3.161 \(\int \frac {x^6}{3+4 x^3+x^6} \, dx\)

Optimal. Leaf size=113 \[ -\frac {1}{12} \log \left (x^2-x+1\right )+\frac {1}{4} \sqrt [3]{3} \log \left (x^2-\sqrt [3]{3} x+3^{2/3}\right )+x+\frac {1}{6} \log (x+1)-\frac {1}{2} \sqrt [3]{3} \log \left (x+\sqrt [3]{3}\right )-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{2} 3^{5/6} \tan ^{-1}\left (\frac {\sqrt [3]{3}-2 x}{3^{5/6}}\right ) \]

[Out]

x+1/2*3^(5/6)*arctan(1/3*(3^(1/3)-2*x)*3^(1/6))+1/6*ln(1+x)-1/2*3^(1/3)*ln(3^(1/3)+x)-1/12*ln(x^2-x+1)+1/4*3^(
1/3)*ln(3^(2/3)-3^(1/3)*x+x^2)-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {1367, 1422, 200, 31, 634, 618, 204, 628, 617} \[ -\frac {1}{12} \log \left (x^2-x+1\right )+\frac {1}{4} \sqrt [3]{3} \log \left (x^2-\sqrt [3]{3} x+3^{2/3}\right )+x+\frac {1}{6} \log (x+1)-\frac {1}{2} \sqrt [3]{3} \log \left (x+\sqrt [3]{3}\right )-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{2} 3^{5/6} \tan ^{-1}\left (\frac {\sqrt [3]{3}-2 x}{3^{5/6}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^6/(3 + 4*x^3 + x^6),x]

[Out]

x - ArcTan[(1 - 2*x)/Sqrt[3]]/(2*Sqrt[3]) + (3^(5/6)*ArcTan[(3^(1/3) - 2*x)/3^(5/6)])/2 + Log[1 + x]/6 - (3^(1
/3)*Log[3^(1/3) + x])/2 - Log[1 - x + x^2]/12 + (3^(1/3)*Log[3^(2/3) - 3^(1/3)*x + x^2])/4

Rule 31

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

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 204

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

Rule 617

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 618

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 628

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 634

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 1367

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

Rule 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 0])

Rubi steps

\begin {align*} \int \frac {x^6}{3+4 x^3+x^6} \, dx &=x-\int \frac {3+4 x^3}{3+4 x^3+x^6} \, dx\\ &=x+\frac {1}{2} \int \frac {1}{1+x^3} \, dx-\frac {9}{2} \int \frac {1}{3+x^3} \, dx\\ &=x+\frac {1}{6} \int \frac {1}{1+x} \, dx+\frac {1}{6} \int \frac {2-x}{1-x+x^2} \, dx-\frac {1}{2} \sqrt [3]{3} \int \frac {1}{\sqrt [3]{3}+x} \, dx-\frac {1}{2} \sqrt [3]{3} \int \frac {2 \sqrt [3]{3}-x}{3^{2/3}-\sqrt [3]{3} x+x^2} \, dx\\ &=x+\frac {1}{6} \log (1+x)-\frac {1}{2} \sqrt [3]{3} \log \left (\sqrt [3]{3}+x\right )-\frac {1}{12} \int \frac {-1+2 x}{1-x+x^2} \, dx+\frac {1}{4} \int \frac {1}{1-x+x^2} \, dx+\frac {1}{4} \sqrt [3]{3} \int \frac {-\sqrt [3]{3}+2 x}{3^{2/3}-\sqrt [3]{3} x+x^2} \, dx-\frac {1}{4} \left (3\ 3^{2/3}\right ) \int \frac {1}{3^{2/3}-\sqrt [3]{3} x+x^2} \, dx\\ &=x+\frac {1}{6} \log (1+x)-\frac {1}{2} \sqrt [3]{3} \log \left (\sqrt [3]{3}+x\right )-\frac {1}{12} \log \left (1-x+x^2\right )+\frac {1}{4} \sqrt [3]{3} \log \left (3^{2/3}-\sqrt [3]{3} x+x^2\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )-\frac {1}{2} \left (3 \sqrt [3]{3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 x}{\sqrt [3]{3}}\right )\\ &=x-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{2} 3^{5/6} \tan ^{-1}\left (\frac {\sqrt [3]{3}-2 x}{3^{5/6}}\right )+\frac {1}{6} \log (1+x)-\frac {1}{2} \sqrt [3]{3} \log \left (\sqrt [3]{3}+x\right )-\frac {1}{12} \log \left (1-x+x^2\right )+\frac {1}{4} \sqrt [3]{3} \log \left (3^{2/3}-\sqrt [3]{3} x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 111, normalized size = 0.98 \[ \frac {1}{12} \left (-\log \left (x^2-x+1\right )+3 \sqrt [3]{3} \log \left (\sqrt [3]{3} x^2-3^{2/3} x+3\right )+12 x+2 \log (x+1)-6 \sqrt [3]{3} \log \left (3^{2/3} x+3\right )+6\ 3^{5/6} \tan ^{-1}\left (\frac {\sqrt [3]{3}-2 x}{3^{5/6}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(3 + 4*x^3 + x^6),x]

[Out]

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

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fricas [A]  time = 1.13, size = 88, normalized size = 0.78 \[ \frac {1}{2} \, \sqrt {3} \left (-3\right )^{\frac {1}{3}} \arctan \left (\frac {1}{9} \, \sqrt {3} {\left (2 \, \left (-3\right )^{\frac {2}{3}} x - 3\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{4} \, \left (-3\right )^{\frac {1}{3}} \log \left (x^{2} + \left (-3\right )^{\frac {1}{3}} x + \left (-3\right )^{\frac {2}{3}}\right ) + \frac {1}{2} \, \left (-3\right )^{\frac {1}{3}} \log \left (x - \left (-3\right )^{\frac {1}{3}}\right ) + x - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(3)*(-3)^(1/3)*arctan(1/9*sqrt(3)*(2*(-3)^(2/3)*x - 3)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) -
1/4*(-3)^(1/3)*log(x^2 + (-3)^(1/3)*x + (-3)^(2/3)) + 1/2*(-3)^(1/3)*log(x - (-3)^(1/3)) + x - 1/12*log(x^2 -
x + 1) + 1/6*log(x + 1)

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giac [A]  time = 0.45, size = 87, normalized size = 0.77 \[ -\frac {1}{2} \cdot 3^{\frac {5}{6}} \arctan \left (\frac {1}{3} \cdot 3^{\frac {1}{6}} {\left (2 \, x - 3^{\frac {1}{3}}\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \cdot 3^{\frac {1}{3}} \log \left (x^{2} - 3^{\frac {1}{3}} x + 3^{\frac {2}{3}}\right ) - \frac {1}{2} \cdot 3^{\frac {1}{3}} \log \left ({\left | x + 3^{\frac {1}{3}} \right |}\right ) + x - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*3^(5/6)*arctan(1/3*3^(1/6)*(2*x - 3^(1/3))) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/4*3^(1/3)*log
(x^2 - 3^(1/3)*x + 3^(2/3)) - 1/2*3^(1/3)*log(abs(x + 3^(1/3))) + x - 1/12*log(x^2 - x + 1) + 1/6*log(abs(x +
1))

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maple [A]  time = 0.01, size = 85, normalized size = 0.75 \[ x -\frac {3^{\frac {5}{6}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \,3^{\frac {2}{3}} x}{3}-1\right )}{3}\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\ln \left (x +1\right )}{6}-\frac {3^{\frac {1}{3}} \ln \left (x +3^{\frac {1}{3}}\right )}{2}+\frac {3^{\frac {1}{3}} \ln \left (x^{2}-3^{\frac {1}{3}} x +3^{\frac {2}{3}}\right )}{4}-\frac {\ln \left (x^{2}-x +1\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^6+4*x^3+3),x)

[Out]

x+1/6*ln(x+1)-1/2*3^(1/3)*ln(x+3^(1/3))+1/4*3^(1/3)*ln(x^2-3^(1/3)*x+3^(2/3))-1/2*3^(5/6)*arctan(1/3*3^(1/2)*(
2/3*3^(2/3)*x-1))-1/12*ln(x^2-x+1)+1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))

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maxima [A]  time = 1.23, size = 85, normalized size = 0.75 \[ -\frac {1}{2} \cdot 3^{\frac {5}{6}} \arctan \left (\frac {1}{3} \cdot 3^{\frac {1}{6}} {\left (2 \, x - 3^{\frac {1}{3}}\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \cdot 3^{\frac {1}{3}} \log \left (x^{2} - 3^{\frac {1}{3}} x + 3^{\frac {2}{3}}\right ) - \frac {1}{2} \cdot 3^{\frac {1}{3}} \log \left (x + 3^{\frac {1}{3}}\right ) + x - \frac {1}{12} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*3^(5/6)*arctan(1/3*3^(1/6)*(2*x - 3^(1/3))) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/4*3^(1/3)*log
(x^2 - 3^(1/3)*x + 3^(2/3)) - 1/2*3^(1/3)*log(x + 3^(1/3)) + x - 1/12*log(x^2 - x + 1) + 1/6*log(x + 1)

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mupad [B]  time = 0.16, size = 104, normalized size = 0.92 \[ x+\frac {\ln \left (x+1\right )}{6}-\frac {3^{1/3}\,\ln \left (x+3^{1/3}\right )}{2}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\ln \left (x-\frac {3^{1/3}}{2}+\frac {3^{5/6}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {3^{1/3}}{4}-\frac {3^{5/6}\,1{}\mathrm {i}}{4}\right )+\frac {{\left (-1\right )}^{1/3}\,3^{1/3}\,\ln \left (x-{\left (-1\right )}^{1/3}\,3^{1/3}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(4*x^3 + x^6 + 3),x)

[Out]

x + log(x + 1)/6 - (3^(1/3)*log(x + 3^(1/3)))/2 - 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) + log(x - 3^(1/3)/2 + (3^(5/6)*1i)/2)*(3^(1/3)/4 - (3^(5/6
)*1i)/4) + ((-1)^(1/3)*3^(1/3)*log(x - (-1)^(1/3)*3^(1/3)))/2

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sympy [C]  time = 0.61, size = 126, normalized size = 1.12 \[ x + \frac {\log {\left (x + 1 \right )}}{6} + \left (- \frac {1}{12} - \frac {\sqrt {3} i}{12}\right ) \log {\left (x - \frac {121}{246} - \frac {121 \sqrt {3} i}{246} + \frac {864 \left (- \frac {1}{12} - \frac {\sqrt {3} i}{12}\right )^{4}}{41} \right )} + \left (- \frac {1}{12} + \frac {\sqrt {3} i}{12}\right ) \log {\left (x - \frac {121}{246} + \frac {864 \left (- \frac {1}{12} + \frac {\sqrt {3} i}{12}\right )^{4}}{41} + \frac {121 \sqrt {3} i}{246} \right )} + \operatorname {RootSum} {\left (8 t^{3} + 3, \left (t \mapsto t \log {\left (\frac {864 t^{4}}{41} + \frac {242 t}{41} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + log(x + 1)/6 + (-1/12 - sqrt(3)*I/12)*log(x - 121/246 - 121*sqrt(3)*I/246 + 864*(-1/12 - sqrt(3)*I/12)**4/
41) + (-1/12 + sqrt(3)*I/12)*log(x - 121/246 + 864*(-1/12 + sqrt(3)*I/12)**4/41 + 121*sqrt(3)*I/246) + RootSum
(8*_t**3 + 3, Lambda(_t, _t*log(864*_t**4/41 + 242*_t/41 + x)))

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