\(\int \frac {x}{1+2 x^4+x^8} \, dx\) [276]

Optimal result

Integrand size = 14, antiderivative size = 23 \[ \int \frac {x}{1+2 x^4+x^8} \, dx=\frac {x^2}{4 \left (1+x^4\right )}+\frac {\arctan \left (x^2\right )}{4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {28, 281, 205, 209} \[ \int \frac {x}{1+2 x^4+x^8} \, dx=\frac {\arctan \left (x^2\right )}{4}+\frac {x^2}{4 \left (x^4+1\right )} \]

[In]

Int[x/(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 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x}{\left (1+x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {x^2}{4 \left (1+x^4\right )}+\frac {1}{4} \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {x}{1+2 x^4+x^8} \, dx=\frac {1}{4} \left (\frac {x^2}{1+x^4}+\arctan \left (x^2\right )\right ) \]

[In]

Integrate[x/(1 + 2*x^4 + x^8),x]

[Out]

(x^2/(1 + x^4) + ArcTan[x^2])/4

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {x^{2}}{4 x^{4}+4}+\frac {\arctan \left (x^{2}\right )}{4}\)	\(20\)
risch	\(\frac {x^{2}}{4 x^{4}+4}+\frac {\arctan \left (x^{2}\right )}{4}\)	\(20\)
parallelrisch	\(-\frac {i \ln \left (x^{2}-i\right ) x^{4}-i \ln \left (x^{2}+i\right ) x^{4}+i \ln \left (x^{2}-i\right )-i \ln \left (x^{2}+i\right )-2 x^{2}}{8 \left (x^{4}+1\right )}\)	\(62\)

[In]

int(x/(x^8+2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x}{1+2 x^4+x^8} \, dx=\frac {x^{2} + {\left (x^{4} + 1\right )} \arctan \left (x^{2}\right )}{4 \, {\left (x^{4} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {x}{1+2 x^4+x^8} \, dx=\frac {x^{2}}{4 x^{4} + 4} + \frac {\operatorname {atan}{\left (x^{2} \right )}}{4} \]

[In]

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

[Out]

x**2/(4*x**4 + 4) + atan(x**2)/4

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x}{1+2 x^4+x^8} \, dx=\frac {x^{2}}{4 \, {\left (x^{4} + 1\right )}} + \frac {1}{4} \, \arctan \left (x^{2}\right ) \]

[In]

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

[Out]

1/4*x^2/(x^4 + 1) + 1/4*arctan(x^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x}{1+2 x^4+x^8} \, dx=\frac {x^{2}}{4 \, {\left (x^{4} + 1\right )}} + \frac {1}{4} \, \arctan \left (x^{2}\right ) \]

[In]

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

[Out]

1/4*x^2/(x^4 + 1) + 1/4*arctan(x^2)

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {x}{1+2 x^4+x^8} \, dx=\frac {\mathrm {atan}\left (x^2\right )}{4}+\frac {x^2}{4\,\left (x^4+1\right )} \]

[In]

int(x/(2*x^4 + x^8 + 1),x)

[Out]

atan(x^2)/4 + x^2/(4*(x^4 + 1))