\(\int \frac {x^3}{1+x^8} \, dx\) [1493]

Optimal result

Integrand size = 11, antiderivative size = 8 \[ \int \frac {x^3}{1+x^8} \, dx=\frac {\arctan \left (x^4\right )}{4} \]

[Out]

1/4*arctan(x^4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {281, 209} \[ \int \frac {x^3}{1+x^8} \, dx=\frac {\arctan \left (x^4\right )}{4} \]

[In]

Int[x^3/(1 + x^8),x]

[Out]

ArcTan[x^4]/4

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{1+x^8} \, dx=\frac {\arctan \left (x^4\right )}{4} \]

[In]

Integrate[x^3/(1 + x^8),x]

[Out]

ArcTan[x^4]/4

Maple [A] (verified)

Time = 3.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {\arctan \left (x^{4}\right )}{4}\)	\(7\)
meijerg	\(\frac {\arctan \left (x^{4}\right )}{4}\)	\(7\)
risch	\(\frac {\arctan \left (x^{4}\right )}{4}\)	\(7\)
parallelrisch	\(\frac {i \ln \left (x^{4}+i\right )}{8}-\frac {i \ln \left (x^{4}-i\right )}{8}\)	\(22\)

[In]

int(x^3/(x^8+1),x,method=_RETURNVERBOSE)

[Out]

1/4*arctan(x^4)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{1+x^8} \, dx=\frac {1}{4} \, \arctan \left (x^{4}\right ) \]

[In]

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

[Out]

1/4*arctan(x^4)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {x^3}{1+x^8} \, dx=\frac {\operatorname {atan}{\left (x^{4} \right )}}{4} \]

[In]

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

[Out]

atan(x**4)/4

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{1+x^8} \, dx=\frac {1}{4} \, \arctan \left (x^{4}\right ) \]

[In]

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

[Out]

1/4*arctan(x^4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{1+x^8} \, dx=\frac {1}{4} \, \arctan \left (x^{4}\right ) \]

[In]

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

[Out]

1/4*arctan(x^4)

Mupad [B] (verification not implemented)

Time = 5.97 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{1+x^8} \, dx=\frac {\mathrm {atan}\left (x^4\right )}{4} \]

[In]

int(x^3/(x^8 + 1),x)

[Out]

atan(x^4)/4