\(\int \frac {x^4}{16+x^{10}} \, dx\) [326]

Optimal result

Integrand size = 11, antiderivative size = 12 \[ \int \frac {x^4}{16+x^{10}} \, dx=\frac {1}{20} \arctan \left (\frac {x^5}{4}\right ) \]

[Out]

1/20*arctan(1/4*x^5)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, 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} \[ \int \frac {x^4}{16+x^{10}} \, dx=\frac {1}{20} \arctan \left (\frac {x^5}{4}\right ) \]

[In]

Int[x^4/(16 + x^10),x]

[Out]

ArcTan[x^5/4]/20

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}{5} \text {Subst}\left (\int \frac {1}{16+x^2} \, dx,x,x^5\right ) \\ & = \frac {1}{20} \arctan \left (\frac {x^5}{4}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {x^4}{16+x^{10}} \, dx=\frac {1}{20} \arctan \left (\frac {x^5}{4}\right ) \]

[In]

Integrate[x^4/(16 + x^10),x]

[Out]

ArcTan[x^5/4]/20

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {\arctan \left (\frac {x^{5}}{4}\right )}{20}\)	\(9\)
meijerg	\(\frac {\arctan \left (\frac {x^{5}}{4}\right )}{20}\)	\(9\)
risch	\(\frac {\arctan \left (\frac {x^{5}}{4}\right )}{20}\)	\(9\)
parallelrisch	\(\frac {i \ln \left (x^{5}+4 i\right )}{40}-\frac {i \ln \left (x^{5}-4 i\right )}{40}\)	\(22\)

[In]

int(x^4/(x^10+16),x,method=_RETURNVERBOSE)

[Out]

1/20*arctan(1/4*x^5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{16+x^{10}} \, dx=\frac {1}{20} \, \arctan \left (\frac {1}{4} \, x^{5}\right ) \]

[In]

integrate(x^4/(x^10+16),x, algorithm="fricas")

[Out]

1/20*arctan(1/4*x^5)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.58 \[ \int \frac {x^4}{16+x^{10}} \, dx=\frac {\operatorname {atan}{\left (\frac {x^{5}}{4} \right )}}{20} \]

[In]

integrate(x**4/(x**10+16),x)

[Out]

atan(x**5/4)/20

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{16+x^{10}} \, dx=\frac {1}{20} \, \arctan \left (\frac {1}{4} \, x^{5}\right ) \]

[In]

integrate(x^4/(x^10+16),x, algorithm="maxima")

[Out]

1/20*arctan(1/4*x^5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{16+x^{10}} \, dx=\frac {1}{20} \, \arctan \left (\frac {1}{4} \, x^{5}\right ) \]

[In]

integrate(x^4/(x^10+16),x, algorithm="giac")

[Out]

1/20*arctan(1/4*x^5)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{16+x^{10}} \, dx=\frac {\mathrm {atan}\left (\frac {x^5}{4}\right )}{20} \]

[In]

int(x^4/(x^10 + 16),x)

[Out]

atan(x^5/4)/20