\(\int \frac {1}{x^2 (3+4 x^3+x^6)} \, dx\) [166]

Optimal result

Integrand size = 16, antiderivative size = 119 \[ \int \frac {1}{x^2 \left (3+4 x^3+x^6\right )} \, dx=-\frac {1}{3 x}+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\arctan \left (\frac {\sqrt [3]{3}-2 x}{3^{5/6}}\right )}{6\ 3^{5/6}}+\frac {1}{6} \log (1+x)-\frac {\log \left (\sqrt [3]{3}+x\right )}{18 \sqrt [3]{3}}-\frac {1}{12} \log \left (1-x+x^2\right )+\frac {\log \left (3^{2/3}-\sqrt [3]{3} x+x^2\right )}{36 \sqrt [3]{3}} \]

[Out]

-1/3/x-1/18*3^(1/6)*arctan(1/3*(3^(1/3)-2*x)*3^(1/6))+1/6*ln(1+x)-1/54*3^(2/3)*ln(3^(1/3)+x)-1/12*ln(x^2-x+1)+
1/108*3^(2/3)*ln(3^(2/3)-3^(1/3)*x+x^2)+1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {1382, 1524, 298, 31, 648, 632, 210, 642, 631} \[ \int \frac {1}{x^2 \left (3+4 x^3+x^6\right )} \, dx=\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\arctan \left (\frac {\sqrt [3]{3}-2 x}{3^{5/6}}\right )}{6\ 3^{5/6}}-\frac {1}{12} \log \left (x^2-x+1\right )+\frac {\log \left (x^2-\sqrt [3]{3} x+3^{2/3}\right )}{36 \sqrt [3]{3}}-\frac {1}{3 x}+\frac {1}{6} \log (x+1)-\frac {\log \left (x+\sqrt [3]{3}\right )}{18 \sqrt [3]{3}} \]

[In]

Int[1/(x^2*(3 + 4*x^3 + x^6)),x]

[Out]

-1/3*1/x + ArcTan[(1 - 2*x)/Sqrt[3]]/(2*Sqrt[3]) - ArcTan[(3^(1/3) - 2*x)/3^(5/6)]/(6*3^(5/6)) + Log[1 + x]/6
- Log[3^(1/3) + x]/(18*3^(1/3)) - Log[1 - x + x^2]/12 + Log[3^(2/3) - 3^(1/3)*x + x^2]/(36*3^(1/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 632

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

Rule 642

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

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), 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 1382

Int[((d_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a +
b*x^n + c*x^(2*n))^(p + 1)/(a*d*(m + 1))), x] - Dist[1/(a*d^n*(m + 1)), Int[(d*x)^(m + n)*(b*(m + n*(p + 1) +
1) + c*(m + 2*n*(p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[n2, 2
*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntegerQ[p]

Rule 1524

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Wi
th[{q = Rt[b^2 - 4*a*c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Dist[e/
2 - (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[n2
, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 x}+\frac {1}{3} \int \frac {x \left (-4-x^3\right )}{3+4 x^3+x^6} \, dx \\ & = -\frac {1}{3 x}+\frac {1}{6} \int \frac {x}{3+x^3} \, dx-\frac {1}{2} \int \frac {x}{1+x^3} \, dx \\ & = -\frac {1}{3 x}+\frac {1}{6} \int \frac {1}{1+x} \, dx-\frac {1}{6} \int \frac {1+x}{1-x+x^2} \, dx-\frac {\int \frac {1}{\sqrt [3]{3}+x} \, dx}{18 \sqrt [3]{3}}+\frac {\int \frac {\sqrt [3]{3}+x}{3^{2/3}-\sqrt [3]{3} x+x^2} \, dx}{18 \sqrt [3]{3}} \\ & = -\frac {1}{3 x}+\frac {1}{6} \log (1+x)-\frac {\log \left (\sqrt [3]{3}+x\right )}{18 \sqrt [3]{3}}-\frac {1}{12} \int \frac {-1+2 x}{1-x+x^2} \, dx+\frac {1}{12} \int \frac {1}{3^{2/3}-\sqrt [3]{3} x+x^2} \, dx-\frac {1}{4} \int \frac {1}{1-x+x^2} \, dx+\frac {\int \frac {-\sqrt [3]{3}+2 x}{3^{2/3}-\sqrt [3]{3} x+x^2} \, dx}{36 \sqrt [3]{3}} \\ & = -\frac {1}{3 x}+\frac {1}{6} \log (1+x)-\frac {\log \left (\sqrt [3]{3}+x\right )}{18 \sqrt [3]{3}}-\frac {1}{12} \log \left (1-x+x^2\right )+\frac {\log \left (3^{2/3}-\sqrt [3]{3} x+x^2\right )}{36 \sqrt [3]{3}}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 x}{\sqrt [3]{3}}\right )}{6 \sqrt [3]{3}} \\ & = -\frac {1}{3 x}+\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{3}-2 x}{3^{5/6}}\right )}{6\ 3^{5/6}}+\frac {1}{6} \log (1+x)-\frac {\log \left (\sqrt [3]{3}+x\right )}{18 \sqrt [3]{3}}-\frac {1}{12} \log \left (1-x+x^2\right )+\frac {\log \left (3^{2/3}-\sqrt [3]{3} x+x^2\right )}{36 \sqrt [3]{3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^2 \left (3+4 x^3+x^6\right )} \, dx=-\frac {36+6 \sqrt [6]{3} x \arctan \left (\frac {\sqrt [3]{3}-2 x}{3^{5/6}}\right )+18 \sqrt {3} x \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )-18 x \log (1+x)+2\ 3^{2/3} x \log \left (3+3^{2/3} x\right )+9 x \log \left (1-x+x^2\right )-3^{2/3} x \log \left (3-3^{2/3} x+\sqrt [3]{3} x^2\right )}{108 x} \]

[In]

Integrate[1/(x^2*(3 + 4*x^3 + x^6)),x]

[Out]

-1/108*(36 + 6*3^(1/6)*x*ArcTan[(3^(1/3) - 2*x)/3^(5/6)] + 18*Sqrt[3]*x*ArcTan[(-1 + 2*x)/Sqrt[3]] - 18*x*Log[
1 + x] + 2*3^(2/3)*x*Log[3 + 3^(2/3)*x] + 9*x*Log[1 - x + x^2] - 3^(2/3)*x*Log[3 - 3^(2/3)*x + 3^(1/3)*x^2])/x

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.51

method	result	size
risch	\(-\frac {1}{3 x}+\frac {\ln \left (x +1\right )}{6}-\frac {\ln \left (x^{2}-x +1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (3 \textit {\_Z}^{3}+1\right )}{\sum }\textit {\_R} \ln \left (3 \textit {\_R}^{2}+x \right )\right )}{18}\)	\(61\)
default	\(\frac {\ln \left (x +1\right )}{6}-\frac {3^{\frac {2}{3}} \ln \left (3^{\frac {1}{3}}+x \right )}{54}+\frac {3^{\frac {2}{3}} \ln \left (3^{\frac {2}{3}}-3^{\frac {1}{3}} x +x^{2}\right )}{108}+\frac {3^{\frac {1}{6}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \,3^{\frac {2}{3}} x}{3}-1\right )}{3}\right )}{18}-\frac {\ln \left (x^{2}-x +1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {1}{3 x}\)	\(89\)

[In]

int(1/x^2/(x^6+4*x^3+3),x,method=_RETURNVERBOSE)

[Out]

-1/3/x+1/6*ln(x+1)-1/12*ln(x^2-x+1)-1/6*3^(1/2)*arctan(2/3*(x-1/2)*3^(1/2))+1/18*sum(_R*ln(3*_R^2+x),_R=RootOf
(3*_Z^3+1))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^2 \left (3+4 x^3+x^6\right )} \, dx=-\frac {3^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x \log \left (-3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} x + x^{2} - 3^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}}\right ) - 2 \cdot 3^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x \log \left (3^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + x\right ) + 18 \, \sqrt {3} x \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 6 \cdot 3^{\frac {1}{6}} \left (-1\right )^{\frac {1}{3}} x \arctan \left (\frac {1}{3} \cdot 3^{\frac {1}{6}} {\left (2 \, \left (-1\right )^{\frac {1}{3}} x + 3^{\frac {1}{3}}\right )}\right ) + 9 \, x \log \left (x^{2} - x + 1\right ) - 18 \, x \log \left (x + 1\right ) + 36}{108 \, x} \]

[In]

integrate(1/x^2/(x^6+4*x^3+3),x, algorithm="fricas")

[Out]

-1/108*(3^(2/3)*(-1)^(1/3)*x*log(-3^(1/3)*(-1)^(2/3)*x + x^2 - 3^(2/3)*(-1)^(1/3)) - 2*3^(2/3)*(-1)^(1/3)*x*lo
g(3^(1/3)*(-1)^(2/3) + x) + 18*sqrt(3)*x*arctan(1/3*sqrt(3)*(2*x - 1)) - 6*3^(1/6)*(-1)^(1/3)*x*arctan(1/3*3^(
1/6)*(2*(-1)^(1/3)*x + 3^(1/3))) + 9*x*log(x^2 - x + 1) - 18*x*log(x + 1) + 36)/x

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.17 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^2 \left (3+4 x^3+x^6\right )} \, dx=\frac {\log {\left (x + 1 \right )}}{6} + \left (- \frac {1}{12} - \frac {\sqrt {3} i}{12}\right ) \log {\left (x - \frac {8188128 \left (- \frac {1}{12} - \frac {\sqrt {3} i}{12}\right )^{5}}{41} + \frac {39384 \left (- \frac {1}{12} - \frac {\sqrt {3} i}{12}\right )^{2}}{41} \right )} + \left (- \frac {1}{12} + \frac {\sqrt {3} i}{12}\right ) \log {\left (x + \frac {39384 \left (- \frac {1}{12} + \frac {\sqrt {3} i}{12}\right )^{2}}{41} - \frac {8188128 \left (- \frac {1}{12} + \frac {\sqrt {3} i}{12}\right )^{5}}{41} \right )} + \operatorname {RootSum} {\left (17496 t^{3} + 1, \left ( t \mapsto t \log {\left (- \frac {8188128 t^{5}}{41} + \frac {39384 t^{2}}{41} + x \right )} \right )\right )} - \frac {1}{3 x} \]

[In]

integrate(1/x**2/(x**6+4*x**3+3),x)

[Out]

log(x + 1)/6 + (-1/12 - sqrt(3)*I/12)*log(x - 8188128*(-1/12 - sqrt(3)*I/12)**5/41 + 39384*(-1/12 - sqrt(3)*I/
12)**2/41) + (-1/12 + sqrt(3)*I/12)*log(x + 39384*(-1/12 + sqrt(3)*I/12)**2/41 - 8188128*(-1/12 + sqrt(3)*I/12
)**5/41) + RootSum(17496*_t**3 + 1, Lambda(_t, _t*log(-8188128*_t**5/41 + 39384*_t**2/41 + x))) - 1/(3*x)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^2 \left (3+4 x^3+x^6\right )} \, dx=\frac {1}{108} \cdot 3^{\frac {2}{3}} \log \left (x^{2} - 3^{\frac {1}{3}} x + 3^{\frac {2}{3}}\right ) - \frac {1}{54} \cdot 3^{\frac {2}{3}} \log \left (x + 3^{\frac {1}{3}}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{18} \cdot 3^{\frac {1}{6}} \arctan \left (\frac {1}{3} \cdot 3^{\frac {1}{6}} {\left (2 \, x - 3^{\frac {1}{3}}\right )}\right ) - \frac {1}{3 \, x} - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) \]

[In]

integrate(1/x^2/(x^6+4*x^3+3),x, algorithm="maxima")

[Out]

1/108*3^(2/3)*log(x^2 - 3^(1/3)*x + 3^(2/3)) - 1/54*3^(2/3)*log(x + 3^(1/3)) - 1/6*sqrt(3)*arctan(1/3*sqrt(3)*
(2*x - 1)) + 1/18*3^(1/6)*arctan(1/3*3^(1/6)*(2*x - 3^(1/3))) - 1/3/x - 1/12*log(x^2 - x + 1) + 1/6*log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^2 \left (3+4 x^3+x^6\right )} \, dx=\frac {1}{108} \cdot 3^{\frac {2}{3}} \log \left (x^{2} - 3^{\frac {1}{3}} x + 3^{\frac {2}{3}}\right ) - \frac {1}{54} \cdot 3^{\frac {2}{3}} \log \left ({\left | x + 3^{\frac {1}{3}} \right |}\right ) - \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{18} \cdot 3^{\frac {1}{6}} \arctan \left (\frac {1}{3} \cdot 3^{\frac {1}{6}} {\left (2 \, x - 3^{\frac {1}{3}}\right )}\right ) - \frac {1}{3 \, x} - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

integrate(1/x^2/(x^6+4*x^3+3),x, algorithm="giac")

[Out]

1/108*3^(2/3)*log(x^2 - 3^(1/3)*x + 3^(2/3)) - 1/54*3^(2/3)*log(abs(x + 3^(1/3))) - 1/6*sqrt(3)*arctan(1/3*sqr
t(3)*(2*x - 1)) + 1/18*3^(1/6)*arctan(1/3*3^(1/6)*(2*x - 3^(1/3))) - 1/3/x - 1/12*log(x^2 - x + 1) + 1/6*log(a
bs(x + 1))

Mupad [B] (verification not implemented)

Time = 8.39 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \left (3+4 x^3+x^6\right )} \, dx=\frac {\ln \left (x+1\right )}{6}-\frac {3^{2/3}\,\ln \left (x+3^{1/3}\right )}{54}+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\frac {1}{3\,x}-\frac {{\left (-1\right )}^{1/3}\,\ln \left (x-\frac {{\left (-1\right )}^{1/3}\,3^{1/3}}{2}-\frac {{\left (-1\right )}^{1/6}\,3^{5/6}}{2}+\frac {3^{1/3}}{2}\right )\,\left (3^{2/3}+3^{1/6}\,3{}\mathrm {i}\right )}{108}+\frac {{\left (-1\right )}^{1/3}\,3^{2/3}\,\ln \left (x+{\left (-1\right )}^{2/3}\,3^{1/3}\right )}{54} \]

[In]

int(1/(x^2*(4*x^3 + x^6 + 3)),x)

[Out]

log(x + 1)/6 - (3^(2/3)*log(x + 3^(1/3)))/54 + log(x - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/12 - 1/12) - log(x
+ (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/12 + 1/12) - 1/(3*x) - ((-1)^(1/3)*log(x - ((-1)^(1/3)*3^(1/3))/2 - ((-1
)^(1/6)*3^(5/6))/2 + 3^(1/3)/2)*(3^(2/3) + 3^(1/6)*3i))/108 + ((-1)^(1/3)*3^(2/3)*log(x + (-1)^(2/3)*3^(1/3)))
/54