Integrand size = 22, antiderivative size = 103 \[ \int \frac {x^8 \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {2 a^2 (A b-a B) \sqrt {a+b x^3}}{3 b^4}-\frac {2 a (2 A b-3 a B) \left (a+b x^3\right )^{3/2}}{9 b^4}+\frac {2 (A b-3 a B) \left (a+b x^3\right )^{5/2}}{15 b^4}+\frac {2 B \left (a+b x^3\right )^{7/2}}{21 b^4} \]
-2/9*a*(2*A*b-3*B*a)*(b*x^3+a)^(3/2)/b^4+2/15*(A*b-3*B*a)*(b*x^3+a)^(5/2)/ b^4+2/21*B*(b*x^3+a)^(7/2)/b^4+2/3*a^2*(A*b-B*a)*(b*x^3+a)^(1/2)/b^4
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int \frac {x^8 \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {2 \sqrt {a+b x^3} \left (56 a^2 A b-48 a^3 B-28 a A b^2 x^3+24 a^2 b B x^3+21 A b^3 x^6-18 a b^2 B x^6+15 b^3 B x^9\right )}{315 b^4} \]
(2*Sqrt[a + b*x^3]*(56*a^2*A*b - 48*a^3*B - 28*a*A*b^2*x^3 + 24*a^2*b*B*x^ 3 + 21*A*b^3*x^6 - 18*a*b^2*B*x^6 + 15*b^3*B*x^9))/(315*b^4)
Time = 0.25 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int \frac {x^8 \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6 \left (B x^3+A\right )}{\sqrt {b x^3+a}}dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{5/2}}{b^3}+\frac {(A b-3 a B) \left (b x^3+a\right )^{3/2}}{b^3}+\frac {a (3 a B-2 A b) \sqrt {b x^3+a}}{b^3}-\frac {a^2 (a B-A b)}{b^3 \sqrt {b x^3+a}}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {2 a^2 \sqrt {a+b x^3} (A b-a B)}{b^4}+\frac {2 \left (a+b x^3\right )^{5/2} (A b-3 a B)}{5 b^4}-\frac {2 a \left (a+b x^3\right )^{3/2} (2 A b-3 a B)}{3 b^4}+\frac {2 B \left (a+b x^3\right )^{7/2}}{7 b^4}\right )\) |
((2*a^2*(A*b - a*B)*Sqrt[a + b*x^3])/b^4 - (2*a*(2*A*b - 3*a*B)*(a + b*x^3 )^(3/2))/(3*b^4) + (2*(A*b - 3*a*B)*(a + b*x^3)^(5/2))/(5*b^4) + (2*B*(a + b*x^3)^(7/2))/(7*b^4))/3
3.3.13.3.1 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 = 4.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {16 \sqrt {b \,x^{3}+a}\, \left (\frac {3 x^{6} \left (\frac {5 x^{3} B}{7}+A \right ) b^{3}}{8}-\frac {x^{3} \left (\frac {9 x^{3} B}{14}+A \right ) a \,b^{2}}{2}+a^{2} \left (\frac {3 x^{3} B}{7}+A \right ) b -\frac {6 a^{3} B}{7}\right )}{45 b^{4}}\) | \(68\) |
gosper | \(\frac {2 \sqrt {b \,x^{3}+a}\, \left (15 b^{3} B \,x^{9}+21 x^{6} b^{3} A -18 B \,x^{6} a \,b^{2}-28 a A \,b^{2} x^{3}+24 B \,a^{2} b \,x^{3}+56 a^{2} b A -48 a^{3} B \right )}{315 b^{4}}\) | \(77\) |
trager | \(\frac {2 \sqrt {b \,x^{3}+a}\, \left (15 b^{3} B \,x^{9}+21 x^{6} b^{3} A -18 B \,x^{6} a \,b^{2}-28 a A \,b^{2} x^{3}+24 B \,a^{2} b \,x^{3}+56 a^{2} b A -48 a^{3} B \right )}{315 b^{4}}\) | \(77\) |
risch | \(\frac {2 \sqrt {b \,x^{3}+a}\, \left (15 b^{3} B \,x^{9}+21 x^{6} b^{3} A -18 B \,x^{6} a \,b^{2}-28 a A \,b^{2} x^{3}+24 B \,a^{2} b \,x^{3}+56 a^{2} b A -48 a^{3} B \right )}{315 b^{4}}\) | \(77\) |
elliptic | \(\frac {2 B \,x^{9} \sqrt {b \,x^{3}+a}}{21 b}+\frac {2 \left (A -\frac {6 a B}{7 b}\right ) x^{6} \sqrt {b \,x^{3}+a}}{15 b}-\frac {8 a \left (A -\frac {6 a B}{7 b}\right ) x^{3} \sqrt {b \,x^{3}+a}}{45 b^{2}}+\frac {16 a^{2} \left (A -\frac {6 a B}{7 b}\right ) \sqrt {b \,x^{3}+a}}{45 b^{3}}\) | \(99\) |
default | \(A \left (\frac {2 x^{6} \sqrt {b \,x^{3}+a}}{15 b}-\frac {8 a \,x^{3} \sqrt {b \,x^{3}+a}}{45 b^{2}}+\frac {16 a^{2} \sqrt {b \,x^{3}+a}}{45 b^{3}}\right )+B \left (\frac {2 x^{9} \sqrt {b \,x^{3}+a}}{21 b}-\frac {4 a \,x^{6} \sqrt {b \,x^{3}+a}}{35 b^{2}}+\frac {16 a^{2} x^{3} \sqrt {b \,x^{3}+a}}{105 b^{3}}-\frac {32 a^{3} \sqrt {b \,x^{3}+a}}{105 b^{4}}\right )\) | \(132\) |
16/45*(b*x^3+a)^(1/2)*(3/8*x^6*(5/7*x^3*B+A)*b^3-1/2*x^3*(9/14*x^3*B+A)*a* b^2+a^2*(3/7*x^3*B+A)*b-6/7*a^3*B)/b^4
Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int \frac {x^8 \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {2 \, {\left (15 \, B b^{3} x^{9} - 3 \, {\left (6 \, B a b^{2} - 7 \, A b^{3}\right )} x^{6} - 48 \, B a^{3} + 56 \, A a^{2} b + 4 \, {\left (6 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{315 \, b^{4}} \]
2/315*(15*B*b^3*x^9 - 3*(6*B*a*b^2 - 7*A*b^3)*x^6 - 48*B*a^3 + 56*A*a^2*b + 4*(6*B*a^2*b - 7*A*a*b^2)*x^3)*sqrt(b*x^3 + a)/b^4
Time = 0.40 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.70 \[ \int \frac {x^8 \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\begin {cases} \frac {16 A a^{2} \sqrt {a + b x^{3}}}{45 b^{3}} - \frac {8 A a x^{3} \sqrt {a + b x^{3}}}{45 b^{2}} + \frac {2 A x^{6} \sqrt {a + b x^{3}}}{15 b} - \frac {32 B a^{3} \sqrt {a + b x^{3}}}{105 b^{4}} + \frac {16 B a^{2} x^{3} \sqrt {a + b x^{3}}}{105 b^{3}} - \frac {4 B a x^{6} \sqrt {a + b x^{3}}}{35 b^{2}} + \frac {2 B x^{9} \sqrt {a + b x^{3}}}{21 b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{9}}{9} + \frac {B x^{12}}{12}}{\sqrt {a}} & \text {otherwise} \end {cases} \]
Piecewise((16*A*a**2*sqrt(a + b*x**3)/(45*b**3) - 8*A*a*x**3*sqrt(a + b*x* *3)/(45*b**2) + 2*A*x**6*sqrt(a + b*x**3)/(15*b) - 32*B*a**3*sqrt(a + b*x* *3)/(105*b**4) + 16*B*a**2*x**3*sqrt(a + b*x**3)/(105*b**3) - 4*B*a*x**6*s qrt(a + b*x**3)/(35*b**2) + 2*B*x**9*sqrt(a + b*x**3)/(21*b), Ne(b, 0)), ( (A*x**9/9 + B*x**12/12)/sqrt(a), True))
Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15 \[ \int \frac {x^8 \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {2}{105} \, B {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}}}{b^{4}} - \frac {21 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a}{b^{4}} + \frac {35 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}{b^{4}} - \frac {35 \, \sqrt {b x^{3} + a} a^{3}}{b^{4}}\right )} + \frac {2}{45} \, A {\left (\frac {3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{b^{3}} - \frac {10 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a}{b^{3}} + \frac {15 \, \sqrt {b x^{3} + a} a^{2}}{b^{3}}\right )} \]
2/105*B*(5*(b*x^3 + a)^(7/2)/b^4 - 21*(b*x^3 + a)^(5/2)*a/b^4 + 35*(b*x^3 + a)^(3/2)*a^2/b^4 - 35*sqrt(b*x^3 + a)*a^3/b^4) + 2/45*A*(3*(b*x^3 + a)^( 5/2)/b^3 - 10*(b*x^3 + a)^(3/2)*a/b^3 + 15*sqrt(b*x^3 + a)*a^2/b^3)
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int \frac {x^8 \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=-\frac {2 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {b x^{3} + a}}{3 \, b^{4}} + \frac {2 \, {\left (15 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} B - 63 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B a + 105 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a^{2} + 21 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A b - 70 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A a b\right )}}{315 \, b^{4}} \]
-2/3*(B*a^3 - A*a^2*b)*sqrt(b*x^3 + a)/b^4 + 2/315*(15*(b*x^3 + a)^(7/2)*B - 63*(b*x^3 + a)^(5/2)*B*a + 105*(b*x^3 + a)^(3/2)*B*a^2 + 21*(b*x^3 + a) ^(5/2)*A*b - 70*(b*x^3 + a)^(3/2)*A*a*b)/b^4
Time = 7.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \frac {x^8 \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {8\,a^2\,\sqrt {b\,x^3+a}\,\left (2\,A-\frac {12\,B\,a}{7\,b}\right )}{45\,b^3}+\frac {x^6\,\sqrt {b\,x^3+a}\,\left (2\,A-\frac {12\,B\,a}{7\,b}\right )}{15\,b}+\frac {2\,B\,x^9\,\sqrt {b\,x^3+a}}{21\,b}-\frac {4\,a\,x^3\,\sqrt {b\,x^3+a}\,\left (2\,A-\frac {12\,B\,a}{7\,b}\right )}{45\,b^2} \]