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

Optimal. Leaf size=41 \[ \frac {2 \sqrt {a x^3-b} \left (3 a^2 x^6+4 a b x^3+8 b^2\right )}{45 a^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.59, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {266, 43} \begin {gather*} \frac {2 b^2 \sqrt {a x^3-b}}{3 a^3}+\frac {2 \left (a x^3-b\right )^{5/2}}{15 a^3}+\frac {4 b \left (a x^3-b\right )^{3/2}}{9 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/Sqrt[-b + a*x^3],x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 41, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {a x^3-b} \left (3 a^2 x^6+4 a b x^3+8 b^2\right )}{45 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/Sqrt[-b + a*x^3],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 41, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {-b+a x^3} \left (8 b^2+4 a b x^3+3 a^2 x^6\right )}{45 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8/Sqrt[-b + a*x^3],x]

[Out]

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

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fricas [A]  time = 0.43, size = 37, normalized size = 0.90 \begin {gather*} \frac {2 \, {\left (3 \, a^{2} x^{6} + 4 \, a b x^{3} + 8 \, b^{2}\right )} \sqrt {a x^{3} - b}}{45 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.30, size = 53, normalized size = 1.29 \begin {gather*} \frac {2 \, \sqrt {a x^{3} - b} b^{2}}{3 \, a^{3}} + \frac {2 \, {\left (3 \, {\left (a x^{3} - b\right )}^{\frac {5}{2}} + 10 \, {\left (a x^{3} - b\right )}^{\frac {3}{2}} b\right )}}{45 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 38, normalized size = 0.93

method	result	size
gosper	\(\frac {2 \sqrt {a \,x^{3}-b}\, \left (3 a^{2} x^{6}+4 a b \,x^{3}+8 b^{2}\right )}{45 a^{3}}\)	\(38\)
trager	\(\frac {2 \sqrt {a \,x^{3}-b}\, \left (3 a^{2} x^{6}+4 a b \,x^{3}+8 b^{2}\right )}{45 a^{3}}\)	\(38\)
risch	\(\frac {2 \sqrt {a \,x^{3}-b}\, \left (3 a^{2} x^{6}+4 a b \,x^{3}+8 b^{2}\right )}{45 a^{3}}\)	\(38\)
default	\(\frac {2 x^{6} \sqrt {a \,x^{3}-b}}{15 a}+\frac {8 b \,x^{3} \sqrt {a \,x^{3}-b}}{45 a^{2}}+\frac {16 b^{2} \sqrt {a \,x^{3}-b}}{45 a^{3}}\)	\(60\)
elliptic	\(\frac {2 x^{6} \sqrt {a \,x^{3}-b}}{15 a}+\frac {8 b \,x^{3} \sqrt {a \,x^{3}-b}}{45 a^{2}}+\frac {16 b^{2} \sqrt {a \,x^{3}-b}}{45 a^{3}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 53, normalized size = 1.29 \begin {gather*} \frac {2 \, {\left (a x^{3} - b\right )}^{\frac {5}{2}}}{15 \, a^{3}} + \frac {4 \, {\left (a x^{3} - b\right )}^{\frac {3}{2}} b}{9 \, a^{3}} + \frac {2 \, \sqrt {a x^{3} - b} b^{2}}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.62, size = 37, normalized size = 0.90 \begin {gather*} \frac {2\,\sqrt {a\,x^3-b}\,\left (3\,a^2\,x^6+4\,a\,b\,x^3+8\,b^2\right )}{45\,a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(a*x^3 - b)^(1/2),x)

[Out]

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

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sympy [A]  time = 1.25, size = 71, normalized size = 1.73 \begin {gather*} \begin {cases} \frac {2 x^{6} \sqrt {a x^{3} - b}}{15 a} + \frac {8 b x^{3} \sqrt {a x^{3} - b}}{45 a^{2}} + \frac {16 b^{2} \sqrt {a x^{3} - b}}{45 a^{3}} & \text {for}\: a \neq 0 \\\frac {x^{9}}{9 \sqrt {- b}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*x**6*sqrt(a*x**3 - b)/(15*a) + 8*b*x**3*sqrt(a*x**3 - b)/(45*a**2) + 16*b**2*sqrt(a*x**3 - b)/(45
*a**3), Ne(a, 0)), (x**9/(9*sqrt(-b)), True))

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