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

Optimal. Leaf size=113 \[ \frac {\log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{32\ 6^{3/4}}-\frac {\log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{32\ 6^{3/4}}+\frac {x^3}{8 \left (3 x^4+2\right )}-\frac {\tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{16\ 6^{3/4}}+\frac {\tan ^{-1}\left (\sqrt [4]{6} x+1\right )}{16\ 6^{3/4}} \]

[Out]

1/8*x^3/(3*x^4+2)+1/96*arctan(-1+6^(1/4)*x)*6^(1/4)+1/96*arctan(1+6^(1/4)*x)*6^(1/4)+1/192*ln(-6^(3/4)*x+3*x^2
+6^(1/2))*6^(1/4)-1/192*ln(6^(3/4)*x+3*x^2+6^(1/2))*6^(1/4)

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Rubi [A]  time = 0.08, antiderivative size = 113, 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 = {290, 297, 1162, 617, 204, 1165, 628} \[ \frac {x^3}{8 \left (3 x^4+2\right )}+\frac {\log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{32\ 6^{3/4}}-\frac {\log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{32\ 6^{3/4}}-\frac {\tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{16\ 6^{3/4}}+\frac {\tan ^{-1}\left (\sqrt [4]{6} x+1\right )}{16\ 6^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(2 + 3*x^4)^2,x]

[Out]

x^3/(8*(2 + 3*x^4)) - ArcTan[1 - 6^(1/4)*x]/(16*6^(3/4)) + ArcTan[1 + 6^(1/4)*x]/(16*6^(3/4)) + Log[Sqrt[6] -
6^(3/4)*x + 3*x^2]/(32*6^(3/4)) - Log[Sqrt[6] + 6^(3/4)*x + 3*x^2]/(32*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 290

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

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 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^2}{\left (2+3 x^4\right )^2} \, dx &=\frac {x^3}{8 \left (2+3 x^4\right )}+\frac {1}{8} \int \frac {x^2}{2+3 x^4} \, dx\\ &=\frac {x^3}{8 \left (2+3 x^4\right )}-\frac {\int \frac {\sqrt {2}-\sqrt {3} x^2}{2+3 x^4} \, dx}{16 \sqrt {3}}+\frac {\int \frac {\sqrt {2}+\sqrt {3} x^2}{2+3 x^4} \, dx}{16 \sqrt {3}}\\ &=\frac {x^3}{8 \left (2+3 x^4\right )}+\frac {1}{96} \int \frac {1}{\sqrt {\frac {2}{3}}-\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx+\frac {1}{96} \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}{32\ 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}{32\ 6^{3/4}}\\ &=\frac {x^3}{8 \left (2+3 x^4\right )}+\frac {\log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{32\ 6^{3/4}}-\frac {\log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{32\ 6^{3/4}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt [4]{6} x\right )}{16\ 6^{3/4}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt [4]{6} x\right )}{16\ 6^{3/4}}\\ &=\frac {x^3}{8 \left (2+3 x^4\right )}-\frac {\tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{16\ 6^{3/4}}+\frac {\tan ^{-1}\left (1+\sqrt [4]{6} x\right )}{16\ 6^{3/4}}+\frac {\log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{32\ 6^{3/4}}-\frac {\log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{32\ 6^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 107, normalized size = 0.95 \[ \frac {1}{192} \left (\sqrt [4]{6} \log \left (\sqrt {6} x^2-2 \sqrt [4]{6} x+2\right )-\sqrt [4]{6} \log \left (\sqrt {6} x^2+2 \sqrt [4]{6} x+2\right )+\frac {24 x^3}{3 x^4+2}-2 \sqrt [4]{6} \tan ^{-1}\left (1-\sqrt [4]{6} x\right )+2 \sqrt [4]{6} \tan ^{-1}\left (\sqrt [4]{6} x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(2 + 3*x^4)^2,x]

[Out]

((24*x^3)/(2 + 3*x^4) - 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 + Sqrt[6]*x^2] - 6^(1/4)*Log[2 + 2*6^(1/4)*x + Sqrt[6]*x^2])/192

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fricas [B]  time = 0.63, size = 202, normalized size = 1.79 \[ -\frac {4 \cdot 54^{\frac {3}{4}} \sqrt {2} {\left (3 \, x^{4} + 2\right )} \arctan \left (-\frac {1}{18} \cdot 54^{\frac {3}{4}} \sqrt {2} x + \frac {1}{54} \cdot 54^{\frac {3}{4}} \sqrt {2} \sqrt {9 \, x^{2} + 3 \cdot 54^{\frac {1}{4}} \sqrt {2} x + 3 \, \sqrt {6}} - 1\right ) + 4 \cdot 54^{\frac {3}{4}} \sqrt {2} {\left (3 \, x^{4} + 2\right )} \arctan \left (-\frac {1}{18} \cdot 54^{\frac {3}{4}} \sqrt {2} x + \frac {1}{54} \cdot 54^{\frac {3}{4}} \sqrt {2} \sqrt {9 \, x^{2} - 3 \cdot 54^{\frac {1}{4}} \sqrt {2} x + 3 \, \sqrt {6}} + 1\right ) + 54^{\frac {3}{4}} \sqrt {2} {\left (3 \, x^{4} + 2\right )} \log \left (9 \, x^{2} + 3 \cdot 54^{\frac {1}{4}} \sqrt {2} x + 3 \, \sqrt {6}\right ) - 54^{\frac {3}{4}} \sqrt {2} {\left (3 \, x^{4} + 2\right )} \log \left (9 \, x^{2} - 3 \cdot 54^{\frac {1}{4}} \sqrt {2} x + 3 \, \sqrt {6}\right ) - 432 \, x^{3}}{3456 \, {\left (3 \, x^{4} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3456*(4*54^(3/4)*sqrt(2)*(3*x^4 + 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) + 4*54^(3/4)*sqrt(2)*(3*x^4 + 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) + 54^(3/4)*sqrt(2)*(3*x^4 + 2)*log(9*x^2
+ 3*54^(1/4)*sqrt(2)*x + 3*sqrt(6)) - 54^(3/4)*sqrt(2)*(3*x^4 + 2)*log(9*x^2 - 3*54^(1/4)*sqrt(2)*x + 3*sqrt(6
)) - 432*x^3)/(3*x^4 + 2)

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giac [A]  time = 0.19, size = 109, normalized size = 0.96 \[ \frac {x^{3}}{8 \, {\left (3 \, x^{4} + 2\right )}} + \frac {1}{96} \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{96} \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) - \frac {1}{192} \cdot 6^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) + \frac {1}{192} \cdot 6^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^3/(3*x^4 + 2) + 1/96*6^(1/4)*arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x + sqrt(2)*(2/3)^(1/4))) + 1/96*6^(1/4)*
arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x - sqrt(2)*(2/3)^(1/4))) - 1/192*6^(1/4)*log(x^2 + sqrt(2)*(2/3)^(1/4)*x +
sqrt(2/3)) + 1/192*6^(1/4)*log(x^2 - sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3))

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maple [A]  time = 0.00, size = 125, normalized size = 1.11 \[ \frac {x^{3}}{24 x^{4}+16}+\frac {\sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}-1\right )}{576}+\frac {\sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}+1\right )}{576}+\frac {\sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}-\frac {\sqrt {3}\, 6^{\frac {1}{4}} \sqrt {2}\, x}{3}+\frac {\sqrt {6}}{3}}{x^{2}+\frac {\sqrt {3}\, 6^{\frac {1}{4}} \sqrt {2}\, x}{3}+\frac {\sqrt {6}}{3}}\right )}{1152} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^3/(3*x^4+2)+1/576*3^(1/2)*6^(3/4)*2^(1/2)*arctan(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x-1)+1/1152*3^(1/2)*6^(3/4)
*2^(1/2)*ln((x^2-1/3*3^(1/2)*6^(1/4)*2^(1/2)*x+1/3*6^(1/2))/(x^2+1/3*3^(1/2)*6^(1/4)*2^(1/2)*x+1/3*6^(1/2)))+1
/576*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 = 3.17, size = 135, normalized size = 1.19 \[ \frac {x^{3}}{8 \, {\left (3 \, x^{4} + 2\right )}} + \frac {1}{96} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x + 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) + \frac {1}{96} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x - 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) - \frac {1}{192} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} + 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) + \frac {1}{192} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} - 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x^3/(3*x^4 + 2) + 1/96*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))) + 1/96*
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))) - 1/192*3^(1/4)*2^(1/4)*log(sqrt(3
)*x^2 + 3^(1/4)*2^(3/4)*x + sqrt(2)) + 1/192*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 = 46, normalized size = 0.41 \[ \frac {x^3}{24\,\left (x^4+\frac {2}{3}\right )}+6^{1/4}\,\mathrm {atan}\left (6^{1/4}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{96}-\frac {1}{96}{}\mathrm {i}\right )+6^{1/4}\,\mathrm {atan}\left (6^{1/4}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{96}+\frac {1}{96}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(3*x^4 + 2)^2,x)

[Out]

6^(1/4)*atan(6^(1/4)*x*(1/2 - 1i/2))*(1/96 - 1i/96) + 6^(1/4)*atan(6^(1/4)*x*(1/2 + 1i/2))*(1/96 + 1i/96) + x^
3/(24*(x^4 + 2/3))

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sympy [A]  time = 0.48, size = 97, normalized size = 0.86 \[ \frac {x^{3}}{24 x^{4} + 16} + \frac {\sqrt [4]{6} \log {\left (x^{2} - \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{192} - \frac {\sqrt [4]{6} \log {\left (x^{2} + \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{192} + \frac {\sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x - 1 \right )}}{96} + \frac {\sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x + 1 \right )}}{96} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3/(24*x**4 + 16) + 6**(1/4)*log(x**2 - 6**(3/4)*x/3 + sqrt(6)/3)/192 - 6**(1/4)*log(x**2 + 6**(3/4)*x/3 + s
qrt(6)/3)/192 + 6**(1/4)*atan(6**(1/4)*x - 1)/96 + 6**(1/4)*atan(6**(1/4)*x + 1)/96

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