\(\int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx\) [256]

Optimal result

Integrand size = 13, antiderivative size = 25 \[ \int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{693} \left (1+x^3\right )^{3/4} \left (32-24 x^3+21 x^6\right ) \]

[Out]

4/693*(x^3+1)^(3/4)*(21*x^6-24*x^3+32)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{33} \left (x^3+1\right )^{11/4}-\frac {8}{21} \left (x^3+1\right )^{7/4}+\frac {4}{9} \left (x^3+1\right )^{3/4} \]

[In]

Int[x^8/(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(3/4))/9 - (8*(1 + x^3)^(7/4))/21 + (4*(1 + x^3)^(11/4))/33

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{1+x}} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (\frac {1}{\sqrt [4]{1+x}}-2 (1+x)^{3/4}+(1+x)^{7/4}\right ) \, dx,x,x^3\right ) \\ & = \frac {4}{9} \left (1+x^3\right )^{3/4}-\frac {8}{21} \left (1+x^3\right )^{7/4}+\frac {4}{33} \left (1+x^3\right )^{11/4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{693} \left (1+x^3\right )^{3/4} \left (32-24 x^3+21 x^6\right ) \]

[In]

Integrate[x^8/(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(3/4)*(32 - 24*x^3 + 21*x^6))/693

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 2.

Time = 0.86 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68

method	result	size
meijerg	\(\frac {x^{9} \operatorname {hypergeom}\left (\left [\frac {1}{4}, 3\right ], \left [4\right ], -x^{3}\right )}{9}\)	\(17\)
trager	\(\left (\frac {4}{33} x^{6}-\frac {32}{231} x^{3}+\frac {128}{693}\right ) \left (x^{3}+1\right )^{\frac {3}{4}}\)	\(21\)
risch	\(\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}} \left (21 x^{6}-24 x^{3}+32\right )}{693}\)	\(22\)
pseudoelliptic	\(\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}} \left (21 x^{6}-24 x^{3}+32\right )}{693}\)	\(22\)
gosper	\(\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (21 x^{6}-24 x^{3}+32\right )}{693 \left (x^{3}+1\right )^{\frac {1}{4}}}\)	\(33\)

[In]

int(x^8/(x^3+1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/9*x^9*hypergeom([1/4,3],[4],-x^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{693} \, {\left (21 \, x^{6} - 24 \, x^{3} + 32\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}} \]

[In]

integrate(x^8/(x^3+1)^(1/4),x, algorithm="fricas")

[Out]

4/693*(21*x^6 - 24*x^3 + 32)*(x^3 + 1)^(3/4)

Sympy [A] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx=\frac {4 x^{6} \left (x^{3} + 1\right )^{\frac {3}{4}}}{33} - \frac {32 x^{3} \left (x^{3} + 1\right )^{\frac {3}{4}}}{231} + \frac {128 \left (x^{3} + 1\right )^{\frac {3}{4}}}{693} \]

[In]

integrate(x**8/(x**3+1)**(1/4),x)

[Out]

4*x**6*(x**3 + 1)**(3/4)/33 - 32*x**3*(x**3 + 1)**(3/4)/231 + 128*(x**3 + 1)**(3/4)/693

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{33} \, {\left (x^{3} + 1\right )}^{\frac {11}{4}} - \frac {8}{21} \, {\left (x^{3} + 1\right )}^{\frac {7}{4}} + \frac {4}{9} \, {\left (x^{3} + 1\right )}^{\frac {3}{4}} \]

[In]

integrate(x^8/(x^3+1)^(1/4),x, algorithm="maxima")

[Out]

4/33*(x^3 + 1)^(11/4) - 8/21*(x^3 + 1)^(7/4) + 4/9*(x^3 + 1)^(3/4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx=\frac {4}{33} \, {\left (x^{3} + 1\right )}^{\frac {11}{4}} - \frac {8}{21} \, {\left (x^{3} + 1\right )}^{\frac {7}{4}} + \frac {4}{9} \, {\left (x^{3} + 1\right )}^{\frac {3}{4}} \]

[In]

integrate(x^8/(x^3+1)^(1/4),x, algorithm="giac")

[Out]

4/33*(x^3 + 1)^(11/4) - 8/21*(x^3 + 1)^(7/4) + 4/9*(x^3 + 1)^(3/4)

Mupad [B] (verification not implemented)

Time = 5.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {x^8}{\sqrt [4]{1+x^3}} \, dx={\left (x^3+1\right )}^{3/4}\,\left (\frac {4\,x^6}{33}-\frac {32\,x^3}{231}+\frac {128}{693}\right ) \]

[In]

int(x^8/(x^3 + 1)^(1/4),x)

[Out]

(x^3 + 1)^(3/4)*((4*x^6)/33 - (32*x^3)/231 + 128/693)