4} + 3\right )} + 7}{x^{8} + 3 \, x^{4} + 1}\right ) - \frac {1}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) + \log \left (x\right ) \]

[In]

integrate(1/x/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

3/40*sqrt(5)*log((2*x^8 + 6*x^4 + sqrt(5)*(2*x^4 + 3) + 7)/(x^8 + 3*x^4 + 1)) - 1/8*log(x^8 + 3*x^4 + 1) + log
(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (1+3 x^4+x^8\right )} \, dx=\log {\left (x \right )} + \left (- \frac {3 \sqrt {5}}{40} - \frac {1}{8}\right ) \log {\left (x^{4} - \frac {\sqrt {5}}{2} + \frac {3}{2} \right )} + \left (- \frac {1}{8} + \frac {3 \sqrt {5}}{40}\right ) \log {\left (x^{4} + \frac {\sqrt {5}}{2} + \frac {3}{2} \right )} \]

[In]

integrate(1/x/(x**8+3*x**4+1),x)

[Out]

log(x) + (-3*sqrt(5)/40 - 1/8)*log(x**4 - sqrt(5)/2 + 3/2) + (-1/8 + 3*sqrt(5)/40)*log(x**4 + sqrt(5)/2 + 3/2)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (1+3 x^4+x^8\right )} \, dx=-\frac {3}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} + 3}{2 \, x^{4} + \sqrt {5} + 3}\right ) - \frac {1}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]

[In]

integrate(1/x/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

-3/40*sqrt(5)*log((2*x^4 - sqrt(5) + 3)/(2*x^4 + sqrt(5) + 3)) - 1/8*log(x^8 + 3*x^4 + 1) + 1/4*log(x^4)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (1+3 x^4+x^8\right )} \, dx=-\frac {3}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - \sqrt {5} + 3}{2 \, x^{4} + \sqrt {5} + 3}\right ) - \frac {1}{8} \, \log \left (x^{8} + 3 \, x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]

[In]

integrate(1/x/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

-3/40*sqrt(5)*log((2*x^4 - sqrt(5) + 3)/(2*x^4 + sqrt(5) + 3)) - 1/8*log(x^8 + 3*x^4 + 1) + 1/4*log(x^4)

Mupad [B] (verification not implemented)

Time = 8.37 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x \left (1+3 x^4+x^8\right )} \, dx=\ln \left (x\right )-\ln \left (x^4-\frac {\sqrt {5}}{2}+\frac {3}{2}\right )\,\left (\frac {3\,\sqrt {5}}{40}+\frac {1}{8}\right )+\ln \left (x^4+\frac {\sqrt {5}}{2}+\frac {3}{2}\right )\,\left (\frac {3\,\sqrt {5}}{40}-\frac {1}{8}\right ) \]

[In]

int(1/(x*(3*x^4 + x^8 + 1)),x)

[Out]

log(x) - log(x^4 - 5^(1/2)/2 + 3/2)*((3*5^(1/2))/40 + 1/8) + log(5^(1/2)/2 + x^4 + 3/2)*((3*5^(1/2))/40 - 1/8)