3.369 \(\int x^8 \sqrt {a+b x^3} \, dx\)

Optimal. Leaf size=59 \[ \frac {2 a^2 \left (a+b x^3\right )^{3/2}}{9 b^3}+\frac {2 \left (a+b x^3\right )^{7/2}}{21 b^3}-\frac {4 a \left (a+b x^3\right )^{5/2}}{15 b^3} \]

[Out]

2/9*a^2*(b*x^3+a)^(3/2)/b^3-4/15*a*(b*x^3+a)^(5/2)/b^3+2/21*(b*x^3+a)^(7/2)/b^3

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Rubi [A]  time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac {2 a^2 \left (a+b x^3\right )^{3/2}}{9 b^3}+\frac {2 \left (a+b x^3\right )^{7/2}}{21 b^3}-\frac {4 a \left (a+b x^3\right )^{5/2}}{15 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^8*Sqrt[a + b*x^3],x]

[Out]

(2*a^2*(a + b*x^3)^(3/2))/(9*b^3) - (4*a*(a + b*x^3)^(5/2))/(15*b^3) + (2*(a + b*x^3)^(7/2))/(21*b^3)

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 {a+b x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int x^2 \sqrt {a+b x} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {a^2 \sqrt {a+b x}}{b^2}-\frac {2 a (a+b x)^{3/2}}{b^2}+\frac {(a+b x)^{5/2}}{b^2}\right ) \, dx,x,x^3\right )\\ &=\frac {2 a^2 \left (a+b x^3\right )^{3/2}}{9 b^3}-\frac {4 a \left (a+b x^3\right )^{5/2}}{15 b^3}+\frac {2 \left (a+b x^3\right )^{7/2}}{21 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 0.66 \[ \frac {2 \left (a+b x^3\right )^{3/2} \left (8 a^2-12 a b x^3+15 b^2 x^6\right )}{315 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8*Sqrt[a + b*x^3],x]

[Out]

(2*(a + b*x^3)^(3/2)*(8*a^2 - 12*a*b*x^3 + 15*b^2*x^6))/(315*b^3)

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fricas [A]  time = 0.75, size = 46, normalized size = 0.78 \[ \frac {2 \, {\left (15 \, b^{3} x^{9} + 3 \, a b^{2} x^{6} - 4 \, a^{2} b x^{3} + 8 \, a^{3}\right )} \sqrt {b x^{3} + a}}{315 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(15*b^3*x^9 + 3*a*b^2*x^6 - 4*a^2*b*x^3 + 8*a^3)*sqrt(b*x^3 + a)/b^3

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giac [A]  time = 0.19, size = 43, normalized size = 0.73 \[ \frac {2 \, {\left (15 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} - 42 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}\right )}}{315 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(15*(b*x^3 + a)^(7/2) - 42*(b*x^3 + a)^(5/2)*a + 35*(b*x^3 + a)^(3/2)*a^2)/b^3

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maple [A]  time = 0.01, size = 36, normalized size = 0.61 \[ \frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}} \left (15 b^{2} x^{6}-12 a b \,x^{3}+8 a^{2}\right )}{315 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(b*x^3+a)^(3/2)*(15*b^2*x^6-12*a*b*x^3+8*a^2)/b^3

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maxima [A]  time = 1.29, size = 47, normalized size = 0.80 \[ \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}}}{21 \, b^{3}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a}{15 \, b^{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}{9 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*(b*x^3 + a)^(7/2)/b^3 - 4/15*(b*x^3 + a)^(5/2)*a/b^3 + 2/9*(b*x^3 + a)^(3/2)*a^2/b^3

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mupad [B]  time = 1.09, size = 70, normalized size = 1.19 \[ \frac {2\,x^9\,\sqrt {b\,x^3+a}}{21}+\frac {16\,a^3\,\sqrt {b\,x^3+a}}{315\,b^3}+\frac {2\,a\,x^6\,\sqrt {b\,x^3+a}}{105\,b}-\frac {8\,a^2\,x^3\,\sqrt {b\,x^3+a}}{315\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a + b*x^3)^(1/2),x)

[Out]

(2*x^9*(a + b*x^3)^(1/2))/21 + (16*a^3*(a + b*x^3)^(1/2))/(315*b^3) + (2*a*x^6*(a + b*x^3)^(1/2))/(105*b) - (8
*a^2*x^3*(a + b*x^3)^(1/2))/(315*b^2)

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sympy [A]  time = 2.39, size = 90, normalized size = 1.53 \[ \begin {cases} \frac {16 a^{3} \sqrt {a + b x^{3}}}{315 b^{3}} - \frac {8 a^{2} x^{3} \sqrt {a + b x^{3}}}{315 b^{2}} + \frac {2 a x^{6} \sqrt {a + b x^{3}}}{105 b} + \frac {2 x^{9} \sqrt {a + b x^{3}}}{21} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{9}}{9} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((16*a**3*sqrt(a + b*x**3)/(315*b**3) - 8*a**2*x**3*sqrt(a + b*x**3)/(315*b**2) + 2*a*x**6*sqrt(a + b
*x**3)/(105*b) + 2*x**9*sqrt(a + b*x**3)/21, Ne(b, 0)), (sqrt(a)*x**9/9, True))

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