Integrand size = 22, antiderivative size = 103 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 a^2 (A b-a B)}{9 b^4 \left (a+b x^3\right )^{3/2}}+\frac {2 a (2 A b-3 a B)}{3 b^4 \sqrt {a+b x^3}}+\frac {2 (A b-3 a B) \sqrt {a+b x^3}}{3 b^4}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^4} \]
-2/9*a^2*(A*b-B*a)/b^4/(b*x^3+a)^(3/2)+2/9*B*(b*x^3+a)^(3/2)/b^4+2/3*a*(2* A*b-3*B*a)/b^4/(b*x^3+a)^(1/2)+2/3*(A*b-3*B*a)*(b*x^3+a)^(1/2)/b^4
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.71 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 \left (-16 a^3 B+8 a^2 b \left (A-3 B x^3\right )-6 a b^2 x^3 \left (-2 A+B x^3\right )+b^3 x^6 \left (3 A+B x^3\right )\right )}{9 b^4 \left (a+b x^3\right )^{3/2}} \]
(2*(-16*a^3*B + 8*a^2*b*(A - 3*B*x^3) - 6*a*b^2*x^3*(-2*A + B*x^3) + b^3*x ^6*(3*A + B*x^3)))/(9*b^4*(a + b*x^3)^(3/2))
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {948, 86, 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.
\(\displaystyle \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6 \left (B x^3+A\right )}{\left (b x^3+a\right )^{5/2}}dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (-\frac {(a B-A b) a^2}{b^3 \left (b x^3+a\right )^{5/2}}+\frac {(3 a B-2 A b) a}{b^3 \left (b x^3+a\right )^{3/2}}+\frac {B \sqrt {b x^3+a}}{b^3}+\frac {A b-3 a B}{b^3 \sqrt {b x^3+a}}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {2 a^2 (A b-a B)}{3 b^4 \left (a+b x^3\right )^{3/2}}+\frac {2 a (2 A b-3 a B)}{b^4 \sqrt {a+b x^3}}+\frac {2 \sqrt {a+b x^3} (A b-3 a B)}{b^4}+\frac {2 B \left (a+b x^3\right )^{3/2}}{3 b^4}\right )\) |
((-2*a^2*(A*b - a*B))/(3*b^4*(a + b*x^3)^(3/2)) + (2*a*(2*A*b - 3*a*B))/(b ^4*Sqrt[a + b*x^3]) + (2*(A*b - 3*a*B)*Sqrt[a + b*x^3])/b^4 + (2*B*(a + b* x^3)^(3/2))/(3*b^4))/3
3.3.44.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 4.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {\left (2 B \,x^{9}+6 A \,x^{6}\right ) b^{3}+24 x^{3} \left (-\frac {x^{3} B}{2}+A \right ) a \,b^{2}+16 a^{2} \left (-3 x^{3} B +A \right ) b -32 a^{3} B}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{4}}\) | \(70\) |
risch | \(\frac {2 \left (b B \,x^{3}+3 A b -8 B a \right ) \sqrt {b \,x^{3}+a}}{9 b^{4}}+\frac {2 a \left (6 A \,b^{2} x^{3}-9 B a b \,x^{3}+5 a b A -8 a^{2} B \right )}{9 b^{4} \left (b \,x^{3}+a \right )^{\frac {3}{2}}}\) | \(75\) |
gosper | \(\frac {\frac {2}{9} b^{3} B \,x^{9}+\frac {2}{3} x^{6} b^{3} A -\frac {4}{3} B \,x^{6} a \,b^{2}+\frac {8}{3} a A \,b^{2} x^{3}-\frac {16}{3} B \,a^{2} b \,x^{3}+\frac {16}{9} a^{2} b A -\frac {32}{9} a^{3} B}{\left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{4}}\) | \(76\) |
trager | \(\frac {\frac {2}{9} b^{3} B \,x^{9}+\frac {2}{3} x^{6} b^{3} A -\frac {4}{3} B \,x^{6} a \,b^{2}+\frac {8}{3} a A \,b^{2} x^{3}-\frac {16}{3} B \,a^{2} b \,x^{3}+\frac {16}{9} a^{2} b A -\frac {32}{9} a^{3} B}{\left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{4}}\) | \(76\) |
elliptic | \(-\frac {2 a^{2} \left (A b -B a \right ) \sqrt {b \,x^{3}+a}}{9 b^{6} \left (x^{3}+\frac {a}{b}\right )^{2}}+\frac {2 \left (2 A b -3 B a \right ) a}{3 b^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 B \,x^{3} \sqrt {b \,x^{3}+a}}{9 b^{3}}+\frac {2 \left (\frac {A b -2 B a}{b^{3}}-\frac {2 B a}{3 b^{3}}\right ) \sqrt {b \,x^{3}+a}}{3 b}\) | \(118\) |
default | \(A \left (-\frac {2 a^{2} \sqrt {b \,x^{3}+a}}{9 b^{5} \left (x^{3}+\frac {a}{b}\right )^{2}}+\frac {4 a}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 \sqrt {b \,x^{3}+a}}{3 b^{3}}\right )+B \left (\frac {2 a^{3} \sqrt {b \,x^{3}+a}}{9 b^{6} \left (x^{3}+\frac {a}{b}\right )^{2}}-\frac {2 a^{2}}{b^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{3} \sqrt {b \,x^{3}+a}}{9 b^{3}}-\frac {16 a \sqrt {b \,x^{3}+a}}{9 b^{4}}\right )\) | \(150\) |
1/9*((2*B*x^9+6*A*x^6)*b^3+24*x^3*(-1/2*x^3*B+A)*a*b^2+16*a^2*(-3*B*x^3+A) *b-32*a^3*B)/(b*x^3+a)^(3/2)/b^4
Time = 0.30 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.95 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 \, {\left (B b^{3} x^{9} - 3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} x^{6} - 16 \, B a^{3} + 8 \, A a^{2} b - 12 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{9 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} \]
2/9*(B*b^3*x^9 - 3*(2*B*a*b^2 - A*b^3)*x^6 - 16*B*a^3 + 8*A*a^2*b - 12*(2* B*a^2*b - A*a*b^2)*x^3)*sqrt(b*x^3 + a)/(b^6*x^6 + 2*a*b^5*x^3 + a^2*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (99) = 198\).
Time = 0.59 (sec) , antiderivative size = 338, normalized size of antiderivative = 3.28 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\begin {cases} \frac {16 A a^{2} b}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} + \frac {24 A a b^{2} x^{3}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} + \frac {6 A b^{3} x^{6}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} - \frac {32 B a^{3}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} - \frac {48 B a^{2} b x^{3}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} - \frac {12 B a b^{2} x^{6}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} + \frac {2 B b^{3} x^{9}}{9 a b^{4} \sqrt {a + b x^{3}} + 9 b^{5} x^{3} \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{9}}{9} + \frac {B x^{12}}{12}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((16*A*a**2*b/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt(a + b *x**3)) + 24*A*a*b**2*x**3/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt(a + b*x**3)) + 6*A*b**3*x**6/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt( a + b*x**3)) - 32*B*a**3/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt(a + b*x**3)) - 48*B*a**2*b*x**3/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3*sqrt (a + b*x**3)) - 12*B*a*b**2*x**6/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3* sqrt(a + b*x**3)) + 2*B*b**3*x**9/(9*a*b**4*sqrt(a + b*x**3) + 9*b**5*x**3 *sqrt(a + b*x**3)), Ne(b, 0)), ((A*x**9/9 + B*x**12/12)/a**(5/2), True))
Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2}{9} \, B {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}}}{b^{4}} - \frac {9 \, \sqrt {b x^{3} + a} a}{b^{4}} - \frac {9 \, a^{2}}{\sqrt {b x^{3} + a} b^{4}} + \frac {a^{3}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{4}}\right )} + \frac {2}{9} \, A {\left (\frac {3 \, \sqrt {b x^{3} + a}}{b^{3}} + \frac {6 \, a}{\sqrt {b x^{3} + a} b^{3}} - \frac {a^{2}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3}}\right )} \]
2/9*B*((b*x^3 + a)^(3/2)/b^4 - 9*sqrt(b*x^3 + a)*a/b^4 - 9*a^2/(sqrt(b*x^3 + a)*b^4) + a^3/((b*x^3 + a)^(3/2)*b^4)) + 2/9*A*(3*sqrt(b*x^3 + a)/b^3 + 6*a/(sqrt(b*x^3 + a)*b^3) - a^2/((b*x^3 + a)^(3/2)*b^3))
Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (b x^{3} + a\right )} B a^{2} - B a^{3} - 6 \, {\left (b x^{3} + a\right )} A a b + A a^{2} b\right )}}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{4}} + \frac {2 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} B b^{8} - 9 \, \sqrt {b x^{3} + a} B a b^{8} + 3 \, \sqrt {b x^{3} + a} A b^{9}\right )}}{9 \, b^{12}} \]
-2/9*(9*(b*x^3 + a)*B*a^2 - B*a^3 - 6*(b*x^3 + a)*A*a*b + A*a^2*b)/((b*x^3 + a)^(3/2)*b^4) + 2/9*((b*x^3 + a)^(3/2)*B*b^8 - 9*sqrt(b*x^3 + a)*B*a*b^ 8 + 3*sqrt(b*x^3 + a)*A*b^9)/b^12
Time = 7.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.41 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {\sqrt {b\,x^3+a}\,\left (\frac {2\,\left (A\,b-2\,B\,a\right )}{b^3}-\frac {4\,B\,a}{3\,b^3}\right )}{3\,b}-\frac {\frac {2\,B\,a^2-2\,A\,a\,b}{3\,b^4}-\frac {a\,\left (\frac {2\,A\,b^2-2\,B\,a\,b}{3\,b^4}-\frac {2\,B\,a}{3\,b^3}\right )}{b}}{\sqrt {b\,x^3+a}}-\frac {a^2\,\left (\frac {2\,A}{9\,b}-\frac {2\,B\,a}{9\,b^2}\right )}{b^2\,{\left (b\,x^3+a\right )}^{3/2}}+\frac {2\,B\,x^3\,\sqrt {b\,x^3+a}}{9\,b^3} \]