\(\int \frac {x^2}{\sqrt [4]{1+x^3}} \, dx\) [19]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int \frac {x^2}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{9} \left (1+x^3\right )^{3/4} \]

[Out]

4/9*(x^3+1)^(3/4)

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} \[ \int \frac {x^2}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{9} \left (x^3+1\right )^{3/4} \]

[In]

Int[x^2/(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(3/4))/9

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 {4}{9} \left (1+x^3\right )^{3/4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{9} \left (1+x^3\right )^{3/4} \]

[In]

Integrate[x^2/(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(3/4))/9

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method result size
derivativedivides \(\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}}}{9}\) \(10\)
default \(\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}}}{9}\) \(10\)
trager \(\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}}}{9}\) \(10\)
risch \(\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}}}{9}\) \(10\)
pseudoelliptic \(\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}}}{9}\) \(10\)
meijerg \(\frac {x^{3} \operatorname {hypergeom}\left (\left [\frac {1}{4}, 1\right ], \left [2\right ], -x^{3}\right )}{3}\) \(17\)
gosper \(\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right )}{9 \left (x^{3}+1\right )^{\frac {1}{4}}}\) \(21\)

[In]

int(x^2/(x^3+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/9*(x^3+1)^(3/4)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^2}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{9} \, {\left (x^{3} + 1\right )}^{\frac {3}{4}} \]

[In]

integrate(x^2/(x^3+1)^(1/4),x, algorithm="fricas")

[Out]

4/9*(x^3 + 1)^(3/4)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {x^2}{\sqrt [4]{1+x^3}} \, dx=\frac {4 \left (x^{3} + 1\right )^{\frac {3}{4}}}{9} \]

[In]

integrate(x**2/(x**3+1)**(1/4),x)

[Out]

4*(x**3 + 1)**(3/4)/9

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^2}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{9} \, {\left (x^{3} + 1\right )}^{\frac {3}{4}} \]

[In]

integrate(x^2/(x^3+1)^(1/4),x, algorithm="maxima")

[Out]

4/9*(x^3 + 1)^(3/4)

Giac [A] (verification not implemented)

none

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

[In]

integrate(x^2/(x^3+1)^(1/4),x, algorithm="giac")

[Out]

4/9*(x^3 + 1)^(3/4)

Mupad [B] (verification not implemented)

Time = 5.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^2}{\sqrt [4]{1+x^3}} \, dx=\frac {4\,{\left (x^3+1\right )}^{3/4}}{9} \]

[In]

int(x^2/(x^3 + 1)^(1/4),x)

[Out]

(4*(x^3 + 1)^(3/4))/9