Integrand size = 17, antiderivative size = 130 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 a^5 \sqrt [3]{a+b x^{3/2}}}{b^6}+\frac {5 a^4 \left (a+b x^{3/2}\right )^{4/3}}{2 b^6}-\frac {20 a^3 \left (a+b x^{3/2}\right )^{7/3}}{7 b^6}+\frac {2 a^2 \left (a+b x^{3/2}\right )^{10/3}}{b^6}-\frac {10 a \left (a+b x^{3/2}\right )^{13/3}}{13 b^6}+\frac {\left (a+b x^{3/2}\right )^{16/3}}{8 b^6} \] Output:
-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
Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {\sqrt [3]{a+b x^{3/2}} \left (-729 a^5+243 a^4 b x^{3/2}-162 a^3 b^2 x^3+126 a^2 b^3 x^{9/2}-105 a b^4 x^6+91 b^5 x^{15/2}\right )}{728 b^6} \] Input:
Integrate[x^8/(a + b*x^(3/2))^(2/3),x]
Output:
((a + b*x^(3/2))^(1/3)*(-729*a^5 + 243*a^4*b*x^(3/2) - 162*a^3*b^2*x^3 + 1 26*a^2*b^3*x^(9/2) - 105*a*b^4*x^6 + 91*b^5*x^(15/2)))/(728*b^6)
Time = 0.39 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {798, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {2}{3} \int \frac {x^{15/2}}{\left (b x^{3/2}+a\right )^{2/3}}dx^{3/2}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {2}{3} \int \left (-\frac {a^5}{b^5 \left (b x^{3/2}+a\right )^{2/3}}+\frac {5 \sqrt [3]{b x^{3/2}+a} a^4}{b^5}-\frac {10 \left (b x^{3/2}+a\right )^{4/3} a^3}{b^5}+\frac {10 \left (b x^{3/2}+a\right )^{7/3} a^2}{b^5}-\frac {5 \left (b x^{3/2}+a\right )^{10/3} a}{b^5}+\frac {\left (b x^{3/2}+a\right )^{13/3}}{b^5}\right )dx^{3/2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{3} \left (-\frac {3 a^5 \sqrt [3]{a+b x^{3/2}}}{b^6}+\frac {15 a^4 \left (a+b x^{3/2}\right )^{4/3}}{4 b^6}-\frac {30 a^3 \left (a+b x^{3/2}\right )^{7/3}}{7 b^6}+\frac {3 a^2 \left (a+b x^{3/2}\right )^{10/3}}{b^6}+\frac {3 \left (a+b x^{3/2}\right )^{16/3}}{16 b^6}-\frac {15 a \left (a+b x^{3/2}\right )^{13/3}}{13 b^6}\right )\) |
Input:
Int[x^8/(a + b*x^(3/2))^(2/3),x]
Output:
(2*((-3*a^5*(a + b*x^(3/2))^(1/3))/b^6 + (15*a^4*(a + b*x^(3/2))^(4/3))/(4 *b^6) - (30*a^3*(a + b*x^(3/2))^(7/3))/(7*b^6) + (3*a^2*(a + b*x^(3/2))^(1 0/3))/b^6 - (15*a*(a + b*x^(3/2))^(13/3))/(13*b^6) + (3*(a + b*x^(3/2))^(1 6/3))/(16*b^6)))/3
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] && 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])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
\[\int \frac {x^{8}}{\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}}d x\]
Input:
int(x^8/(a+b*x^(3/2))^(2/3),x)
Output:
int(x^8/(a+b*x^(3/2))^(2/3),x)
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.55 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {{\left (105 \, a b^{4} x^{6} + 162 \, a^{3} b^{2} x^{3} + 729 \, a^{5} - {\left (91 \, b^{5} x^{7} + 126 \, a^{2} b^{3} x^{4} + 243 \, a^{4} b x\right )} \sqrt {x}\right )} {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}}}{728 \, b^{6}} \] Input:
integrate(x^8/(a+b*x^(3/2))^(2/3),x, algorithm="fricas")
Output:
-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
Time = 12.96 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.19 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\begin {cases} - \frac {729 a^{5} \sqrt [3]{a + b x^{\frac {3}{2}}}}{728 b^{6}} + \frac {243 a^{4} x^{\frac {3}{2}} \sqrt [3]{a + b x^{\frac {3}{2}}}}{728 b^{5}} - \frac {81 a^{3} x^{3} \sqrt [3]{a + b x^{\frac {3}{2}}}}{364 b^{4}} + \frac {9 a^{2} x^{\frac {9}{2}} \sqrt [3]{a + b x^{\frac {3}{2}}}}{52 b^{3}} - \frac {15 a x^{6} \sqrt [3]{a + b x^{\frac {3}{2}}}}{104 b^{2}} + \frac {x^{\frac {15}{2}} \sqrt [3]{a + b x^{\frac {3}{2}}}}{8 b} & \text {for}\: b \neq 0 \\\frac {x^{9}}{9 a^{\frac {2}{3}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**8/(a+b*x**(3/2))**(2/3),x)
Output:
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))
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {16}{3}}}{8 \, b^{6}} - \frac {10 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {13}{3}} a}{13 \, b^{6}} + \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {10}{3}} a^{2}}{b^{6}} - \frac {20 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {7}{3}} a^{3}}{7 \, b^{6}} + \frac {5 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}} a^{4}}{2 \, b^{6}} - \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{5}}{b^{6}} \] Input:
integrate(x^8/(a+b*x^(3/2))^(2/3),x, algorithm="maxima")
Output:
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
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{5}}{b^{6}} + \frac {91 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {16}{3}} - 560 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {13}{3}} a + 1456 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {10}{3}} a^{2} - 2080 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {7}{3}} a^{3} + 1820 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}} a^{4}}{728 \, b^{6}} \] Input:
integrate(x^8/(a+b*x^(3/2))^(2/3),x, algorithm="giac")
Output:
-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
Time = 0.67 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {{\left (a+b\,x^{3/2}\right )}^{16/3}}{8\,b^6}-\frac {10\,a\,{\left (a+b\,x^{3/2}\right )}^{13/3}}{13\,b^6}-\frac {2\,a^5\,{\left (a+b\,x^{3/2}\right )}^{1/3}}{b^6}+\frac {5\,a^4\,{\left (a+b\,x^{3/2}\right )}^{4/3}}{2\,b^6}-\frac {20\,a^3\,{\left (a+b\,x^{3/2}\right )}^{7/3}}{7\,b^6}+\frac {2\,a^2\,{\left (a+b\,x^{3/2}\right )}^{10/3}}{b^6} \] Input:
int(x^8/(a + b*x^(3/2))^(2/3),x)
Output:
(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
\[ \int \frac {x^8}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\int \frac {x^{8}}{\left (\sqrt {x}\, b x +a \right )^{\frac {2}{3}}}d x \] Input:
int(x^8/(a+b*x^(3/2))^(2/3),x)
Output:
int(x**8/(sqrt(x)*b*x + a)**(2/3),x)