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

Optimal. Leaf size=47 \[ \frac{1}{8 x^2 \left (3 x^4+2\right )}-\frac{3}{16 x^2}-\frac{3}{16} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0204528, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {275, 290, 325, 203} \[ \frac{1}{8 x^2 \left (3 x^4+2\right )}-\frac{3}{16 x^2}-\frac{3}{16} \sqrt{\frac{3}{2}} \tan ^{-1}\left (\sqrt{\frac{3}{2}} x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(2 + 3*x^4)^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 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 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 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 )^2} \, dx &=\frac{1}{2} \operatorname{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} \operatorname{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} \operatorname{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]  time = 0.0356645, size = 59, normalized size = 1.26 \[ \frac{1}{32} \left (-\frac{6 x^2}{3 x^4+2}-\frac{4}{x^2}+3 \sqrt{6} \tan ^{-1}\left (1-\sqrt [4]{6} x\right )+3 \sqrt{6} \tan ^{-1}\left (\sqrt [4]{6} x+1\right )\right ) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.01, size = 33, normalized size = 0.7 \begin{align*} -{\frac{1}{8\,{x}^{2}}}-{\frac{{x}^{2}}{16} \left ({x}^{4}+{\frac{2}{3}} \right ) ^{-1}}-{\frac{3\,\sqrt{6}}{32}\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)^2,x)

[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)

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Maxima [A]  time = 1.53441, size = 50, normalized size = 1.06 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.66629, size = 140, normalized size = 2.98 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.252969, size = 37, normalized size = 0.79 \begin{align*} - \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.13009, size = 50, normalized size = 1.06 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)