Integrand size = 22, antiderivative size = 103 \[ \int x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2 a^2 (A b-a B) \left (a+b x^3\right )^{5/2}}{15 b^4}-\frac {2 a (2 A b-3 a B) \left (a+b x^3\right )^{7/2}}{21 b^4}+\frac {2 (A b-3 a B) \left (a+b x^3\right )^{9/2}}{27 b^4}+\frac {2 B \left (a+b x^3\right )^{11/2}}{33 b^4} \]
2/15*a^2*(A*b-B*a)*(b*x^3+a)^(5/2)/b^4-2/21*a*(2*A*b-3*B*a)*(b*x^3+a)^(7/2 )/b^4+2/27*(A*b-3*B*a)*(b*x^3+a)^(9/2)/b^4+2/33*B*(b*x^3+a)^(11/2)/b^4
Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2 \left (a+b x^3\right )^{5/2} \left (88 a^2 A b-48 a^3 B-220 a A b^2 x^3+120 a^2 b B x^3+385 A b^3 x^6-210 a b^2 B x^6+315 b^3 B x^9\right )}{10395 b^4} \]
(2*(a + b*x^3)^(5/2)*(88*a^2*A*b - 48*a^3*B - 220*a*A*b^2*x^3 + 120*a^2*b* B*x^3 + 385*A*b^3*x^6 - 210*a*b^2*B*x^6 + 315*b^3*B*x^9))/(10395*b^4)
Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04, 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.
\(\displaystyle \int x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int x^6 \left (b x^3+a\right )^{3/2} \left (B x^3+A\right )dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{9/2}}{b^3}+\frac {(A b-3 a B) \left (b x^3+a\right )^{7/2}}{b^3}+\frac {a (3 a B-2 A b) \left (b x^3+a\right )^{5/2}}{b^3}-\frac {a^2 (a B-A b) \left (b x^3+a\right )^{3/2}}{b^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {2 a^2 \left (a+b x^3\right )^{5/2} (A b-a B)}{5 b^4}+\frac {2 \left (a+b x^3\right )^{9/2} (A b-3 a B)}{9 b^4}-\frac {2 a \left (a+b x^3\right )^{7/2} (2 A b-3 a B)}{7 b^4}+\frac {2 B \left (a+b x^3\right )^{11/2}}{11 b^4}\right )\) |
((2*a^2*(A*b - a*B)*(a + b*x^3)^(5/2))/(5*b^4) - (2*a*(2*A*b - 3*a*B)*(a + b*x^3)^(7/2))/(7*b^4) + (2*(A*b - 3*a*B)*(a + b*x^3)^(9/2))/(9*b^4) + (2* B*(a + b*x^3)^(11/2))/(11*b^4))/3
3.2.96.3.1 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 = 4.31 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {16 \left (b \,x^{3}+a \right )^{\frac {5}{2}} \left (\frac {35 x^{6} \left (\frac {9 x^{3} B}{11}+A \right ) b^{3}}{8}-\frac {5 x^{3} \left (\frac {21 x^{3} B}{22}+A \right ) a \,b^{2}}{2}+a^{2} \left (\frac {15 x^{3} B}{11}+A \right ) b -\frac {6 a^{3} B}{11}\right )}{945 b^{4}}\) | \(68\) |
gosper | \(\frac {2 \left (b \,x^{3}+a \right )^{\frac {5}{2}} \left (315 b^{3} B \,x^{9}+385 x^{6} b^{3} A -210 B \,x^{6} a \,b^{2}-220 a A \,b^{2} x^{3}+120 B \,a^{2} b \,x^{3}+88 a^{2} b A -48 a^{3} B \right )}{10395 b^{4}}\) | \(77\) |
trager | \(\frac {2 \left (315 b^{5} B \,x^{15}+385 b^{5} A \,x^{12}+420 a \,b^{4} B \,x^{12}+550 a \,b^{4} A \,x^{9}+15 a^{2} b^{3} B \,x^{9}+33 a^{2} A \,b^{3} x^{6}-18 a^{3} b^{2} B \,x^{6}-44 a^{3} A \,b^{2} x^{3}+24 B \,a^{4} b \,x^{3}+88 a^{4} b A -48 a^{5} B \right ) \sqrt {b \,x^{3}+a}}{10395 b^{4}}\) | \(125\) |
risch | \(\frac {2 \left (315 b^{5} B \,x^{15}+385 b^{5} A \,x^{12}+420 a \,b^{4} B \,x^{12}+550 a \,b^{4} A \,x^{9}+15 a^{2} b^{3} B \,x^{9}+33 a^{2} A \,b^{3} x^{6}-18 a^{3} b^{2} B \,x^{6}-44 a^{3} A \,b^{2} x^{3}+24 B \,a^{4} b \,x^{3}+88 a^{4} b A -48 a^{5} B \right ) \sqrt {b \,x^{3}+a}}{10395 b^{4}}\) | \(125\) |
default | \(A \left (\frac {2 b \,x^{12} \sqrt {b \,x^{3}+a}}{27}+\frac {20 a \,x^{9} \sqrt {b \,x^{3}+a}}{189}+\frac {2 a^{2} x^{6} \sqrt {b \,x^{3}+a}}{315 b}-\frac {8 a^{3} x^{3} \sqrt {b \,x^{3}+a}}{945 b^{2}}+\frac {16 a^{4} \sqrt {b \,x^{3}+a}}{945 b^{3}}\right )+B \left (\frac {2 b \,x^{15} \sqrt {b \,x^{3}+a}}{33}+\frac {8 a \,x^{12} \sqrt {b \,x^{3}+a}}{99}+\frac {2 a^{2} x^{9} \sqrt {b \,x^{3}+a}}{693 b}-\frac {4 a^{3} x^{6} \sqrt {b \,x^{3}+a}}{1155 b^{2}}+\frac {16 a^{4} x^{3} \sqrt {b \,x^{3}+a}}{3465 b^{3}}-\frac {32 a^{5} \sqrt {b \,x^{3}+a}}{3465 b^{4}}\right )\) | \(202\) |
elliptic | \(\frac {2 B b \,x^{15} \sqrt {b \,x^{3}+a}}{33}+\frac {2 \left (b^{2} A +\frac {12}{11} a b B \right ) x^{12} \sqrt {b \,x^{3}+a}}{27 b}+\frac {2 \left (2 a b A +a^{2} B -\frac {8 a \left (b^{2} A +\frac {12}{11} a b B \right )}{9 b}\right ) x^{9} \sqrt {b \,x^{3}+a}}{21 b}+\frac {2 \left (a^{2} A -\frac {6 a \left (2 a b A +a^{2} B -\frac {8 a \left (b^{2} A +\frac {12}{11} a b B \right )}{9 b}\right )}{7 b}\right ) x^{6} \sqrt {b \,x^{3}+a}}{15 b}-\frac {8 a \left (a^{2} A -\frac {6 a \left (2 a b A +a^{2} B -\frac {8 a \left (b^{2} A +\frac {12}{11} a b B \right )}{9 b}\right )}{7 b}\right ) x^{3} \sqrt {b \,x^{3}+a}}{45 b^{2}}+\frac {16 a^{2} \left (a^{2} A -\frac {6 a \left (2 a b A +a^{2} B -\frac {8 a \left (b^{2} A +\frac {12}{11} a b B \right )}{9 b}\right )}{7 b}\right ) \sqrt {b \,x^{3}+a}}{45 b^{3}}\) | \(263\) |
16/945*(b*x^3+a)^(5/2)*(35/8*x^6*(9/11*x^3*B+A)*b^3-5/2*x^3*(21/22*x^3*B+A )*a*b^2+a^2*(15/11*x^3*B+A)*b-6/11*a^3*B)/b^4
Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.20 \[ \int x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2 \, {\left (315 \, B b^{5} x^{15} + 35 \, {\left (12 \, B a b^{4} + 11 \, A b^{5}\right )} x^{12} + 5 \, {\left (3 \, B a^{2} b^{3} + 110 \, A a b^{4}\right )} x^{9} - 3 \, {\left (6 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{6} - 48 \, B a^{5} + 88 \, A a^{4} b + 4 \, {\left (6 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{10395 \, b^{4}} \]
2/10395*(315*B*b^5*x^15 + 35*(12*B*a*b^4 + 11*A*b^5)*x^12 + 5*(3*B*a^2*b^3 + 110*A*a*b^4)*x^9 - 3*(6*B*a^3*b^2 - 11*A*a^2*b^3)*x^6 - 48*B*a^5 + 88*A *a^4*b + 4*(6*B*a^4*b - 11*A*a^3*b^2)*x^3)*sqrt(b*x^3 + a)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (100) = 200\).
Time = 0.58 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.59 \[ \int x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\begin {cases} \frac {16 A a^{4} \sqrt {a + b x^{3}}}{945 b^{3}} - \frac {8 A a^{3} x^{3} \sqrt {a + b x^{3}}}{945 b^{2}} + \frac {2 A a^{2} x^{6} \sqrt {a + b x^{3}}}{315 b} + \frac {20 A a x^{9} \sqrt {a + b x^{3}}}{189} + \frac {2 A b x^{12} \sqrt {a + b x^{3}}}{27} - \frac {32 B a^{5} \sqrt {a + b x^{3}}}{3465 b^{4}} + \frac {16 B a^{4} x^{3} \sqrt {a + b x^{3}}}{3465 b^{3}} - \frac {4 B a^{3} x^{6} \sqrt {a + b x^{3}}}{1155 b^{2}} + \frac {2 B a^{2} x^{9} \sqrt {a + b x^{3}}}{693 b} + \frac {8 B a x^{12} \sqrt {a + b x^{3}}}{99} + \frac {2 B b x^{15} \sqrt {a + b x^{3}}}{33} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{9}}{9} + \frac {B x^{12}}{12}\right ) & \text {otherwise} \end {cases} \]
Piecewise((16*A*a**4*sqrt(a + b*x**3)/(945*b**3) - 8*A*a**3*x**3*sqrt(a + b*x**3)/(945*b**2) + 2*A*a**2*x**6*sqrt(a + b*x**3)/(315*b) + 20*A*a*x**9* sqrt(a + b*x**3)/189 + 2*A*b*x**12*sqrt(a + b*x**3)/27 - 32*B*a**5*sqrt(a + b*x**3)/(3465*b**4) + 16*B*a**4*x**3*sqrt(a + b*x**3)/(3465*b**3) - 4*B* a**3*x**6*sqrt(a + b*x**3)/(1155*b**2) + 2*B*a**2*x**9*sqrt(a + b*x**3)/(6 93*b) + 8*B*a*x**12*sqrt(a + b*x**3)/99 + 2*B*b*x**15*sqrt(a + b*x**3)/33, Ne(b, 0)), (a**(3/2)*(A*x**9/9 + B*x**12/12), True))
Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15 \[ \int x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2}{945} \, {\left (\frac {35 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}}}{b^{3}} - \frac {90 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} a}{b^{3}} + \frac {63 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{2}}{b^{3}}\right )} A + \frac {2}{3465} \, {\left (\frac {105 \, {\left (b x^{3} + a\right )}^{\frac {11}{2}}}{b^{4}} - \frac {385 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} a}{b^{4}} + \frac {495 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} a^{2}}{b^{4}} - \frac {231 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{3}}{b^{4}}\right )} B \]
2/945*(35*(b*x^3 + a)^(9/2)/b^3 - 90*(b*x^3 + a)^(7/2)*a/b^3 + 63*(b*x^3 + a)^(5/2)*a^2/b^3)*A + 2/3465*(105*(b*x^3 + a)^(11/2)/b^4 - 385*(b*x^3 + a )^(9/2)*a/b^4 + 495*(b*x^3 + a)^(7/2)*a^2/b^4 - 231*(b*x^3 + a)^(5/2)*a^3/ b^4)*B
Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2 \, {\left (315 \, {\left (b x^{3} + a\right )}^{\frac {11}{2}} B - 1155 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} B a + 1485 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} B a^{2} - 693 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B a^{3} + 385 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} A b - 990 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} A a b + 693 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A a^{2} b\right )}}{10395 \, b^{4}} \]
2/10395*(315*(b*x^3 + a)^(11/2)*B - 1155*(b*x^3 + a)^(9/2)*B*a + 1485*(b*x ^3 + a)^(7/2)*B*a^2 - 693*(b*x^3 + a)^(5/2)*B*a^3 + 385*(b*x^3 + a)^(9/2)* A*b - 990*(b*x^3 + a)^(7/2)*A*a*b + 693*(b*x^3 + a)^(5/2)*A*a^2*b)/b^4
Time = 6.99 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.00 \[ \int x^8 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {20\,A\,a\,x^9\,\sqrt {b\,x^3+a}}{189}+\frac {2\,A\,b\,x^{12}\,\sqrt {b\,x^3+a}}{27}+\frac {8\,B\,a\,x^{12}\,\sqrt {b\,x^3+a}}{99}+\frac {2\,B\,b\,x^{15}\,\sqrt {b\,x^3+a}}{33}+\frac {16\,A\,a^4\,\sqrt {b\,x^3+a}}{945\,b^3}-\frac {32\,B\,a^5\,\sqrt {b\,x^3+a}}{3465\,b^4}-\frac {8\,A\,a^3\,x^3\,\sqrt {b\,x^3+a}}{945\,b^2}+\frac {2\,A\,a^2\,x^6\,\sqrt {b\,x^3+a}}{315\,b}+\frac {16\,B\,a^4\,x^3\,\sqrt {b\,x^3+a}}{3465\,b^3}-\frac {4\,B\,a^3\,x^6\,\sqrt {b\,x^3+a}}{1155\,b^2}+\frac {2\,B\,a^2\,x^9\,\sqrt {b\,x^3+a}}{693\,b} \]
(20*A*a*x^9*(a + b*x^3)^(1/2))/189 + (2*A*b*x^12*(a + b*x^3)^(1/2))/27 + ( 8*B*a*x^12*(a + b*x^3)^(1/2))/99 + (2*B*b*x^15*(a + b*x^3)^(1/2))/33 + (16 *A*a^4*(a + b*x^3)^(1/2))/(945*b^3) - (32*B*a^5*(a + b*x^3)^(1/2))/(3465*b ^4) - (8*A*a^3*x^3*(a + b*x^3)^(1/2))/(945*b^2) + (2*A*a^2*x^6*(a + b*x^3) ^(1/2))/(315*b) + (16*B*a^4*x^3*(a + b*x^3)^(1/2))/(3465*b^3) - (4*B*a^3*x ^6*(a + b*x^3)^(1/2))/(1155*b^2) + (2*B*a^2*x^9*(a + b*x^3)^(1/2))/(693*b)