3.706 \(\int \frac {1}{x^2 (2+3 x^4)^2} \, dx\)

Optimal. Leaf size=140 \[ \frac {1}{8 x \left (3 x^4+2\right )}-\frac {5 \sqrt [4]{3} \log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{64\ 2^{3/4}}+\frac {5 \sqrt [4]{3} \log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{64\ 2^{3/4}}-\frac {5}{16 x}+\frac {5 \sqrt [4]{3} \tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{32\ 2^{3/4}}-\frac {5 \sqrt [4]{3} \tan ^{-1}\left (\sqrt [4]{6} x+1\right )}{32\ 2^{3/4}} \]

[Out]

-5/16/x+1/8/x/(3*x^4+2)-5/64*3^(1/4)*arctan(-1+6^(1/4)*x)*2^(1/4)-5/64*3^(1/4)*arctan(1+6^(1/4)*x)*2^(1/4)-5/1
28*3^(1/4)*ln(-6^(3/4)*x+3*x^2+6^(1/2))*2^(1/4)+5/128*3^(1/4)*ln(6^(3/4)*x+3*x^2+6^(1/2))*2^(1/4)

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Rubi [A]  time = 0.08, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {290, 325, 297, 1162, 617, 204, 1165, 628} \[ \frac {1}{8 x \left (3 x^4+2\right )}-\frac {5 \sqrt [4]{3} \log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{64\ 2^{3/4}}+\frac {5 \sqrt [4]{3} \log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{64\ 2^{3/4}}-\frac {5}{16 x}+\frac {5 \sqrt [4]{3} \tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{32\ 2^{3/4}}-\frac {5 \sqrt [4]{3} \tan ^{-1}\left (\sqrt [4]{6} x+1\right )}{32\ 2^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(2 + 3*x^4)^2),x]

[Out]

-5/(16*x) + 1/(8*x*(2 + 3*x^4)) + (5*3^(1/4)*ArcTan[1 - 6^(1/4)*x])/(32*2^(3/4)) - (5*3^(1/4)*ArcTan[1 + 6^(1/
4)*x])/(32*2^(3/4)) - (5*3^(1/4)*Log[Sqrt[6] - 6^(3/4)*x + 3*x^2])/(64*2^(3/4)) + (5*3^(1/4)*Log[Sqrt[6] + 6^(
3/4)*x + 3*x^2])/(64*2^(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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 {1}{x^2 \left (2+3 x^4\right )^2} \, dx &=\frac {1}{8 x \left (2+3 x^4\right )}+\frac {5}{8} \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx\\ &=-\frac {5}{16 x}+\frac {1}{8 x \left (2+3 x^4\right )}-\frac {15}{16} \int \frac {x^2}{2+3 x^4} \, dx\\ &=-\frac {5}{16 x}+\frac {1}{8 x \left (2+3 x^4\right )}+\frac {1}{32} \left (5 \sqrt {3}\right ) \int \frac {\sqrt {2}-\sqrt {3} x^2}{2+3 x^4} \, dx-\frac {1}{32} \left (5 \sqrt {3}\right ) \int \frac {\sqrt {2}+\sqrt {3} x^2}{2+3 x^4} \, dx\\ &=-\frac {5}{16 x}+\frac {1}{8 x \left (2+3 x^4\right )}-\frac {5}{64} \int \frac {1}{\sqrt {\frac {2}{3}}-\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx-\frac {5}{64} \int \frac {1}{\sqrt {\frac {2}{3}}+\frac {2^{3/4} x}{\sqrt [4]{3}}+x^2} \, dx-\frac {\left (5 \sqrt [4]{3}\right ) \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}{64\ 2^{3/4}}-\frac {\left (5 \sqrt [4]{3}\right ) \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}{64\ 2^{3/4}}\\ &=-\frac {5}{16 x}+\frac {1}{8 x \left (2+3 x^4\right )}-\frac {5 \sqrt [4]{3} \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{64\ 2^{3/4}}+\frac {5 \sqrt [4]{3} \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{64\ 2^{3/4}}-\frac {\left (5 \sqrt [4]{3}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt [4]{6} x\right )}{32\ 2^{3/4}}+\frac {\left (5 \sqrt [4]{3}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt [4]{6} x\right )}{32\ 2^{3/4}}\\ &=-\frac {5}{16 x}+\frac {1}{8 x \left (2+3 x^4\right )}+\frac {5 \sqrt [4]{3} \tan ^{-1}\left (1-\sqrt [4]{6} x\right )}{32\ 2^{3/4}}-\frac {5 \sqrt [4]{3} \tan ^{-1}\left (1+\sqrt [4]{6} x\right )}{32\ 2^{3/4}}-\frac {5 \sqrt [4]{3} \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{64\ 2^{3/4}}+\frac {5 \sqrt [4]{3} \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{64\ 2^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 113, normalized size = 0.81 \[ \frac {1}{128} \left (-5 \sqrt [4]{6} \log \left (\sqrt {6} x^2-2 \sqrt [4]{6} x+2\right )+5 \sqrt [4]{6} \log \left (\sqrt {6} x^2+2 \sqrt [4]{6} x+2\right )-\frac {24 x^3}{3 x^4+2}-\frac {32}{x}+10 \sqrt [4]{6} \tan ^{-1}\left (1-\sqrt [4]{6} x\right )-10 \sqrt [4]{6} \tan ^{-1}\left (\sqrt [4]{6} x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(2 + 3*x^4)^2),x]

[Out]

(-32/x - (24*x^3)/(2 + 3*x^4) + 10*6^(1/4)*ArcTan[1 - 6^(1/4)*x] - 10*6^(1/4)*ArcTan[1 + 6^(1/4)*x] - 5*6^(1/4
)*Log[2 - 2*6^(1/4)*x + Sqrt[6]*x^2] + 5*6^(1/4)*Log[2 + 2*6^(1/4)*x + Sqrt[6]*x^2])/128

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fricas [B]  time = 0.72, size = 220, normalized size = 1.57 \[ -\frac {120 \, x^{4} - 20 \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} {\left (3 \, x^{5} + 2 \, x\right )} \arctan \left (\frac {1}{3} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} \sqrt {3^{\frac {3}{4}} 2^{\frac {3}{4}} x + 3 \, x^{2} + \sqrt {3} \sqrt {2}} - 3^{\frac {1}{4}} 2^{\frac {1}{4}} x - 1\right ) - 20 \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} {\left (3 \, x^{5} + 2 \, x\right )} \arctan \left (\frac {1}{3} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} \sqrt {-3^{\frac {3}{4}} 2^{\frac {3}{4}} x + 3 \, x^{2} + \sqrt {3} \sqrt {2}} - 3^{\frac {1}{4}} 2^{\frac {1}{4}} x + 1\right ) - 5 \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} {\left (3 \, x^{5} + 2 \, x\right )} \log \left (3^{\frac {3}{4}} 2^{\frac {3}{4}} x + 3 \, x^{2} + \sqrt {3} \sqrt {2}\right ) + 5 \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} {\left (3 \, x^{5} + 2 \, x\right )} \log \left (-3^{\frac {3}{4}} 2^{\frac {3}{4}} x + 3 \, x^{2} + \sqrt {3} \sqrt {2}\right ) + 64}{128 \, {\left (3 \, x^{5} + 2 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(120*x^4 - 20*3^(1/4)*2^(1/4)*(3*x^5 + 2*x)*arctan(1/3*3^(3/4)*2^(1/4)*sqrt(3^(3/4)*2^(3/4)*x + 3*x^2 +
sqrt(3)*sqrt(2)) - 3^(1/4)*2^(1/4)*x - 1) - 20*3^(1/4)*2^(1/4)*(3*x^5 + 2*x)*arctan(1/3*3^(3/4)*2^(1/4)*sqrt(
-3^(3/4)*2^(3/4)*x + 3*x^2 + sqrt(3)*sqrt(2)) - 3^(1/4)*2^(1/4)*x + 1) - 5*3^(1/4)*2^(1/4)*(3*x^5 + 2*x)*log(3
^(3/4)*2^(3/4)*x + 3*x^2 + sqrt(3)*sqrt(2)) + 5*3^(1/4)*2^(1/4)*(3*x^5 + 2*x)*log(-3^(3/4)*2^(3/4)*x + 3*x^2 +
sqrt(3)*sqrt(2)) + 64)/(3*x^5 + 2*x)

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giac [A]  time = 0.19, size = 115, normalized size = 0.82 \[ -\frac {5}{64} \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) - \frac {5}{64} \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) + \frac {5}{128} \cdot 6^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) - \frac {5}{128} \cdot 6^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) - \frac {15 \, x^{4} + 8}{16 \, {\left (3 \, x^{5} + 2 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*6^(1/4)*arctan(3/4*sqrt(2)*(2/3)^(3/4)*(2*x + sqrt(2)*(2/3)^(1/4))) - 5/64*6^(1/4)*arctan(3/4*sqrt(2)*(2
/3)^(3/4)*(2*x - sqrt(2)*(2/3)^(1/4))) + 5/128*6^(1/4)*log(x^2 + sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3)) - 5/128*6^
(1/4)*log(x^2 - sqrt(2)*(2/3)^(1/4)*x + sqrt(2/3)) - 1/16*(15*x^4 + 8)/(3*x^5 + 2*x)

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maple [A]  time = 0.01, size = 128, normalized size = 0.91 \[ -\frac {x^{3}}{16 \left (x^{4}+\frac {2}{3}\right )}-\frac {5 \sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}-1\right )}{384}-\frac {5 \sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}+1\right )}{384}-\frac {5 \sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}-\frac {\sqrt {3}\, 6^{\frac {1}{4}} \sqrt {2}\, x}{3}+\frac {\sqrt {6}}{3}}{x^{2}+\frac {\sqrt {3}\, 6^{\frac {1}{4}} \sqrt {2}\, x}{3}+\frac {\sqrt {6}}{3}}\right )}{768}-\frac {1}{4 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(3*x^4+2)^2,x)

[Out]

-1/16*x^3/(x^4+2/3)-5/768*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)))-5/384*3^(1/2)*6^(3/4)*2^(1/2)*arctan(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x+1)-
5/384*3^(1/2)*6^(3/4)*2^(1/2)*arctan(1/6*2^(1/2)*3^(1/2)*6^(3/4)*x-1)-1/4/x

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maxima [A]  time = 3.06, size = 141, normalized size = 1.01 \[ -\frac {5}{64} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x + 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) - \frac {5}{64} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x - 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) + \frac {5}{128} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} + 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) - \frac {5}{128} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} - 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) - \frac {15 \, x^{4} + 8}{16 \, {\left (3 \, x^{5} + 2 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*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))) - 5/64*3^(1/4)*2^(1/4)*arcta
n(1/6*3^(3/4)*2^(1/4)*(2*sqrt(3)*x - 3^(1/4)*2^(3/4))) + 5/128*3^(1/4)*2^(1/4)*log(sqrt(3)*x^2 + 3^(1/4)*2^(3/
4)*x + sqrt(2)) - 5/128*3^(1/4)*2^(1/4)*log(sqrt(3)*x^2 - 3^(1/4)*2^(3/4)*x + sqrt(2)) - 1/16*(15*x^4 + 8)/(3*
x^5 + 2*x)

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mupad [B]  time = 1.10, size = 51, normalized size = 0.36 \[ -\frac {\frac {5\,x^4}{16}+\frac {1}{6}}{x^5+\frac {2\,x}{3}}+6^{1/4}\,\mathrm {atan}\left (6^{1/4}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{64}+\frac {5}{64}{}\mathrm {i}\right )+6^{1/4}\,\mathrm {atan}\left (6^{1/4}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{64}-\frac {5}{64}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(3*x^4 + 2)^2),x)

[Out]

- 6^(1/4)*atan(6^(1/4)*x*(1/2 - 1i/2))*(5/64 - 5i/64) - 6^(1/4)*atan(6^(1/4)*x*(1/2 + 1i/2))*(5/64 + 5i/64) -
((5*x^4)/16 + 1/6)/((2*x)/3 + x^5)

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sympy [A]  time = 0.67, size = 110, normalized size = 0.79 \[ \frac {- 15 x^{4} - 8}{48 x^{5} + 32 x} - \frac {5 \sqrt [4]{6} \log {\left (x^{2} - \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{128} + \frac {5 \sqrt [4]{6} \log {\left (x^{2} + \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{128} - \frac {5 \sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x - 1 \right )}}{64} - \frac {5 \sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x + 1 \right )}}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-15*x**4 - 8)/(48*x**5 + 32*x) - 5*6**(1/4)*log(x**2 - 6**(3/4)*x/3 + sqrt(6)/3)/128 + 5*6**(1/4)*log(x**2 +
6**(3/4)*x/3 + sqrt(6)/3)/128 - 5*6**(1/4)*atan(6**(1/4)*x - 1)/64 - 5*6**(1/4)*atan(6**(1/4)*x + 1)/64

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