\(\int \frac {x^8}{(a+b x^{3/2})^{2/3}} \, dx\) [2276]

Optimal result

Integrand size = 17, antiderivative size = 130 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 a^5 \sqrt [3]{a+b x^{3/2}}}{b^6}+\frac {5 a^4 \left (a+b x^{3/2}\right )^{4/3}}{2 b^6}-\frac {20 a^3 \left (a+b x^{3/2}\right )^{7/3}}{7 b^6}+\frac {2 a^2 \left (a+b x^{3/2}\right )^{10/3}}{b^6}-\frac {10 a \left (a+b x^{3/2}\right )^{13/3}}{13 b^6}+\frac {\left (a+b x^{3/2}\right )^{16/3}}{8 b^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {272, 45} \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 a^5 \sqrt [3]{a+b x^{3/2}}}{b^6}+\frac {5 a^4 \left (a+b x^{3/2}\right )^{4/3}}{2 b^6}-\frac {20 a^3 \left (a+b x^{3/2}\right )^{7/3}}{7 b^6}+\frac {2 a^2 \left (a+b x^{3/2}\right )^{10/3}}{b^6}+\frac {\left (a+b x^{3/2}\right )^{16/3}}{8 b^6}-\frac {10 a \left (a+b x^{3/2}\right )^{13/3}}{13 b^6} \]

[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 45

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 272

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*} \text {integral}& = \frac {2}{3} \text {Subst}\left (\int \frac {x^5}{(a+b x)^{2/3}} \, dx,x,x^{3/2}\right ) \\ & = \frac {2}{3} \text {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] (verified)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {\sqrt [3]{a+b x^{3/2}} \left (-729 a^5+243 a^4 b x^{3/2}-162 a^3 b^2 x^3+126 a^2 b^3 x^{9/2}-105 a b^4 x^6+91 b^5 x^{15/2}\right )}{728 b^6} \]

[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)

Maple [F]

\[\int \frac {x^{8}}{\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.55 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {{\left (105 \, a b^{4} x^{6} + 162 \, a^{3} b^{2} x^{3} + 729 \, a^{5} - {\left (91 \, b^{5} x^{7} + 126 \, a^{2} b^{3} x^{4} + 243 \, a^{4} b x\right )} \sqrt {x}\right )} {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}}}{728 \, b^{6}} \]

[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

Sympy [A] (verification not implemented)

Time = 12.78 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.19 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\begin {cases} - \frac {729 a^{5} \sqrt [3]{a + b x^{\frac {3}{2}}}}{728 b^{6}} + \frac {243 a^{4} x^{\frac {3}{2}} \sqrt [3]{a + b x^{\frac {3}{2}}}}{728 b^{5}} - \frac {81 a^{3} x^{3} \sqrt [3]{a + b x^{\frac {3}{2}}}}{364 b^{4}} + \frac {9 a^{2} x^{\frac {9}{2}} \sqrt [3]{a + b x^{\frac {3}{2}}}}{52 b^{3}} - \frac {15 a x^{6} \sqrt [3]{a + b x^{\frac {3}{2}}}}{104 b^{2}} + \frac {x^{\frac {15}{2}} \sqrt [3]{a + b x^{\frac {3}{2}}}}{8 b} & \text {for}\: b \neq 0 \\\frac {x^{9}}{9 a^{\frac {2}{3}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-729*a**5*(a + b*x**(3/2))**(1/3)/(728*b**6) + 243*a**4*x**(3/2)*(a + b*x**(3/2))**(1/3)/(728*b**5)
- 81*a**3*x**3*(a + b*x**(3/2))**(1/3)/(364*b**4) + 9*a**2*x**(9/2)*(a + b*x**(3/2))**(1/3)/(52*b**3) - 15*a*
x**6*(a + b*x**(3/2))**(1/3)/(104*b**2) + x**(15/2)*(a + b*x**(3/2))**(1/3)/(8*b), Ne(b, 0)), (x**9/(9*a**(2/3
)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {16}{3}}}{8 \, b^{6}} - \frac {10 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {13}{3}} a}{13 \, b^{6}} + \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {10}{3}} a^{2}}{b^{6}} - \frac {20 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {7}{3}} a^{3}}{7 \, b^{6}} + \frac {5 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}} a^{4}}{2 \, b^{6}} - \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{5}}{b^{6}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{5}}{b^{6}} + \frac {91 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {16}{3}} - 560 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {13}{3}} a + 1456 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {10}{3}} a^{2} - 2080 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {7}{3}} a^{3} + 1820 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}} a^{4}}{728 \, b^{6}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.82 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {{\left (a+b\,x^{3/2}\right )}^{16/3}}{8\,b^6}-\frac {10\,a\,{\left (a+b\,x^{3/2}\right )}^{13/3}}{13\,b^6}-\frac {2\,a^5\,{\left (a+b\,x^{3/2}\right )}^{1/3}}{b^6}+\frac {5\,a^4\,{\left (a+b\,x^{3/2}\right )}^{4/3}}{2\,b^6}-\frac {20\,a^3\,{\left (a+b\,x^{3/2}\right )}^{7/3}}{7\,b^6}+\frac {2\,a^2\,{\left (a+b\,x^{3/2}\right )}^{10/3}}{b^6} \]

[In]

int(x^8/(a + b*x^(3/2))^(2/3),x)

[Out]

(a + b*x^(3/2))^(16/3)/(8*b^6) - (10*a*(a + b*x^(3/2))^(13/3))/(13*b^6) - (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