\(\int \frac {1}{x^3 (2+3 x^4)^2} \, dx\) [702]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )^2} \, dx=-\frac {3}{16 x^2}+\frac {1}{8 x^2 \left (2+3 x^4\right )}-\frac {3}{16} \sqrt {\frac {3}{2}} \arctan \left (\sqrt {\frac {3}{2}} x^2\right ) \]

[Out]

-3/16/x^2+1/8/x^2/(3*x^4+2)-3/32*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 296, 331, 209} \[ \int \frac {1}{x^3 \left (2+3 x^4\right )^2} \, dx=-\frac {3}{16} \sqrt {\frac {3}{2}} \arctan \left (\sqrt {\frac {3}{2}} x^2\right )-\frac {3}{16 x^2}+\frac {1}{8 x^2 \left (3 x^4+2\right )} \]

[In]

Int[1/(x^3*(2 + 3*x^4)^2),x]

[Out]

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

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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 )^2} \, dx,x,x^2\right ) \\ & = \frac {1}{8 x^2 \left (2+3 x^4\right )}+\frac {3}{8} \text {Subst}\left (\int \frac {1}{x^2 \left (2+3 x^2\right )} \, dx,x,x^2\right ) \\ & = -\frac {3}{16 x^2}+\frac {1}{8 x^2 \left (2+3 x^4\right )}-\frac {9}{16} \text {Subst}\left (\int \frac {1}{2+3 x^2} \, dx,x,x^2\right ) \\ & = -\frac {3}{16 x^2}+\frac {1}{8 x^2 \left (2+3 x^4\right )}-\frac {3}{16} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\sqrt {\frac {3}{2}} x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )^2} \, dx=\frac {1}{32} \left (-\frac {4}{x^2}-\frac {6 x^2}{2+3 x^4}+3 \sqrt {6} \arctan \left (1-\sqrt [4]{6} x\right )+3 \sqrt {6} \arctan \left (1+\sqrt [4]{6} x\right )\right ) \]

[In]

Integrate[1/(x^3*(2 + 3*x^4)^2),x]

[Out]

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

Maple [A] (verified)

Time = 3.92 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.70

method	result	size
default	\(-\frac {1}{8 x^{2}}-\frac {x^{2}}{16 \left (x^{4}+\frac {2}{3}\right )}-\frac {3 \arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{32}\)	\(33\)
risch	\(\frac {-\frac {9 x^{4}}{16}-\frac {1}{4}}{x^{2} \left (3 x^{4}+2\right )}-\frac {3 \arctan \left (\frac {x^{2} \sqrt {6}}{2}\right ) \sqrt {6}}{32}\)	\(37\)
meijerg	\(\frac {\sqrt {3}\, \sqrt {2}\, \left (-\frac {2 \sqrt {3}\, \sqrt {2}\, \left (\frac {9 x^{4}}{2}+2\right )}{3 x^{2} \left (3 x^{4}+2\right )}-3 \arctan \left (\frac {\sqrt {2}\, \sqrt {3}\, x^{2}}{2}\right )\right )}{32}\)	\(51\)

[In]

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

[Out]

-1/8/x^2-1/16*x^2/(x^4+2/3)-3/32*arctan(1/2*x^2*6^(1/2))*6^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )^2} \, dx=-\frac {18 \, x^{4} + 3 \, \sqrt {3} \sqrt {2} {\left (3 \, x^{6} + 2 \, x^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {3} \sqrt {2} x^{2}\right ) + 8}{32 \, {\left (3 \, x^{6} + 2 \, x^{2}\right )}} \]

[In]

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

[Out]

-1/32*(18*x^4 + 3*sqrt(3)*sqrt(2)*(3*x^6 + 2*x^2)*arctan(1/2*sqrt(3)*sqrt(2)*x^2) + 8)/(3*x^6 + 2*x^2)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )^2} \, dx=\frac {- 9 x^{4} - 4}{48 x^{6} + 32 x^{2}} - \frac {3 \sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x^{2}}{2} \right )}}{32} \]

[In]

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

[Out]

(-9*x**4 - 4)/(48*x**6 + 32*x**2) - 3*sqrt(6)*atan(sqrt(6)*x**2/2)/32

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )^2} \, dx=-\frac {3}{32} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) - \frac {9 \, x^{4} + 4}{16 \, {\left (3 \, x^{6} + 2 \, x^{2}\right )}} \]

[In]

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

[Out]

-3/32*sqrt(6)*arctan(1/2*sqrt(6)*x^2) - 1/16*(9*x^4 + 4)/(3*x^6 + 2*x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )^2} \, dx=-\frac {3}{32} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x^{2}\right ) - \frac {9 \, x^{4} + 4}{16 \, {\left (3 \, x^{6} + 2 \, x^{2}\right )}} \]

[In]

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

[Out]

-3/32*sqrt(6)*arctan(1/2*sqrt(6)*x^2) - 1/16*(9*x^4 + 4)/(3*x^6 + 2*x^2)

Mupad [B] (verification not implemented)

Time = 5.43 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^3 \left (2+3 x^4\right )^2} \, dx=-\frac {\frac {3\,x^4}{16}+\frac {1}{12}}{x^6+\frac {2\,x^2}{3}}-\frac {3\,\sqrt {2}\,\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {3}\,x^2}{2}\right )}{32} \]

[In]

int(1/(x^3*(3*x^4 + 2)^2),x)

[Out]

- ((3*x^4)/16 + 1/12)/((2*x^2)/3 + x^6) - (3*2^(1/2)*3^(1/2)*atan((2^(1/2)*3^(1/2)*x^2)/2))/32