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

Optimal. Leaf size=31 \[ -\frac{1}{4 x^2}-\frac{1}{4} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0118667, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {275, 325, 203} \[ -\frac{1}{4 x^2}-\frac{1}{4} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(2 + 3*x^4)),x]

[Out]

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

Rule 275

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 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 203

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

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (2+3 x^4\right )} \, dx &=\frac{1}{2} \operatorname{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} \operatorname{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]  time = 0.0184258, size = 48, normalized size = 1.55 \[ \frac{\sqrt{6} x^2 \tan ^{-1}\left (1-\sqrt [4]{6} x\right )+\sqrt{6} x^2 \tan ^{-1}\left (\sqrt [4]{6} x+1\right )-2}{8 x^2} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.004, size = 21, normalized size = 0.7 \begin{align*} -{\frac{1}{4\,{x}^{2}}}-{\frac{\sqrt{6}}{8}\arctan \left ({\frac{{x}^{2}\sqrt{6}}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.49816, size = 27, normalized size = 0.87 \begin{align*} -\frac{1}{8} \, \sqrt{6} \arctan \left (\frac{1}{2} \, \sqrt{6} x^{2}\right ) - \frac{1}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.69073, size = 92, normalized size = 2.97 \begin{align*} -\frac{\sqrt{3} \sqrt{2} x^{2} \arctan \left (\frac{1}{2} \, \sqrt{3} \sqrt{2} x^{2}\right ) + 2}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.150799, size = 26, normalized size = 0.84 \begin{align*} - \frac{\sqrt{6} \operatorname{atan}{\left (\frac{\sqrt{6} x^{2}}{2} \right )}}{8} - \frac{1}{4 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14018, size = 27, normalized size = 0.87 \begin{align*} -\frac{1}{8} \, \sqrt{6} \arctan \left (\frac{1}{2} \, \sqrt{6} x^{2}\right ) - \frac{1}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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