3.694 \(\int \frac {x^6}{2+3 x^4} \, dx\)

Optimal. Leaf size=104 \[ \frac {x^3}{9}-\frac {\log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{6\ 6^{3/4}}+\frac {\log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{6\ 6^{3/4}}+\frac {\tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{3\ 6^{3/4}}-\frac {\tan ^{-1}\left (\sqrt [4]{6} x+1\right )}{3\ 6^{3/4}} \]

[Out]

1/9*x^3-1/18*arctan(-1+6^(1/4)*x)*6^(1/4)-1/18*arctan(1+6^(1/4)*x)*6^(1/4)-1/36*ln(-6^(3/4)*x+3*x^2+6^(1/2))*6
^(1/4)+1/36*ln(6^(3/4)*x+3*x^2+6^(1/2))*6^(1/4)

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Rubi [A]  time = 0.10, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {321, 297, 1162, 617, 204, 1165, 628} \[ \frac {x^3}{9}-\frac {\log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{6\ 6^{3/4}}+\frac {\log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{6\ 6^{3/4}}+\frac {\tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{3\ 6^{3/4}}-\frac {\tan ^{-1}\left (\sqrt [4]{6} x+1\right )}{3\ 6^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(2 + 3*x^4),x]

[Out]

x^3/9 + ArcTan[1 - 6^(1/4)*x]/(3*6^(3/4)) - ArcTan[1 + 6^(1/4)*x]/(3*6^(3/4)) - Log[Sqrt[6] - 6^(3/4)*x + 3*x^
2]/(6*6^(3/4)) + Log[Sqrt[6] + 6^(3/4)*x + 3*x^2]/(6*6^(3/4))

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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {x^6}{2+3 x^4} \, dx &=\frac {x^3}{9}-\frac {2}{3} \int \frac {x^2}{2+3 x^4} \, dx\\ &=\frac {x^3}{9}+\frac {\int \frac {\sqrt {2}-\sqrt {3} x^2}{2+3 x^4} \, dx}{3 \sqrt {3}}-\frac {\int \frac {\sqrt {2}+\sqrt {3} x^2}{2+3 x^4} \, dx}{3 \sqrt {3}}\\ &=\frac {x^3}{9}-\frac {1}{18} \int \frac {1}{\sqrt {\frac {2}{3}}-\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx-\frac {1}{18} \int \frac {1}{\sqrt {\frac {2}{3}}+\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx-\frac {\int \frac {\frac {2^{3/4}}{\sqrt [4]{3}}+2 x}{-\sqrt {\frac {2}{3}}-\frac {2^{3/4} x}{\sqrt [4]{3}}-x^2} \, dx}{6\ 6^{3/4}}-\frac {\int \frac {\frac {2^{3/4}}{\sqrt [4]{3}}-2 x}{-\sqrt {\frac {2}{3}}+\frac {2^{3/4} x}{\sqrt [4]{3}}-x^2} \, dx}{6\ 6^{3/4}}\\ &=\frac {x^3}{9}-\frac {\log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{6\ 6^{3/4}}+\frac {\log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{6\ 6^{3/4}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt [4]{6} x\right )}{3\ 6^{3/4}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt [4]{6} x\right )}{3\ 6^{3/4}}\\ &=\frac {x^3}{9}+\frac {\tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{3\ 6^{3/4}}-\frac {\tan ^{-1}\left (1+\sqrt [4]{6} x\right )}{3\ 6^{3/4}}-\frac {\log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{6\ 6^{3/4}}+\frac {\log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{6\ 6^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 98, normalized size = 0.94 \[ \frac {1}{36} \left (4 x^3-\sqrt [4]{6} \log \left (\sqrt {6} x^2-2 \sqrt [4]{6} x+2\right )+\sqrt [4]{6} \log \left (\sqrt {6} x^2+2 \sqrt [4]{6} x+2\right )+2 \sqrt [4]{6} \tan ^{-1}\left (1-\sqrt [4]{6} x\right )-2 \sqrt [4]{6} \tan ^{-1}\left (\sqrt [4]{6} x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(2 + 3*x^4),x]

[Out]

(4*x^3 + 2*6^(1/4)*ArcTan[1 - 6^(1/4)*x] - 2*6^(1/4)*ArcTan[1 + 6^(1/4)*x] - 6^(1/4)*Log[2 - 2*6^(1/4)*x + Sqr
t[6]*x^2] + 6^(1/4)*Log[2 + 2*6^(1/4)*x + Sqrt[6]*x^2])/36

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fricas [B]  time = 0.65, size = 164, normalized size = 1.58 \[ \frac {1}{9} \, x^{3} + \frac {1}{162} \cdot 54^{\frac {3}{4}} \sqrt {2} \arctan \left (-\frac {1}{18} \cdot 54^{\frac {3}{4}} \sqrt {2} x + \frac {1}{54} \cdot 54^{\frac {3}{4}} \sqrt {2} \sqrt {9 \, x^{2} + 3 \cdot 54^{\frac {1}{4}} \sqrt {2} x + 3 \, \sqrt {6}} - 1\right ) + \frac {1}{162} \cdot 54^{\frac {3}{4}} \sqrt {2} \arctan \left (-\frac {1}{18} \cdot 54^{\frac {3}{4}} \sqrt {2} x + \frac {1}{54} \cdot 54^{\frac {3}{4}} \sqrt {2} \sqrt {9 \, x^{2} - 3 \cdot 54^{\frac {1}{4}} \sqrt {2} x + 3 \, \sqrt {6}} + 1\right ) + \frac {1}{648} \cdot 54^{\frac {3}{4}} \sqrt {2} \log \left (9 \, x^{2} + 3 \cdot 54^{\frac {1}{4}} \sqrt {2} x + 3 \, \sqrt {6}\right ) - \frac {1}{648} \cdot 54^{\frac {3}{4}} \sqrt {2} \log \left (9 \, x^{2} - 3 \cdot 54^{\frac {1}{4}} \sqrt {2} x + 3 \, \sqrt {6}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*x^3 + 1/162*54^(3/4)*sqrt(2)*arctan(-1/18*54^(3/4)*sqrt(2)*x + 1/54*54^(3/4)*sqrt(2)*sqrt(9*x^2 + 3*54^(1/
4)*sqrt(2)*x + 3*sqrt(6)) - 1) + 1/162*54^(3/4)*sqrt(2)*arctan(-1/18*54^(3/4)*sqrt(2)*x + 1/54*54^(3/4)*sqrt(2
)*sqrt(9*x^2 - 3*54^(1/4)*sqrt(2)*x + 3*sqrt(6)) + 1) + 1/648*54^(3/4)*sqrt(2)*log(9*x^2 + 3*54^(1/4)*sqrt(2)*
x + 3*sqrt(6)) - 1/648*54^(3/4)*sqrt(2)*log(9*x^2 - 3*54^(1/4)*sqrt(2)*x + 3*sqrt(6))

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giac [A]  time = 0.19, size = 100, normalized size = 0.96 \[ \frac {1}{9} \, x^{3} - \frac {1}{18} \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) - \frac {1}{18} \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{36} \cdot 6^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) - \frac {1}{36} \cdot 6^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*x^3 - 1/18*6^(1/4)*arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x + sqrt(2)*(2/3)^(1/4))) - 1/18*6^(1/4)*arctan(3/4*s
qrt(2)*(2/3)^(3/4)*(2*x - sqrt(2)*(2/3)^(1/4))) + 1/36*6^(1/4)*log(x^2 + sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3)) -
1/36*6^(1/4)*log(x^2 - sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3))

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maple [A]  time = 0.01, size = 116, normalized size = 1.12 \[ \frac {x^{3}}{9}-\frac {\sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}-1\right )}{108}-\frac {\sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}+1\right )}{108}-\frac {\sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}-\frac {\sqrt {3}\, 6^{\frac {1}{4}} \sqrt {2}\, x}{3}+\frac {\sqrt {6}}{3}}{x^{2}+\frac {\sqrt {3}\, 6^{\frac {1}{4}} \sqrt {2}\, x}{3}+\frac {\sqrt {6}}{3}}\right )}{216} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(3*x^4+2),x)

[Out]

1/9*x^3-1/216*3^(1/2)*6^(3/4)*2^(1/2)*ln((x^2-1/3*3^(1/2)*6^(1/4)*x*2^(1/2)+1/3*6^(1/2))/(x^2+1/3*3^(1/2)*6^(1
/4)*x*2^(1/2)+1/3*6^(1/2)))-1/108*3^(1/2)*6^(3/4)*2^(1/2)*arctan(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x+1)-1/108*3^(1/2
)*6^(3/4)*2^(1/2)*arctan(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x-1)

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maxima [A]  time = 2.94, size = 126, normalized size = 1.21 \[ \frac {1}{9} \, x^{3} - \frac {1}{18} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x + 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) - \frac {1}{18} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x - 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) + \frac {1}{36} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} + 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) - \frac {1}{36} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} - 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*x^3 - 1/18*3^(1/4)*2^(1/4)*arctan(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x + 3^(1/4)*2^(3/4))) - 1/18*3^(1/4)*2^(1
/4)*arctan(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x - 3^(1/4)*2^(3/4))) + 1/36*3^(1/4)*2^(1/4)*log(sqrt(3)*x^2 + 3^(1/
4)*2^(3/4)*x + sqrt(2)) - 1/36*3^(1/4)*2^(1/4)*log(sqrt(3)*x^2 - 3^(1/4)*2^(3/4)*x + sqrt(2))

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mupad [B]  time = 0.12, size = 38, normalized size = 0.37 \[ \frac {x^3}{9}+6^{1/4}\,\mathrm {atan}\left (6^{1/4}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{18}+\frac {1}{18}{}\mathrm {i}\right )+6^{1/4}\,\mathrm {atan}\left (6^{1/4}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{18}-\frac {1}{18}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(3*x^4 + 2),x)

[Out]

x^3/9 - 6^(1/4)*atan(6^(1/4)*x*(1/2 + 1i/2))*(1/18 + 1i/18) - 6^(1/4)*atan(6^(1/4)*x*(1/2 - 1i/2))*(1/18 - 1i/
18)

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sympy [A]  time = 0.47, size = 92, normalized size = 0.88 \[ \frac {x^{3}}{9} - \frac {\sqrt [4]{6} \log {\left (x^{2} - \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{36} + \frac {\sqrt [4]{6} \log {\left (x^{2} + \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{36} - \frac {\sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x - 1 \right )}}{18} - \frac {\sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x + 1 \right )}}{18} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3/9 - 6**(1/4)*log(x**2 - 6**(3/4)*x/3 + sqrt(6)/3)/36 + 6**(1/4)*log(x**2 + 6**(3/4)*x/3 + sqrt(6)/3)/36 -
6**(1/4)*atan(6**(1/4)*x - 1)/18 - 6**(1/4)*atan(6**(1/4)*x + 1)/18

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