3.5.5 \(\int \frac {x^8}{\sqrt {a+b x^3}} \, dx\) [405]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {272, 45} \begin {gather*} \frac {2 a^2 \sqrt {a+b x^3}}{3 b^3}+\frac {2 \left (a+b x^3\right )^{5/2}}{15 b^3}-\frac {4 a \left (a+b x^3\right )^{3/2}}{9 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/Sqrt[a + b*x^3],x]

[Out]

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

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Mathematica [A]
time = 0.03, size = 39, normalized size = 0.66 \begin {gather*} \frac {2 \sqrt {a+b x^3} \left (8 a^2-4 a b x^3+3 b^2 x^6\right )}{45 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/Sqrt[a + b*x^3],x]

[Out]

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

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Maple [A]
time = 0.14, size = 54, normalized size = 0.92

method	result	size
gosper	\(\frac {2 \sqrt {b \,x^{3}+a}\, \left (3 b^{2} x^{6}-4 a b \,x^{3}+8 a^{2}\right )}{45 b^{3}}\)	\(36\)
trager	\(\frac {2 \sqrt {b \,x^{3}+a}\, \left (3 b^{2} x^{6}-4 a b \,x^{3}+8 a^{2}\right )}{45 b^{3}}\)	\(36\)
risch	\(\frac {2 \sqrt {b \,x^{3}+a}\, \left (3 b^{2} x^{6}-4 a b \,x^{3}+8 a^{2}\right )}{45 b^{3}}\)	\(36\)
default	\(\frac {2 x^{6} \sqrt {b \,x^{3}+a}}{15 b}-\frac {8 a \,x^{3} \sqrt {b \,x^{3}+a}}{45 b^{2}}+\frac {16 a^{2} \sqrt {b \,x^{3}+a}}{45 b^{3}}\)	\(54\)
elliptic	\(\frac {2 x^{6} \sqrt {b \,x^{3}+a}}{15 b}-\frac {8 a \,x^{3} \sqrt {b \,x^{3}+a}}{45 b^{2}}+\frac {16 a^{2} \sqrt {b \,x^{3}+a}}{45 b^{3}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*x^6*(b*x^3+a)^(1/2)/b-8/45*a/b^2*x^3*(b*x^3+a)^(1/2)+16/45*a^2*(b*x^3+a)^(1/2)/b^3

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Maxima [A]
time = 0.29, size = 47, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{15 \, b^{3}} - \frac {4 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a}{9 \, b^{3}} + \frac {2 \, \sqrt {b x^{3} + a} a^{2}}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 35, normalized size = 0.59 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} x^{6} - 4 \, a b x^{3} + 8 \, a^{2}\right )} \sqrt {b x^{3} + a}}{45 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.35, size = 70, normalized size = 1.19 \begin {gather*} \begin {cases} \frac {16 a^{2} \sqrt {a + b x^{3}}}{45 b^{3}} - \frac {8 a x^{3} \sqrt {a + b x^{3}}}{45 b^{2}} + \frac {2 x^{6} \sqrt {a + b x^{3}}}{15 b} & \text {for}\: b \neq 0 \\\frac {x^{9}}{9 \sqrt {a}} & \text {otherwise} \end {cases} \end {gather*}

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**2*sqrt(a + b*x**3)/(45*b**3) - 8*a*x**3*sqrt(a + b*x**3)/(45*b**2) + 2*x**6*sqrt(a + b*x**3)/
(15*b), Ne(b, 0)), (x**9/(9*sqrt(a)), True))

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Giac [A]
time = 0.97, size = 47, normalized size = 0.80 \begin {gather*} \frac {2 \, \sqrt {b x^{3} + a} a^{2}}{3 \, b^{3}} + \frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a\right )}}{45 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(b*x^3 + a)*a^2/b^3 + 2/45*(3*(b*x^3 + a)^(5/2) - 10*(b*x^3 + a)^(3/2)*a)/b^3

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Mupad [B]
time = 1.11, size = 35, normalized size = 0.59 \begin {gather*} \frac {2\,\sqrt {b\,x^3+a}\,\left (8\,a^2-4\,a\,b\,x^3+3\,b^2\,x^6\right )}{45\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(a + b*x^3)^(1/2),x)

[Out]

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

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