3.406 \(\int \frac {x^5}{\sqrt {a+b x^3}} \, dx\)

Optimal. Leaf size=38 \[ \frac {2 \left (a+b x^3\right )^{3/2}}{9 b^2}-\frac {2 a \sqrt {a+b x^3}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, 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} \[ \frac {2 \left (a+b x^3\right )^{3/2}}{9 b^2}-\frac {2 a \sqrt {a+b x^3}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[a + b*x^3],x]

[Out]

(-2*a*Sqrt[a + b*x^3])/(3*b^2) + (2*(a + b*x^3)^(3/2))/(9*b^2)

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.71 \[ \frac {2 \left (b x^3-2 a\right ) \sqrt {a+b x^3}}{9 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/Sqrt[a + b*x^3],x]

[Out]

(2*(-2*a + b*x^3)*Sqrt[a + b*x^3])/(9*b^2)

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fricas [A]  time = 0.81, size = 23, normalized size = 0.61 \[ \frac {2 \, \sqrt {b x^{3} + a} {\left (b x^{3} - 2 \, a\right )}}{9 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

2/9*sqrt(b*x^3 + a)*(b*x^3 - 2*a)/b^2

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giac [A]  time = 0.15, size = 30, normalized size = 0.79 \[ \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{9 \, b^{2}} - \frac {2 \, \sqrt {b x^{3} + a} a}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 25, normalized size = 0.66 \[ -\frac {2 \sqrt {b \,x^{3}+a}\, \left (-b \,x^{3}+2 a \right )}{9 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^3+a)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.32, size = 30, normalized size = 0.79 \[ \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{9 \, b^{2}} - \frac {2 \, \sqrt {b x^{3} + a} a}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.09, size = 24, normalized size = 0.63 \[ -\frac {2\,\sqrt {b\,x^3+a}\,\left (2\,a-b\,x^3\right )}{9\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^3)^(1/2),x)

[Out]

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

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sympy [A]  time = 1.73, size = 46, normalized size = 1.21 \[ \begin {cases} - \frac {4 a \sqrt {a + b x^{3}}}{9 b^{2}} + \frac {2 x^{3} \sqrt {a + b x^{3}}}{9 b} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 \sqrt {a}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**3+a)**(1/2),x)

[Out]

Piecewise((-4*a*sqrt(a + b*x**3)/(9*b**2) + 2*x**3*sqrt(a + b*x**3)/(9*b), Ne(b, 0)), (x**6/(6*sqrt(a)), True)
)

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