\(\int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx\) [545]

Optimal result

Integrand size = 15, antiderivative size = 59 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx=\frac {a^2 \left (a+b x^3\right )^{2/3}}{2 b^3}-\frac {2 a \left (a+b x^3\right )^{5/3}}{5 b^3}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3} \]

[Out]

1/2*a^2*(b*x^3+a)^(2/3)/b^3-2/5*a*(b*x^3+a)^(5/3)/b^3+1/8*(b*x^3+a)^(8/3)/b^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx=\frac {a^2 \left (a+b x^3\right )^{2/3}}{2 b^3}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3}-\frac {2 a \left (a+b x^3\right )^{5/3}}{5 b^3} \]

[In]

Int[x^8/(a + b*x^3)^(1/3),x]

[Out]

(a^2*(a + b*x^3)^(2/3))/(2*b^3) - (2*a*(a + b*x^3)^(5/3))/(5*b^3) + (a + b*x^3)^(8/3)/(8*b^3)

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 [3]{a+b x}} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (\frac {a^2}{b^2 \sqrt [3]{a+b x}}-\frac {2 a (a+b x)^{2/3}}{b^2}+\frac {(a+b x)^{5/3}}{b^2}\right ) \, dx,x,x^3\right ) \\ & = \frac {a^2 \left (a+b x^3\right )^{2/3}}{2 b^3}-\frac {2 a \left (a+b x^3\right )^{5/3}}{5 b^3}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (9 a^2-6 a b x^3+5 b^2 x^6\right )}{40 b^3} \]

[In]

Integrate[x^8/(a + b*x^3)^(1/3),x]

[Out]

((a + b*x^3)^(2/3)*(9*a^2 - 6*a*b*x^3 + 5*b^2*x^6))/(40*b^3)

Maple [A] (verified)

Time = 3.84 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.61

method	result	size
gosper	\(\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (5 b^{2} x^{6}-6 a b \,x^{3}+9 a^{2}\right )}{40 b^{3}}\)	\(36\)
trager	\(\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (5 b^{2} x^{6}-6 a b \,x^{3}+9 a^{2}\right )}{40 b^{3}}\)	\(36\)
risch	\(\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (5 b^{2} x^{6}-6 a b \,x^{3}+9 a^{2}\right )}{40 b^{3}}\)	\(36\)
pseudoelliptic	\(\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (5 b^{2} x^{6}-6 a b \,x^{3}+9 a^{2}\right )}{40 b^{3}}\)	\(36\)

[In]

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

[Out]

1/40*(b*x^3+a)^(2/3)*(5*b^2*x^6-6*a*b*x^3+9*a^2)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.59 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx=\frac {{\left (5 \, b^{2} x^{6} - 6 \, a b x^{3} + 9 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{40 \, b^{3}} \]

[In]

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

[Out]

1/40*(5*b^2*x^6 - 6*a*b*x^3 + 9*a^2)*(b*x^3 + a)^(2/3)/b^3

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx=\begin {cases} \frac {9 a^{2} \left (a + b x^{3}\right )^{\frac {2}{3}}}{40 b^{3}} - \frac {3 a x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{20 b^{2}} + \frac {x^{6} \left (a + b x^{3}\right )^{\frac {2}{3}}}{8 b} & \text {for}\: b \neq 0 \\\frac {x^{9}}{9 \sqrt [3]{a}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**8/(b*x**3+a)**(1/3),x)

[Out]

Piecewise((9*a**2*(a + b*x**3)**(2/3)/(40*b**3) - 3*a*x**3*(a + b*x**3)**(2/3)/(20*b**2) + x**6*(a + b*x**3)**
(2/3)/(8*b), Ne(b, 0)), (x**9/(9*a**(1/3)), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx=\frac {{\left (b x^{3} + a\right )}^{\frac {8}{3}}}{8 \, b^{3}} - \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a}{5 \, b^{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2}}{2 \, b^{3}} \]

[In]

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

[Out]

1/8*(b*x^3 + a)^(8/3)/b^3 - 2/5*(b*x^3 + a)^(5/3)*a/b^3 + 1/2*(b*x^3 + a)^(2/3)*a^2/b^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx=\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2}}{2 \, b^{3}} + \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} - 16 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a}{40 \, b^{3}} \]

[In]

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

[Out]

1/2*(b*x^3 + a)^(2/3)*a^2/b^3 + 1/40*(5*(b*x^3 + a)^(8/3) - 16*(b*x^3 + a)^(5/3)*a)/b^3

Mupad [B] (verification not implemented)

Time = 5.69 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.61 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3}} \, dx={\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {9\,a^2}{40\,b^3}+\frac {x^6}{8\,b}-\frac {3\,a\,x^3}{20\,b^2}\right ) \]

[In]

int(x^8/(a + b*x^3)^(1/3),x)

[Out]

(a + b*x^3)^(2/3)*((9*a^2)/(40*b^3) + x^6/(8*b) - (3*a*x^3)/(20*b^2))