3.2276 \(\int \frac{x^8}{(a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=130 \[ \frac{2 a^2 \left (a+b x^{3/2}\right )^{10/3}}{b^6}-\frac{20 a^3 \left (a+b x^{3/2}\right )^{7/3}}{7 b^6}+\frac{5 a^4 \left (a+b x^{3/2}\right )^{4/3}}{2 b^6}-\frac{2 a^5 \sqrt [3]{a+b x^{3/2}}}{b^6}+\frac{\left (a+b x^{3/2}\right )^{16/3}}{8 b^6}-\frac{10 a \left (a+b x^{3/2}\right )^{13/3}}{13 b^6} \]

[Out]

(-2*a^5*(a + b*x^(3/2))^(1/3))/b^6 + (5*a^4*(a + b*x^(3/2))^(4/3))/(2*b^6) - (20*a^3*(a + b*x^(3/2))^(7/3))/(7
*b^6) + (2*a^2*(a + b*x^(3/2))^(10/3))/b^6 - (10*a*(a + b*x^(3/2))^(13/3))/(13*b^6) + (a + b*x^(3/2))^(16/3)/(
8*b^6)

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Rubi [A]  time = 0.0614772, antiderivative size = 130, 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} \[ \frac{2 a^2 \left (a+b x^{3/2}\right )^{10/3}}{b^6}-\frac{20 a^3 \left (a+b x^{3/2}\right )^{7/3}}{7 b^6}+\frac{5 a^4 \left (a+b x^{3/2}\right )^{4/3}}{2 b^6}-\frac{2 a^5 \sqrt [3]{a+b x^{3/2}}}{b^6}+\frac{\left (a+b x^{3/2}\right )^{16/3}}{8 b^6}-\frac{10 a \left (a+b x^{3/2}\right )^{13/3}}{13 b^6} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^(3/2))^(2/3),x]

[Out]

(-2*a^5*(a + b*x^(3/2))^(1/3))/b^6 + (5*a^4*(a + b*x^(3/2))^(4/3))/(2*b^6) - (20*a^3*(a + b*x^(3/2))^(7/3))/(7
*b^6) + (2*a^2*(a + b*x^(3/2))^(10/3))/b^6 - (10*a*(a + b*x^(3/2))^(13/3))/(13*b^6) + (a + b*x^(3/2))^(16/3)/(
8*b^6)

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^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx &=\frac{2}{3} \operatorname{Subst}\left (\int \frac{x^5}{(a+b x)^{2/3}} \, dx,x,x^{3/2}\right )\\ &=\frac{2}{3} \operatorname{Subst}\left (\int \left (-\frac{a^5}{b^5 (a+b x)^{2/3}}+\frac{5 a^4 \sqrt [3]{a+b x}}{b^5}-\frac{10 a^3 (a+b x)^{4/3}}{b^5}+\frac{10 a^2 (a+b x)^{7/3}}{b^5}-\frac{5 a (a+b x)^{10/3}}{b^5}+\frac{(a+b x)^{13/3}}{b^5}\right ) \, dx,x,x^{3/2}\right )\\ &=-\frac{2 a^5 \sqrt [3]{a+b x^{3/2}}}{b^6}+\frac{5 a^4 \left (a+b x^{3/2}\right )^{4/3}}{2 b^6}-\frac{20 a^3 \left (a+b x^{3/2}\right )^{7/3}}{7 b^6}+\frac{2 a^2 \left (a+b x^{3/2}\right )^{10/3}}{b^6}-\frac{10 a \left (a+b x^{3/2}\right )^{13/3}}{13 b^6}+\frac{\left (a+b x^{3/2}\right )^{16/3}}{8 b^6}\\ \end{align*}

Mathematica [A]  time = 0.0355528, size = 80, normalized size = 0.62 \[ \frac{\sqrt [3]{a+b x^{3/2}} \left (126 a^2 b^3 x^{9/2}-162 a^3 b^2 x^3+243 a^4 b x^{3/2}-729 a^5-105 a b^4 x^6+91 b^5 x^{15/2}\right )}{728 b^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^(3/2))^(2/3),x]

[Out]

((a + b*x^(3/2))^(1/3)*(-729*a^5 + 243*a^4*b*x^(3/2) - 162*a^3*b^2*x^3 + 126*a^2*b^3*x^(9/2) - 105*a*b^4*x^6 +
 91*b^5*x^(15/2)))/(728*b^6)

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Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int{{x}^{8} \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^8/(a+b*x^(3/2))^(2/3),x)

[Out]

int(x^8/(a+b*x^(3/2))^(2/3),x)

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Maxima [A]  time = 0.974046, size = 132, normalized size = 1.02 \begin{align*} \frac{{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{16}{3}}}{8 \, b^{6}} - \frac{10 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{13}{3}} a}{13 \, b^{6}} + \frac{2 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{10}{3}} a^{2}}{b^{6}} - \frac{20 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{7}{3}} a^{3}}{7 \, b^{6}} + \frac{5 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{4}{3}} a^{4}}{2 \, b^{6}} - \frac{2 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}} a^{5}}{b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x^(3/2) + a)^(16/3)/b^6 - 10/13*(b*x^(3/2) + a)^(13/3)*a/b^6 + 2*(b*x^(3/2) + a)^(10/3)*a^2/b^6 - 20/7*
(b*x^(3/2) + a)^(7/3)*a^3/b^6 + 5/2*(b*x^(3/2) + a)^(4/3)*a^4/b^6 - 2*(b*x^(3/2) + a)^(1/3)*a^5/b^6

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Fricas [A]  time = 3.93249, size = 180, normalized size = 1.38 \begin{align*} -\frac{{\left (105 \, a b^{4} x^{6} + 162 \, a^{3} b^{2} x^{3} + 729 \, a^{5} -{\left (91 \, b^{5} x^{7} + 126 \, a^{2} b^{3} x^{4} + 243 \, a^{4} b x\right )} \sqrt{x}\right )}{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}}}{728 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/728*(105*a*b^4*x^6 + 162*a^3*b^2*x^3 + 729*a^5 - (91*b^5*x^7 + 126*a^2*b^3*x^4 + 243*a^4*b*x)*sqrt(x))*(b*x
^(3/2) + a)^(1/3)/b^6

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.13806, size = 115, normalized size = 0.88 \begin{align*} \frac{91 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{16}{3}} - 560 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{13}{3}} a + 1456 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{10}{3}} a^{2} - 2080 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{7}{3}} a^{3} + 1820 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{4}{3}} a^{4} - 1456 \,{\left (b x^{\frac{3}{2}} + a\right )}^{\frac{1}{3}} a^{5}}{728 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/728*(91*(b*x^(3/2) + a)^(16/3) - 560*(b*x^(3/2) + a)^(13/3)*a + 1456*(b*x^(3/2) + a)^(10/3)*a^2 - 2080*(b*x^
(3/2) + a)^(7/3)*a^3 + 1820*(b*x^(3/2) + a)^(4/3)*a^4 - 1456*(b*x^(3/2) + a)^(1/3)*a^5)/b^6