\(\int \frac {1}{x^3 (2+3 x^4)} \, dx\) [693]

Optimal result

Integrand size = 13, antiderivative size = 31 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )} \, dx=-\frac {1}{4 x^2}-\frac {1}{4} \sqrt {\frac {3}{2}} \arctan \left (\sqrt {\frac {3}{2}} x^2\right ) \]

[Out]

-1/4/x^2-1/8*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {281, 331, 209} \[ \int \frac {1}{x^3 \left (2+3 x^4\right )} \, dx=-\frac {1}{4} \sqrt {\frac {3}{2}} \arctan \left (\sqrt {\frac {3}{2}} x^2\right )-\frac {1}{4 x^2} \]

[In]

Int[1/(x^3*(2 + 3*x^4)),x]

[Out]

-1/4*1/x^2 - (Sqrt[3/2]*ArcTan[Sqrt[3/2]*x^2])/4

Rule 209

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

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (2+3 x^2\right )} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^2}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{2+3 x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^2}-\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )} \, dx=\frac {-2+\sqrt {6} x^2 \arctan \left (1-\sqrt [4]{6} x\right )+\sqrt {6} x^2 \arctan \left (1+\sqrt [4]{6} x\right )}{8 x^2} \]

[In]

Integrate[1/(x^3*(2 + 3*x^4)),x]

[Out]

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

Maple [A] (verified)

Time = 3.94 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68

method	result	size
default	\(-\frac {1}{4 x^{2}}-\frac {\arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{8}\)	\(21\)
risch	\(-\frac {1}{4 x^{2}}-\frac {\arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{8}\)	\(21\)
meijerg	\(\frac {\sqrt {3}\, \sqrt {2}\, \left (-\frac {2 \sqrt {3}\, \sqrt {2}}{3 x^{2}}-2 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, x^{2}}{2}\right )\right )}{16}\)	\(35\)

[In]

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

[Out]

-1/4/x^2-1/8*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )} \, dx=-\frac {\sqrt {3} \sqrt {2} x^{2} \arctan \left (\frac {1}{2} \, \sqrt {3} \sqrt {2} x^{2}\right ) + 2}{8 \, x^{2}} \]

[In]

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

[Out]

-1/8*(sqrt(3)*sqrt(2)*x^2*arctan(1/2*sqrt(3)*sqrt(2)*x^2) + 2)/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )} \, dx=- \frac {\sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x^{2}}{2} \right )}}{8} - \frac {1}{4 x^{2}} \]

[In]

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

[Out]

-sqrt(6)*atan(sqrt(6)*x**2/2)/8 - 1/(4*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )} \, dx=-\frac {1}{8} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) - \frac {1}{4 \, x^{2}} \]

[In]

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

[Out]

-1/8*sqrt(6)*arctan(1/2*sqrt(6)*x^2) - 1/4/x^2

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )} \, dx=-\frac {1}{8} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) - \frac {1}{4 \, x^{2}} \]

[In]

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

[Out]

-1/8*sqrt(6)*arctan(1/2*sqrt(6)*x^2) - 1/4/x^2

Mupad [B] (verification not implemented)

Time = 5.49 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )} \, dx=-\frac {\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x^2}{2}\right )}{8}-\frac {1}{4\,x^2} \]

[In]

int(1/(x^3*(3*x^4 + 2)),x)

[Out]

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