3.274 \(\int \frac{x^5}{1+2 x^4+x^8} \, dx\)

Optimal. Leaf size=23 \[ \frac{1}{4} \tan ^{-1}\left (x^2\right )-\frac{x^2}{4 \left (x^4+1\right )} \]

[Out]

-x^2/(4*(1 + x^4)) + ArcTan[x^2]/4

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Rubi [A]  time = 0.0093071, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {28, 275, 288, 203} \[ \frac{1}{4} \tan ^{-1}\left (x^2\right )-\frac{x^2}{4 \left (x^4+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(1 + 2*x^4 + x^8),x]

[Out]

-x^2/(4*(1 + x^4)) + 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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& 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{x^5}{1+2 x^4+x^8} \, dx &=\int \frac{x^5}{\left (1+x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac{x^2}{4 \left (1+x^4\right )}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,x^2\right )\\ &=-\frac{x^2}{4 \left (1+x^4\right )}+\frac{1}{4} \tan ^{-1}\left (x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0096704, size = 23, normalized size = 1. \[ \frac{1}{4} \tan ^{-1}\left (x^2\right )-\frac{x^2}{4 \left (x^4+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(1 + 2*x^4 + x^8),x]

[Out]

-x^2/(4*(1 + x^4)) + ArcTan[x^2]/4

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Maple [A]  time = 0.004, size = 20, normalized size = 0.9 \begin{align*} -{\frac{{x}^{2}}{4\,{x}^{4}+4}}+{\frac{\arctan \left ({x}^{2} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(x^8+2*x^4+1),x)

[Out]

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

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Maxima [A]  time = 1.53714, size = 26, normalized size = 1.13 \begin{align*} -\frac{x^{2}}{4 \,{\left (x^{4} + 1\right )}} + \frac{1}{4} \, \arctan \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

-1/4*x^2/(x^4 + 1) + 1/4*arctan(x^2)

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Fricas [A]  time = 1.42827, size = 62, normalized size = 2.7 \begin{align*} -\frac{x^{2} -{\left (x^{4} + 1\right )} \arctan \left (x^{2}\right )}{4 \,{\left (x^{4} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.121301, size = 15, normalized size = 0.65 \begin{align*} - \frac{x^{2}}{4 x^{4} + 4} + \frac{\operatorname{atan}{\left (x^{2} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(x**8+2*x**4+1),x)

[Out]

-x**2/(4*x**4 + 4) + atan(x**2)/4

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Giac [A]  time = 1.09713, size = 26, normalized size = 1.13 \begin{align*} -\frac{x^{2}}{4 \,{\left (x^{4} + 1\right )}} + \frac{1}{4} \, \arctan \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

-1/4*x^2/(x^4 + 1) + 1/4*arctan(x^2)