\(\int \frac {1}{x^2 (2+3 x^4)} \, dx\) [698]

Optimal result

Integrand size = 13, antiderivative size = 124 \[ \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx=-\frac {1}{2 x}+\frac {\sqrt [4]{3} \arctan \left (1-\sqrt [4]{6} x\right )}{4\ 2^{3/4}}-\frac {\sqrt [4]{3} \arctan \left (1+\sqrt [4]{6} x\right )}{4\ 2^{3/4}}-\frac {\sqrt [4]{3} \log \left (\sqrt {6}-6^{3/4} x+3 x^2\right )}{8\ 2^{3/4}}+\frac {\sqrt [4]{3} \log \left (\sqrt {6}+6^{3/4} x+3 x^2\right )}{8\ 2^{3/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {331, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx=\frac {\sqrt [4]{3} \arctan \left (1-\sqrt [4]{6} x\right )}{4\ 2^{3/4}}-\frac {\sqrt [4]{3} \arctan \left (\sqrt [4]{6} x+1\right )}{4\ 2^{3/4}}-\frac {\sqrt [4]{3} \log \left (3 x^2-6^{3/4} x+\sqrt {6}\right )}{8\ 2^{3/4}}+\frac {\sqrt [4]{3} \log \left (3 x^2+6^{3/4} x+\sqrt {6}\right )}{8\ 2^{3/4}}-\frac {1}{2 x} \]

[In]

Int[1/(x^2*(2 + 3*x^4)),x]

[Out]

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

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 303

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 331

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 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 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 1176

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 1179

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx=-\frac {8-2 \sqrt [4]{6} x \arctan \left (1-\sqrt [4]{6} x\right )+2 \sqrt [4]{6} x \arctan \left (1+\sqrt [4]{6} x\right )+\sqrt [4]{6} x \log \left (2-2 \sqrt [4]{6} x+\sqrt {6} x^2\right )-\sqrt [4]{6} x \log \left (2+2 \sqrt [4]{6} x+\sqrt {6} x^2\right )}{16 x} \]

[In]

Integrate[1/(x^2*(2 + 3*x^4)),x]

[Out]

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

Maple [C] (verified)

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

Time = 3.99 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.24

method	result	size
risch	\(-\frac {1}{2 x}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (54 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (-18 \textit {\_R}^{3}+x \right )\right )}{8}\)	\(30\)
default	\(-\frac {\sqrt {3}\, 6^{\frac {3}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\frac {\sqrt {3}\, 6^{\frac {1}{4}} x \sqrt {2}}{3}+\frac {\sqrt {6}}{3}}{x^{2}+\frac {\sqrt {3}\, 6^{\frac {1}{4}} x \sqrt {2}}{3}+\frac {\sqrt {6}}{3}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, 6^{\frac {3}{4}} x}{6}-1\right )\right )}{96}-\frac {1}{2 x}\)	\(99\)
meijerg	\(\frac {2^{\frac {3}{4}} 3^{\frac {1}{4}} \left (-\frac {4 \,3^{\frac {3}{4}} 2^{\frac {1}{4}}}{3 x}-\frac {x^{3} 3^{\frac {3}{4}} 2^{\frac {1}{4}} \left (\frac {2^{\frac {1}{4}} 27^{\frac {3}{4}} \ln \left (1-6^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}+\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {x^{4}}}{2}\right )}{27 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {2 \,2^{\frac {1}{4}} 27^{\frac {3}{4}} \arctan \left (\frac {3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{8-3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}\right )}{27 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {2^{\frac {1}{4}} 27^{\frac {3}{4}} \ln \left (1+6^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}+\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {x^{4}}}{2}\right )}{27 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {2 \,2^{\frac {1}{4}} 27^{\frac {3}{4}} \arctan \left (\frac {3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{8+3^{\frac {1}{4}} 8^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}\right )}{27 \left (x^{4}\right )^{\frac {3}{4}}}\right )}{2}\right )}{16}\)	\(198\)

[In]

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

[Out]

-1/2/x+3/8*sum(_R*ln(-18*_R^3+x),_R=RootOf(54*_Z^4+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx=-\frac {2^{\frac {3}{4}} \left (-3\right )^{\frac {1}{4}} x \log \left (2^{\frac {1}{4}} \left (-3\right )^{\frac {3}{4}} + 3 \, x\right ) - i \cdot 2^{\frac {3}{4}} \left (-3\right )^{\frac {1}{4}} x \log \left (i \cdot 2^{\frac {1}{4}} \left (-3\right )^{\frac {3}{4}} + 3 \, x\right ) + i \cdot 2^{\frac {3}{4}} \left (-3\right )^{\frac {1}{4}} x \log \left (-i \cdot 2^{\frac {1}{4}} \left (-3\right )^{\frac {3}{4}} + 3 \, x\right ) - 2^{\frac {3}{4}} \left (-3\right )^{\frac {1}{4}} x \log \left (-2^{\frac {1}{4}} \left (-3\right )^{\frac {3}{4}} + 3 \, x\right ) + 8}{16 \, x} \]

[In]

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

[Out]

-1/16*(2^(3/4)*(-3)^(1/4)*x*log(2^(1/4)*(-3)^(3/4) + 3*x) - I*2^(3/4)*(-3)^(1/4)*x*log(I*2^(1/4)*(-3)^(3/4) +
3*x) + I*2^(3/4)*(-3)^(1/4)*x*log(-I*2^(1/4)*(-3)^(3/4) + 3*x) - 2^(3/4)*(-3)^(1/4)*x*log(-2^(1/4)*(-3)^(3/4)
+ 3*x) + 8)/x

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx=- \frac {\sqrt [4]{6} \log {\left (x^{2} - \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{16} + \frac {\sqrt [4]{6} \log {\left (x^{2} + \frac {6^{\frac {3}{4}} x}{3} + \frac {\sqrt {6}}{3} \right )}}{16} - \frac {\sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x - 1 \right )}}{8} - \frac {\sqrt [4]{6} \operatorname {atan}{\left (\sqrt [4]{6} x + 1 \right )}}{8} - \frac {1}{2 x} \]

[In]

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

[Out]

-6**(1/4)*log(x**2 - 6**(3/4)*x/3 + sqrt(6)/3)/16 + 6**(1/4)*log(x**2 + 6**(3/4)*x/3 + sqrt(6)/3)/16 - 6**(1/4
)*atan(6**(1/4)*x - 1)/8 - 6**(1/4)*atan(6**(1/4)*x + 1)/8 - 1/(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx=-\frac {1}{8} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x + 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) - \frac {1}{8} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} 2^{\frac {1}{4}} {\left (2 \, \sqrt {3} x - 3^{\frac {1}{4}} 2^{\frac {3}{4}}\right )}\right ) + \frac {1}{16} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} + 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) - \frac {1}{16} \cdot 3^{\frac {1}{4}} 2^{\frac {1}{4}} \log \left (\sqrt {3} x^{2} - 3^{\frac {1}{4}} 2^{\frac {3}{4}} x + \sqrt {2}\right ) - \frac {1}{2 \, x} \]

[In]

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

[Out]

-1/8*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))) - 1/8*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))) + 1/16*3^(1/4)*2^(1/4)*log(sqrt(3)*x^2 + 3^(1/4)*2^(3/4)*
x + sqrt(2)) - 1/16*3^(1/4)*2^(1/4)*log(sqrt(3)*x^2 - 3^(1/4)*2^(3/4)*x + sqrt(2)) - 1/2/x

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx=-\frac {1}{8} \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) - \frac {1}{8} \cdot 6^{\frac {1}{4}} \arctan \left (\frac {3}{4} \, \sqrt {2} \left (\frac {2}{3}\right )^{\frac {3}{4}} {\left (2 \, x - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}}\right )}\right ) + \frac {1}{16} \cdot 6^{\frac {1}{4}} \log \left (x^{2} + \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) - \frac {1}{16} \cdot 6^{\frac {1}{4}} \log \left (x^{2} - \sqrt {2} \left (\frac {2}{3}\right )^{\frac {1}{4}} x + \sqrt {\frac {2}{3}}\right ) - \frac {1}{2 \, x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^2 \left (2+3 x^4\right )} \, dx=-\frac {1}{2\,x}+6^{1/4}\,\mathrm {atan}\left (6^{1/4}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+6^{1/4}\,\mathrm {atan}\left (6^{1/4}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \]

[In]

int(1/(x^2*(3*x^4 + 2)),x)

[Out]

- 6^(1/4)*atan(6^(1/4)*x*(1/2 - 1i/2))*(1/8 - 1i/8) - 6^(1/4)*atan(6^(1/4)*x*(1/2 + 1i/2))*(1/8 + 1i/8) - 1/(2
*x)