3.2277 \(\int \frac{x^5}{(a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=86 \[ \frac{3 a^2 \left (a+b x^{3/2}\right )^{4/3}}{2 b^4}-\frac{2 a^3 \sqrt [3]{a+b x^{3/2}}}{b^4}+\frac{\left (a+b x^{3/2}\right )^{10/3}}{5 b^4}-\frac{6 a \left (a+b x^{3/2}\right )^{7/3}}{7 b^4} \]

[Out]

(-2*a^3*(a + b*x^(3/2))^(1/3))/b^4 + (3*a^2*(a + b*x^(3/2))^(4/3))/(2*b^4) - (6*a*(a + b*x^(3/2))^(7/3))/(7*b^
4) + (a + b*x^(3/2))^(10/3)/(5*b^4)

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Rubi [A]  time = 0.0408598, antiderivative size = 86, normalized size of antiderivative = 1., 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} \[ \frac{3 a^2 \left (a+b x^{3/2}\right )^{4/3}}{2 b^4}-\frac{2 a^3 \sqrt [3]{a+b x^{3/2}}}{b^4}+\frac{\left (a+b x^{3/2}\right )^{10/3}}{5 b^4}-\frac{6 a \left (a+b x^{3/2}\right )^{7/3}}{7 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^(3/2))^(2/3),x]

[Out]

(-2*a^3*(a + b*x^(3/2))^(1/3))/b^4 + (3*a^2*(a + b*x^(3/2))^(4/3))/(2*b^4) - (6*a*(a + b*x^(3/2))^(7/3))/(7*b^
4) + (a + b*x^(3/2))^(10/3)/(5*b^4)

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^{3/2}\right )^{2/3}} \, dx &=\frac{2}{3} \operatorname{Subst}\left (\int \frac{x^3}{(a+b x)^{2/3}} \, dx,x,x^{3/2}\right )\\ &=\frac{2}{3} \operatorname{Subst}\left (\int \left (-\frac{a^3}{b^3 (a+b x)^{2/3}}+\frac{3 a^2 \sqrt [3]{a+b x}}{b^3}-\frac{3 a (a+b x)^{4/3}}{b^3}+\frac{(a+b x)^{7/3}}{b^3}\right ) \, dx,x,x^{3/2}\right )\\ &=-\frac{2 a^3 \sqrt [3]{a+b x^{3/2}}}{b^4}+\frac{3 a^2 \left (a+b x^{3/2}\right )^{4/3}}{2 b^4}-\frac{6 a \left (a+b x^{3/2}\right )^{7/3}}{7 b^4}+\frac{\left (a+b x^{3/2}\right )^{10/3}}{5 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0238537, size = 56, normalized size = 0.65 \[ \frac{\sqrt [3]{a+b x^{3/2}} \left (27 a^2 b x^{3/2}-81 a^3-18 a b^2 x^3+14 b^3 x^{9/2}\right )}{70 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^(3/2))^(2/3),x]

[Out]

((a + b*x^(3/2))^(1/3)*(-81*a^3 + 27*a^2*b*x^(3/2) - 18*a*b^2*x^3 + 14*b^3*x^(9/2)))/(70*b^4)

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{{x}^{5} \left ( a+b{x}^{{\frac{3}{2}}} \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.961197, size = 86, normalized size = 1. \begin{align*} \frac{{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{10}{3}}}{5 \, b^{4}} - \frac{6 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{7}{3}} a}{7 \, b^{4}} + \frac{3 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{4}{3}} a^{2}}{2 \, b^{4}} - \frac{2 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}} a^{3}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b*x^(3/2) + a)^(10/3)/b^4 - 6/7*(b*x^(3/2) + a)^(7/3)*a/b^4 + 3/2*(b*x^(3/2) + a)^(4/3)*a^2/b^4 - 2*(b*x^
(3/2) + a)^(1/3)*a^3/b^4

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Fricas [A]  time = 3.70544, size = 126, normalized size = 1.47 \begin{align*} -\frac{{\left (18 \, a b^{2} x^{3} + 81 \, a^{3} -{\left (14 \, b^{3} x^{4} + 27 \, a^{2} b x\right )} \sqrt{x}\right )}{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}}}{70 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/70*(18*a*b^2*x^3 + 81*a^3 - (14*b^3*x^4 + 27*a^2*b*x)*sqrt(x))*(b*x^(3/2) + a)^(1/3)/b^4

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Sympy [A]  time = 42.394, size = 102, normalized size = 1.19 \begin{align*} \begin{cases} - \frac{81 a^{3} \sqrt [3]{a + b x^{\frac{3}{2}}}}{70 b^{4}} + \frac{27 a^{2} x^{\frac{3}{2}} \sqrt [3]{a + b x^{\frac{3}{2}}}}{70 b^{3}} - \frac{9 a x^{3} \sqrt [3]{a + b x^{\frac{3}{2}}}}{35 b^{2}} + \frac{x^{\frac{9}{2}} \sqrt [3]{a + b x^{\frac{3}{2}}}}{5 b} & \text{for}\: b \neq 0 \\\frac{x^{6}}{6 a^{\frac{2}{3}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-81*a**3*(a + b*x**(3/2))**(1/3)/(70*b**4) + 27*a**2*x**(3/2)*(a + b*x**(3/2))**(1/3)/(70*b**3) - 9
*a*x**3*(a + b*x**(3/2))**(1/3)/(35*b**2) + x**(9/2)*(a + b*x**(3/2))**(1/3)/(5*b), Ne(b, 0)), (x**6/(6*a**(2/
3)), True))

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Giac [A]  time = 1.10293, size = 77, normalized size = 0.9 \begin{align*} \frac{14 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{10}{3}} - 60 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{7}{3}} a + 105 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{4}{3}} a^{2} - 140 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}} a^{3}}{70 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/70*(14*(b*x^(3/2) + a)^(10/3) - 60*(b*x^(3/2) + a)^(7/3)*a + 105*(b*x^(3/2) + a)^(4/3)*a^2 - 140*(b*x^(3/2)
+ a)^(1/3)*a^3)/b^4