3.423 \(\int \frac{x^5}{(a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=38 \[ \frac{2 a}{3 b^2 \sqrt{a+b x^3}}+\frac{2 \sqrt{a+b x^3}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0225487, 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} \[ \frac{2 a}{3 b^2 \sqrt{a+b x^3}}+\frac{2 \sqrt{a+b x^3}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^3)^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.011241, size = 27, normalized size = 0.71 \[ \frac{2 \left (2 a+b x^3\right )}{3 b^2 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^3)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.004, size = 24, normalized size = 0.6 \begin{align*}{\frac{2\,b{x}^{3}+4\,a}{3\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{3}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.957168, size = 41, normalized size = 1.08 \begin{align*} \frac{2 \, \sqrt{b x^{3} + a}}{3 \, b^{2}} + \frac{2 \, a}{3 \, \sqrt{b x^{3} + a} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.40859, size = 72, normalized size = 1.89 \begin{align*} \frac{2 \,{\left (b x^{3} + 2 \, a\right )} \sqrt{b x^{3} + a}}{3 \,{\left (b^{3} x^{3} + a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.11768, size = 46, normalized size = 1.21 \begin{align*} \begin{cases} \frac{4 a}{3 b^{2} \sqrt{a + b x^{3}}} + \frac{2 x^{3}}{3 b \sqrt{a + b x^{3}}} & \text{for}\: b \neq 0 \\\frac{x^{6}}{6 a^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12257, size = 35, normalized size = 0.92 \begin{align*} \frac{2 \,{\left (\sqrt{b x^{3} + a} + \frac{a}{\sqrt{b x^{3} + a}}\right )}}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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