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

Optimal. Leaf size=30 \[ \frac {4 \sqrt [4]{x^3+1} \left (45 x^9+5 x^6-8 x^3+32\right )}{1755} \]

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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 43} \begin {gather*} \frac {4}{39} \left (x^3+1\right )^{13/4}-\frac {8}{27} \left (x^3+1\right )^{9/4}+\frac {4}{15} \left (x^3+1\right )^{5/4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8*(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(5/4))/15 - (8*(1 + x^3)^(9/4))/27 + (4*(1 + x^3)^(13/4))/39

Rule 43

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 266

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*} \int x^8 \sqrt [4]{1+x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int x^2 \sqrt [4]{1+x} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\sqrt [4]{1+x}-2 (1+x)^{5/4}+(1+x)^{9/4}\right ) \, dx,x,x^3\right )\\ &=\frac {4}{15} \left (1+x^3\right )^{5/4}-\frac {8}{27} \left (1+x^3\right )^{9/4}+\frac {4}{39} \left (1+x^3\right )^{13/4}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 0.83 \begin {gather*} \frac {4 \left (x^3+1\right )^{5/4} \left (45 x^6-40 x^3+32\right )}{1755} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8*(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(5/4)*(32 - 40*x^3 + 45*x^6))/1755

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IntegrateAlgebraic [A]  time = 0.02, size = 30, normalized size = 1.00 \begin {gather*} \frac {4 \sqrt [4]{1+x^3} \left (32-8 x^3+5 x^6+45 x^9\right )}{1755} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8*(1 + x^3)^(1/4),x]

[Out]

(4*(1 + x^3)^(1/4)*(32 - 8*x^3 + 5*x^6 + 45*x^9))/1755

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fricas [A]  time = 0.46, size = 26, normalized size = 0.87 \begin {gather*} \frac {4}{1755} \, {\left (45 \, x^{9} + 5 \, x^{6} - 8 \, x^{3} + 32\right )} {\left (x^{3} + 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/1755*(45*x^9 + 5*x^6 - 8*x^3 + 32)*(x^3 + 1)^(1/4)

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giac [A]  time = 0.45, size = 28, normalized size = 0.93 \begin {gather*} \frac {4}{39} \, {\left (x^{3} + 1\right )}^{\frac {13}{4}} - \frac {8}{27} \, {\left (x^{3} + 1\right )}^{\frac {9}{4}} + \frac {4}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/39*(x^3 + 1)^(13/4) - 8/27*(x^3 + 1)^(9/4) + 4/15*(x^3 + 1)^(5/4)

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maple [C]  time = 0.07, size = 17, normalized size = 0.57

method	result	size
meijerg	\(\frac {\hypergeom \left (\left [-\frac {1}{4}, 3\right ], \relax [4], -x^{3}\right ) x^{9}}{9}\)	\(17\)
trager	\(\left (\frac {4}{39} x^{9}+\frac {4}{351} x^{6}-\frac {32}{1755} x^{3}+\frac {128}{1755}\right ) \left (x^{3}+1\right )^{\frac {1}{4}}\)	\(26\)
risch	\(\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}} \left (45 x^{9}+5 x^{6}-8 x^{3}+32\right )}{1755}\)	\(27\)
gosper	\(\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (45 x^{6}-40 x^{3}+32\right ) \left (x^{3}+1\right )^{\frac {1}{4}}}{1755}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 28, normalized size = 0.93 \begin {gather*} \frac {4}{39} \, {\left (x^{3} + 1\right )}^{\frac {13}{4}} - \frac {8}{27} \, {\left (x^{3} + 1\right )}^{\frac {9}{4}} + \frac {4}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/39*(x^3 + 1)^(13/4) - 8/27*(x^3 + 1)^(9/4) + 4/15*(x^3 + 1)^(5/4)

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mupad [B]  time = 0.21, size = 25, normalized size = 0.83 \begin {gather*} {\left (x^3+1\right )}^{1/4}\,\left (\frac {4\,x^9}{39}+\frac {4\,x^6}{351}-\frac {32\,x^3}{1755}+\frac {128}{1755}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(x^3 + 1)^(1/4),x)

[Out]

(x^3 + 1)^(1/4)*((4*x^6)/351 - (32*x^3)/1755 + (4*x^9)/39 + 128/1755)

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sympy [B]  time = 1.41, size = 56, normalized size = 1.87 \begin {gather*} \frac {4 x^{9} \sqrt [4]{x^{3} + 1}}{39} + \frac {4 x^{6} \sqrt [4]{x^{3} + 1}}{351} - \frac {32 x^{3} \sqrt [4]{x^{3} + 1}}{1755} + \frac {128 \sqrt [4]{x^{3} + 1}}{1755} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x**9*(x**3 + 1)**(1/4)/39 + 4*x**6*(x**3 + 1)**(1/4)/351 - 32*x**3*(x**3 + 1)**(1/4)/1755 + 128*(x**3 + 1)**
(1/4)/1755

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