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

Optimal. Leaf size=38 \[ \frac{2 \left (a+b x^3\right )^{5/2}}{15 b^2}-\frac{2 a \left (a+b x^3\right )^{3/2}}{9 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0224393, antiderivative size = 38, normalized size of antiderivative = 1., 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} \[ \frac{2 \left (a+b x^3\right )^{5/2}}{15 b^2}-\frac{2 a \left (a+b x^3\right )^{3/2}}{9 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Sqrt[a + b*x^3],x]

[Out]

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

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]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0132593, size = 28, normalized size = 0.74 \[ \frac{2 \left (a+b x^3\right )^{3/2} \left (3 b x^3-2 a\right )}{45 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[a + b*x^3],x]

[Out]

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

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Maple [A]  time = 0.005, size = 25, normalized size = 0.7 \begin{align*} -{\frac{-6\,b{x}^{3}+4\,a}{45\,{b}^{2}} \left ( b{x}^{3}+a \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.987786, size = 41, normalized size = 1.08 \begin{align*} \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}}}{15 \, b^{2}} - \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a}{9 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.4963, size = 76, normalized size = 2. \begin{align*} \frac{2 \,{\left (3 \, b^{2} x^{6} + a b x^{3} - 2 \, a^{2}\right )} \sqrt{b x^{3} + a}}{45 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.767954, size = 66, normalized size = 1.74 \begin{align*} \begin{cases} - \frac{4 a^{2} \sqrt{a + b x^{3}}}{45 b^{2}} + \frac{2 a x^{3} \sqrt{a + b x^{3}}}{45 b} + \frac{2 x^{6} \sqrt{a + b x^{3}}}{15} & \text{for}\: b \neq 0 \\\frac{\sqrt{a} x^{6}}{6} & \text{otherwise} \end{cases} \end{align*}

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

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Giac [A]  time = 1.10482, size = 39, normalized size = 1.03 \begin{align*} \frac{2 \,{\left (3 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a\right )}}{45 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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