\(\int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx\) [32]

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx=\frac {3}{8} \left (1+x^4\right )^{2/3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \[ \int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx=\frac {3}{8} \left (x^4+1\right )^{2/3} \]

[In]

Int[x^3/(1 + x^4)^(1/3),x]

[Out]

(3*(1 + x^4)^(2/3))/8

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {3}{8} \left (1+x^4\right )^{2/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx=\frac {3}{8} \left (1+x^4\right )^{2/3} \]

[In]

Integrate[x^3/(1 + x^4)^(1/3),x]

[Out]

(3*(1 + x^4)^(2/3))/8

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
gosper	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}\)	\(10\)
derivativedivides	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}\)	\(10\)
default	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}\)	\(10\)
trager	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}\)	\(10\)
risch	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}\)	\(10\)
pseudoelliptic	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}\)	\(10\)
meijerg	\(\frac {x^{4} \operatorname {hypergeom}\left (\left [\frac {1}{3}, 1\right ], \left [2\right ], -x^{4}\right )}{4}\)	\(17\)

[In]

int(x^3/(x^4+1)^(1/3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx=\frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} \]

[In]

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

[Out]

3/8*(x^4 + 1)^(2/3)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx=\frac {3 \left (x^{4} + 1\right )^{\frac {2}{3}}}{8} \]

[In]

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

[Out]

3*(x**4 + 1)**(2/3)/8

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx=\frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} \]

[In]

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

[Out]

3/8*(x^4 + 1)^(2/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx=\frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} \]

[In]

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

[Out]

3/8*(x^4 + 1)^(2/3)

Mupad [B] (verification not implemented)

Time = 4.83 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\sqrt [3]{1+x^4}} \, dx=\frac {3\,{\left (x^4+1\right )}^{2/3}}{8} \]

[In]

int(x^3/(x^4 + 1)^(1/3),x)

[Out]

(3*(x^4 + 1)^(2/3))/8