3.439 \(\int \frac{x^8}{\sqrt{1+x^3}} \, dx\)

Optimal. Leaf size=40 \[ \frac{2}{15} \left (x^3+1\right )^{5/2}-\frac{4}{9} \left (x^3+1\right )^{3/2}+\frac{2 \sqrt{x^3+1}}{3} \]

[Out]

(2*Sqrt[1 + x^3])/3 - (4*(1 + x^3)^(3/2))/9 + (2*(1 + x^3)^(5/2))/15

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Rubi [A]  time = 0.0146081, antiderivative size = 40, normalized size of antiderivative = 1., 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} \[ \frac{2}{15} \left (x^3+1\right )^{5/2}-\frac{4}{9} \left (x^3+1\right )^{3/2}+\frac{2 \sqrt{x^3+1}}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^8/Sqrt[1 + x^3],x]

[Out]

(2*Sqrt[1 + x^3])/3 - (4*(1 + x^3)^(3/2))/9 + (2*(1 + x^3)^(5/2))/15

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]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0068485, size = 25, normalized size = 0.62 \[ \frac{2}{45} \sqrt{x^3+1} \left (3 x^6-4 x^3+8\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/Sqrt[1 + x^3],x]

[Out]

(2*Sqrt[1 + x^3]*(8 - 4*x^3 + 3*x^6))/45

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Maple [A]  time = 0.004, size = 33, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2+2\,x \right ) \left ({x}^{2}-x+1 \right ) \left ( 3\,{x}^{6}-4\,{x}^{3}+8 \right ) }{45}{\frac{1}{\sqrt{{x}^{3}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(x^3+1)^(1/2),x)

[Out]

2/45*(1+x)*(x^2-x+1)*(3*x^6-4*x^3+8)/(x^3+1)^(1/2)

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Maxima [A]  time = 0.954821, size = 38, normalized size = 0.95 \begin{align*} \frac{2}{15} \,{\left (x^{3} + 1\right )}^{\frac{5}{2}} - \frac{4}{9} \,{\left (x^{3} + 1\right )}^{\frac{3}{2}} + \frac{2}{3} \, \sqrt{x^{3} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^3+1)^(1/2),x, algorithm="maxima")

[Out]

2/15*(x^3 + 1)^(5/2) - 4/9*(x^3 + 1)^(3/2) + 2/3*sqrt(x^3 + 1)

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Fricas [A]  time = 1.73187, size = 54, normalized size = 1.35 \begin{align*} \frac{2}{45} \,{\left (3 \, x^{6} - 4 \, x^{3} + 8\right )} \sqrt{x^{3} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^3+1)^(1/2),x, algorithm="fricas")

[Out]

2/45*(3*x^6 - 4*x^3 + 8)*sqrt(x^3 + 1)

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Sympy [A]  time = 0.743006, size = 41, normalized size = 1.02 \begin{align*} \frac{2 x^{6} \sqrt{x^{3} + 1}}{15} - \frac{8 x^{3} \sqrt{x^{3} + 1}}{45} + \frac{16 \sqrt{x^{3} + 1}}{45} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(x**3+1)**(1/2),x)

[Out]

2*x**6*sqrt(x**3 + 1)/15 - 8*x**3*sqrt(x**3 + 1)/45 + 16*sqrt(x**3 + 1)/45

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Giac [A]  time = 1.11367, size = 38, normalized size = 0.95 \begin{align*} \frac{2}{15} \,{\left (x^{3} + 1\right )}^{\frac{5}{2}} - \frac{4}{9} \,{\left (x^{3} + 1\right )}^{\frac{3}{2}} + \frac{2}{3} \, \sqrt{x^{3} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^3+1)^(1/2),x, algorithm="giac")

[Out]

2/15*(x^3 + 1)^(5/2) - 4/9*(x^3 + 1)^(3/2) + 2/3*sqrt(x^3 + 1)