\(\int x^7 (1+x^4)^{2/3} \, dx\) [259]

Optimal result

Integrand size = 13, antiderivative size = 25 \[ \int x^7 \left (1+x^4\right )^{2/3} \, dx=\frac {3}{160} \left (1+x^4\right )^{2/3} \left (-3+2 x^4+5 x^8\right ) \]

[Out]

3/160*(x^4+1)^(2/3)*(5*x^8+2*x^4-3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int x^7 \left (1+x^4\right )^{2/3} \, dx=\frac {3}{32} \left (x^4+1\right )^{8/3}-\frac {3}{20} \left (x^4+1\right )^{5/3} \]

[In]

Int[x^7*(1 + x^4)^(2/3),x]

[Out]

(-3*(1 + x^4)^(5/3))/20 + (3*(1 + x^4)^(8/3))/32

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}{4} \text {Subst}\left (\int x (1+x)^{2/3} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (-(1+x)^{2/3}+(1+x)^{5/3}\right ) \, dx,x,x^4\right ) \\ & = -\frac {3}{20} \left (1+x^4\right )^{5/3}+\frac {3}{32} \left (1+x^4\right )^{8/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int x^7 \left (1+x^4\right )^{2/3} \, dx=\frac {3}{160} \left (1+x^4\right )^{5/3} \left (-3+5 x^4\right ) \]

[In]

Integrate[x^7*(1 + x^4)^(2/3),x]

[Out]

(3*(1 + x^4)^(5/3)*(-3 + 5*x^4))/160

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68

method	result	size
gosper	\(\frac {3 \left (x^{4}+1\right )^{\frac {5}{3}} \left (5 x^{4}-3\right )}{160}\)	\(17\)
meijerg	\(\frac {x^{8} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, 2\right ], \left [3\right ], -x^{4}\right )}{8}\)	\(17\)
pseudoelliptic	\(\frac {3 \left (x^{4}+1\right )^{\frac {5}{3}} \left (5 x^{4}-3\right )}{160}\)	\(17\)
trager	\(\left (\frac {3}{32} x^{8}+\frac {3}{80} x^{4}-\frac {9}{160}\right ) \left (x^{4}+1\right )^{\frac {2}{3}}\)	\(21\)
risch	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}} \left (5 x^{8}+2 x^{4}-3\right )}{160}\)	\(22\)

[In]

int(x^7*(x^4+1)^(2/3),x,method=_RETURNVERBOSE)

[Out]

3/160*(x^4+1)^(5/3)*(5*x^4-3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int x^7 \left (1+x^4\right )^{2/3} \, dx=\frac {3}{160} \, {\left (5 \, x^{8} + 2 \, x^{4} - 3\right )} {\left (x^{4} + 1\right )}^{\frac {2}{3}} \]

[In]

integrate(x^7*(x^4+1)^(2/3),x, algorithm="fricas")

[Out]

3/160*(5*x^8 + 2*x^4 - 3)*(x^4 + 1)^(2/3)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int x^7 \left (1+x^4\right )^{2/3} \, dx=\frac {3 x^{8} \left (x^{4} + 1\right )^{\frac {2}{3}}}{32} + \frac {3 x^{4} \left (x^{4} + 1\right )^{\frac {2}{3}}}{80} - \frac {9 \left (x^{4} + 1\right )^{\frac {2}{3}}}{160} \]

[In]

integrate(x**7*(x**4+1)**(2/3),x)

[Out]

3*x**8*(x**4 + 1)**(2/3)/32 + 3*x**4*(x**4 + 1)**(2/3)/80 - 9*(x**4 + 1)**(2/3)/160

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x^7 \left (1+x^4\right )^{2/3} \, dx=\frac {3}{32} \, {\left (x^{4} + 1\right )}^{\frac {8}{3}} - \frac {3}{20} \, {\left (x^{4} + 1\right )}^{\frac {5}{3}} \]

[In]

integrate(x^7*(x^4+1)^(2/3),x, algorithm="maxima")

[Out]

3/32*(x^4 + 1)^(8/3) - 3/20*(x^4 + 1)^(5/3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int x^7 \left (1+x^4\right )^{2/3} \, dx=\frac {3}{32} \, {\left (x^{4} + 1\right )}^{\frac {8}{3}} - \frac {3}{20} \, {\left (x^{4} + 1\right )}^{\frac {5}{3}} \]

[In]

integrate(x^7*(x^4+1)^(2/3),x, algorithm="giac")

[Out]

3/32*(x^4 + 1)^(8/3) - 3/20*(x^4 + 1)^(5/3)

Mupad [B] (verification not implemented)

Time = 5.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int x^7 \left (1+x^4\right )^{2/3} \, dx={\left (x^4+1\right )}^{2/3}\,\left (\frac {3\,x^8}{32}+\frac {3\,x^4}{80}-\frac {9}{160}\right ) \]

[In]

int(x^7*(x^4 + 1)^(2/3),x)

[Out]

(x^4 + 1)^(2/3)*((3*x^4)/80 + (3*x^8)/32 - 9/160)