3.278 \(\int \frac{1}{x^3 (1+2 x^4+x^8)} \, dx\)

Optimal. Leaf size=30 \[ \frac{1}{4 x^2 \left (x^4+1\right )}-\frac{3}{4 x^2}-\frac{3}{4} \tan ^{-1}\left (x^2\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0123191, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {28, 275, 290, 325, 203} \[ \frac{1}{4 x^2 \left (x^4+1\right )}-\frac{3}{4 x^2}-\frac{3}{4} \tan ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + 2*x^4 + x^8)),x]

[Out]

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

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

Mathematica [A]  time = 0.0115794, size = 30, normalized size = 1. \[ -\frac{x^2}{4 \left (x^4+1\right )}-\frac{1}{2 x^2}+\frac{3}{4} \tan ^{-1}\left (\frac{1}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 + 2*x^4 + x^8)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 25, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,{x}^{2}}}-{\frac{{x}^{2}}{4\,{x}^{4}+4}}-{\frac{3\,\arctan \left ({x}^{2} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(x^8+2*x^4+1),x)

[Out]

-1/2/x^2-1/4*x^2/(x^4+1)-3/4*arctan(x^2)

________________________________________________________________________________________

Maxima [A]  time = 1.49736, size = 34, normalized size = 1.13 \begin{align*} -\frac{3 \, x^{4} + 2}{4 \,{\left (x^{6} + x^{2}\right )}} - \frac{3}{4} \, \arctan \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.4231, size = 78, normalized size = 2.6 \begin{align*} -\frac{3 \, x^{4} + 3 \,{\left (x^{6} + x^{2}\right )} \arctan \left (x^{2}\right ) + 2}{4 \,{\left (x^{6} + x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.1484, size = 26, normalized size = 0.87 \begin{align*} - \frac{3 x^{4} + 2}{4 x^{6} + 4 x^{2}} - \frac{3 \operatorname{atan}{\left (x^{2} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(x**8+2*x**4+1),x)

[Out]

-(3*x**4 + 2)/(4*x**6 + 4*x**2) - 3*atan(x**2)/4

________________________________________________________________________________________

Giac [A]  time = 1.11739, size = 34, normalized size = 1.13 \begin{align*} -\frac{3 \, x^{4} + 2}{4 \,{\left (x^{6} + x^{2}\right )}} - \frac{3}{4} \, \arctan \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

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