\(\int x^4 (1+x^5)^{2/3} \, dx\) [37]

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int x^4 \left (1+x^5\right )^{2/3} \, dx=\frac {3}{25} \left (1+x^5\right )^{5/3} \]

[Out]

3/25*(x^5+1)^(5/3)

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} \[ \int x^4 \left (1+x^5\right )^{2/3} \, dx=\frac {3}{25} \left (x^5+1\right )^{5/3} \]

[In]

Int[x^4*(1 + x^5)^(2/3),x]

[Out]

(3*(1 + x^5)^(5/3))/25

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}{25} \left (1+x^5\right )^{5/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int x^4 \left (1+x^5\right )^{2/3} \, dx=\frac {3}{25} \left (1+x^5\right )^{5/3} \]

[In]

Integrate[x^4*(1 + x^5)^(2/3),x]

[Out]

(3*(1 + x^5)^(5/3))/25

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {3 \left (x^{5}+1\right )^{\frac {5}{3}}}{25}\)	\(10\)
default	\(\frac {3 \left (x^{5}+1\right )^{\frac {5}{3}}}{25}\)	\(10\)
risch	\(\frac {3 \left (x^{5}+1\right )^{\frac {5}{3}}}{25}\)	\(10\)
pseudoelliptic	\(\frac {3 \left (x^{5}+1\right )^{\frac {5}{3}}}{25}\)	\(10\)
trager	\(\left (\frac {3}{25}+\frac {3 x^{5}}{25}\right ) \left (x^{5}+1\right )^{\frac {2}{3}}\)	\(16\)
meijerg	\(\frac {x^{5} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, 1\right ], \left [2\right ], -x^{5}\right )}{5}\)	\(17\)
gosper	\(\frac {3 \left (1+x \right ) \left (x^{4}-x^{3}+x^{2}-x +1\right ) \left (x^{5}+1\right )^{\frac {2}{3}}}{25}\)	\(29\)

[In]

int(x^4*(x^5+1)^(2/3),x,method=_RETURNVERBOSE)

[Out]

3/25*(x^5+1)^(5/3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^4 \left (1+x^5\right )^{2/3} \, dx=\frac {3}{25} \, {\left (x^{5} + 1\right )}^{\frac {5}{3}} \]

[In]

integrate(x^4*(x^5+1)^(2/3),x, algorithm="fricas")

[Out]

3/25*(x^5 + 1)^(5/3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.00 \[ \int x^4 \left (1+x^5\right )^{2/3} \, dx=\frac {3 x^{5} \left (x^{5} + 1\right )^{\frac {2}{3}}}{25} + \frac {3 \left (x^{5} + 1\right )^{\frac {2}{3}}}{25} \]

[In]

integrate(x**4*(x**5+1)**(2/3),x)

[Out]

3*x**5*(x**5 + 1)**(2/3)/25 + 3*(x**5 + 1)**(2/3)/25

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^4 \left (1+x^5\right )^{2/3} \, dx=\frac {3}{25} \, {\left (x^{5} + 1\right )}^{\frac {5}{3}} \]

[In]

integrate(x^4*(x^5+1)^(2/3),x, algorithm="maxima")

[Out]

3/25*(x^5 + 1)^(5/3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^4 \left (1+x^5\right )^{2/3} \, dx=\frac {3}{25} \, {\left (x^{5} + 1\right )}^{\frac {5}{3}} \]

[In]

integrate(x^4*(x^5+1)^(2/3),x, algorithm="giac")

[Out]

3/25*(x^5 + 1)^(5/3)

Mupad [B] (verification not implemented)

Time = 4.96 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int x^4 \left (1+x^5\right )^{2/3} \, dx=\frac {3\,{\left (x^5+1\right )}^{5/3}}{25} \]

[In]

int(x^4*(x^5 + 1)^(2/3),x)

[Out]

(3*(x^5 + 1)^(5/3))/25