Integrand size = 22, antiderivative size = 73 \[ \int x^5 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=-\frac {2 a (A b-a B) \left (a+b x^3\right )^{5/2}}{15 b^3}+\frac {2 (A b-2 a B) \left (a+b x^3\right )^{7/2}}{21 b^3}+\frac {2 B \left (a+b x^3\right )^{9/2}}{27 b^3} \]
-2/15*a*(A*b-B*a)*(b*x^3+a)^(5/2)/b^3+2/21*(A*b-2*B*a)*(b*x^3+a)^(7/2)/b^3 +2/27*B*(b*x^3+a)^(9/2)/b^3
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int x^5 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2 \left (a+b x^3\right )^{5/2} \left (-18 a A b+8 a^2 B+45 A b^2 x^3-20 a b B x^3+35 b^2 B x^6\right )}{945 b^3} \]
(2*(a + b*x^3)^(5/2)*(-18*a*A*b + 8*a^2*B + 45*A*b^2*x^3 - 20*a*b*B*x^3 + 35*b^2*B*x^6))/(945*b^3)
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int x^5 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int x^3 \left (b x^3+a\right )^{3/2} \left (B x^3+A\right )dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{7/2}}{b^2}+\frac {(A b-2 a B) \left (b x^3+a\right )^{5/2}}{b^2}+\frac {a (a B-A b) \left (b x^3+a\right )^{3/2}}{b^2}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \left (a+b x^3\right )^{7/2} (A b-2 a B)}{7 b^3}-\frac {2 a \left (a+b x^3\right )^{5/2} (A b-a B)}{5 b^3}+\frac {2 B \left (a+b x^3\right )^{9/2}}{9 b^3}\right )\) |
((-2*a*(A*b - a*B)*(a + b*x^3)^(5/2))/(5*b^3) + (2*(A*b - 2*a*B)*(a + b*x^ 3)^(7/2))/(7*b^3) + (2*B*(a + b*x^3)^(9/2))/(9*b^3))/3
3.2.97.3.1 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 = 4.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {4 \left (b \,x^{3}+a \right )^{\frac {5}{2}} \left (-\frac {5 x^{3} \left (\frac {7 x^{3} B}{9}+A \right ) b^{2}}{2}+a \left (\frac {10 x^{3} B}{9}+A \right ) b -\frac {4 a^{2} B}{9}\right )}{105 b^{3}}\) | \(49\) |
gosper | \(-\frac {2 \left (b \,x^{3}+a \right )^{\frac {5}{2}} \left (-35 b^{2} B \,x^{6}-45 A \,b^{2} x^{3}+20 B a b \,x^{3}+18 a b A -8 a^{2} B \right )}{945 b^{3}}\) | \(53\) |
trager | \(-\frac {2 \left (-35 B \,b^{4} x^{12}-45 A \,b^{4} x^{9}-50 B a \,b^{3} x^{9}-72 A a \,b^{3} x^{6}-3 B \,a^{2} b^{2} x^{6}-9 A \,a^{2} b^{2} x^{3}+4 B \,a^{3} b \,x^{3}+18 A \,a^{3} b -8 B \,a^{4}\right ) \sqrt {b \,x^{3}+a}}{945 b^{3}}\) | \(101\) |
risch | \(-\frac {2 \left (-35 B \,b^{4} x^{12}-45 A \,b^{4} x^{9}-50 B a \,b^{3} x^{9}-72 A a \,b^{3} x^{6}-3 B \,a^{2} b^{2} x^{6}-9 A \,a^{2} b^{2} x^{3}+4 B \,a^{3} b \,x^{3}+18 A \,a^{3} b -8 B \,a^{4}\right ) \sqrt {b \,x^{3}+a}}{945 b^{3}}\) | \(101\) |
default | \(B \left (\frac {2 b \,x^{12} \sqrt {b \,x^{3}+a}}{27}+\frac {20 a \,x^{9} \sqrt {b \,x^{3}+a}}{189}+\frac {2 a^{2} x^{6} \sqrt {b \,x^{3}+a}}{315 b}-\frac {8 a^{3} x^{3} \sqrt {b \,x^{3}+a}}{945 b^{2}}+\frac {16 a^{4} \sqrt {b \,x^{3}+a}}{945 b^{3}}\right )+A \left (\frac {2 b \,x^{9} \sqrt {b \,x^{3}+a}}{21}+\frac {16 a \,x^{6} \sqrt {b \,x^{3}+a}}{105}+\frac {2 a^{2} x^{3} \sqrt {b \,x^{3}+a}}{105 b}-\frac {4 a^{3} \sqrt {b \,x^{3}+a}}{105 b^{2}}\right )\) | \(162\) |
elliptic | \(\frac {2 B b \,x^{12} \sqrt {b \,x^{3}+a}}{27}+\frac {2 \left (b^{2} A +\frac {10}{9} a b B \right ) x^{9} \sqrt {b \,x^{3}+a}}{21 b}+\frac {2 \left (2 a b A +a^{2} B -\frac {6 a \left (b^{2} A +\frac {10}{9} a b B \right )}{7 b}\right ) x^{6} \sqrt {b \,x^{3}+a}}{15 b}+\frac {2 \left (a^{2} A -\frac {4 a \left (2 a b A +a^{2} B -\frac {6 a \left (b^{2} A +\frac {10}{9} a b B \right )}{7 b}\right )}{5 b}\right ) x^{3} \sqrt {b \,x^{3}+a}}{9 b}-\frac {4 a \left (a^{2} A -\frac {4 a \left (2 a b A +a^{2} B -\frac {6 a \left (b^{2} A +\frac {10}{9} a b B \right )}{7 b}\right )}{5 b}\right ) \sqrt {b \,x^{3}+a}}{9 b^{2}}\) | \(203\) |
Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.36 \[ \int x^5 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2 \, {\left (35 \, B b^{4} x^{12} + 5 \, {\left (10 \, B a b^{3} + 9 \, A b^{4}\right )} x^{9} + 3 \, {\left (B a^{2} b^{2} + 24 \, A a b^{3}\right )} x^{6} + 8 \, B a^{4} - 18 \, A a^{3} b - {\left (4 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{945 \, b^{3}} \]
2/945*(35*B*b^4*x^12 + 5*(10*B*a*b^3 + 9*A*b^4)*x^9 + 3*(B*a^2*b^2 + 24*A* a*b^3)*x^6 + 8*B*a^4 - 18*A*a^3*b - (4*B*a^3*b - 9*A*a^2*b^2)*x^3)*sqrt(b* x^3 + a)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (70) = 140\).
Time = 0.42 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.96 \[ \int x^5 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\begin {cases} - \frac {4 A a^{3} \sqrt {a + b x^{3}}}{105 b^{2}} + \frac {2 A a^{2} x^{3} \sqrt {a + b x^{3}}}{105 b} + \frac {16 A a x^{6} \sqrt {a + b x^{3}}}{105} + \frac {2 A b x^{9} \sqrt {a + b x^{3}}}{21} + \frac {16 B a^{4} \sqrt {a + b x^{3}}}{945 b^{3}} - \frac {8 B a^{3} x^{3} \sqrt {a + b x^{3}}}{945 b^{2}} + \frac {2 B a^{2} x^{6} \sqrt {a + b x^{3}}}{315 b} + \frac {20 B a x^{9} \sqrt {a + b x^{3}}}{189} + \frac {2 B b x^{12} \sqrt {a + b x^{3}}}{27} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{6}}{6} + \frac {B x^{9}}{9}\right ) & \text {otherwise} \end {cases} \]
Piecewise((-4*A*a**3*sqrt(a + b*x**3)/(105*b**2) + 2*A*a**2*x**3*sqrt(a + b*x**3)/(105*b) + 16*A*a*x**6*sqrt(a + b*x**3)/105 + 2*A*b*x**9*sqrt(a + b *x**3)/21 + 16*B*a**4*sqrt(a + b*x**3)/(945*b**3) - 8*B*a**3*x**3*sqrt(a + b*x**3)/(945*b**2) + 2*B*a**2*x**6*sqrt(a + b*x**3)/(315*b) + 20*B*a*x**9 *sqrt(a + b*x**3)/189 + 2*B*b*x**12*sqrt(a + b*x**3)/27, Ne(b, 0)), (a**(3 /2)*(A*x**6/6 + B*x**9/9), True))
Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int x^5 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2}{105} \, {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}}}{b^{2}} - \frac {7 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a}{b^{2}}\right )} A + \frac {2}{945} \, {\left (\frac {35 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}}}{b^{3}} - \frac {90 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} a}{b^{3}} + \frac {63 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{2}}{b^{3}}\right )} B \]
2/105*(5*(b*x^3 + a)^(7/2)/b^2 - 7*(b*x^3 + a)^(5/2)*a/b^2)*A + 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)*B
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int x^5 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2 \, {\left (35 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} B - 90 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} B a + 63 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B a^{2} + 45 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} A b - 63 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A a b\right )}}{945 \, b^{3}} \]
2/945*(35*(b*x^3 + a)^(9/2)*B - 90*(b*x^3 + a)^(7/2)*B*a + 63*(b*x^3 + a)^ (5/2)*B*a^2 + 45*(b*x^3 + a)^(7/2)*A*b - 63*(b*x^3 + a)^(5/2)*A*a*b)/b^3
Time = 6.83 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.89 \[ \int x^5 \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {x^6\,\sqrt {b\,x^3+a}\,\left (2\,B\,a^2+4\,A\,a\,b-\frac {6\,a\,\left (2\,A\,b^2+\frac {20\,B\,a\,b}{9}\right )}{7\,b}\right )}{15\,b}-\frac {2\,a\,\left (2\,A\,a^2-\frac {4\,a\,\left (2\,B\,a^2+4\,A\,a\,b-\frac {6\,a\,\left (2\,A\,b^2+\frac {20\,B\,a\,b}{9}\right )}{7\,b}\right )}{5\,b}\right )\,\sqrt {b\,x^3+a}}{9\,b^2}+\frac {2\,B\,b\,x^{12}\,\sqrt {b\,x^3+a}}{27}+\frac {x^3\,\left (2\,A\,a^2-\frac {4\,a\,\left (2\,B\,a^2+4\,A\,a\,b-\frac {6\,a\,\left (2\,A\,b^2+\frac {20\,B\,a\,b}{9}\right )}{7\,b}\right )}{5\,b}\right )\,\sqrt {b\,x^3+a}}{9\,b}+\frac {x^9\,\left (2\,A\,b^2+\frac {20\,B\,a\,b}{9}\right )\,\sqrt {b\,x^3+a}}{21\,b} \]
(x^6*(a + b*x^3)^(1/2)*(2*B*a^2 + 4*A*a*b - (6*a*(2*A*b^2 + (20*B*a*b)/9)) /(7*b)))/(15*b) - (2*a*(2*A*a^2 - (4*a*(2*B*a^2 + 4*A*a*b - (6*a*(2*A*b^2 + (20*B*a*b)/9))/(7*b)))/(5*b))*(a + b*x^3)^(1/2))/(9*b^2) + (2*B*b*x^12*( a + b*x^3)^(1/2))/27 + (x^3*(2*A*a^2 - (4*a*(2*B*a^2 + 4*A*a*b - (6*a*(2*A *b^2 + (20*B*a*b)/9))/(7*b)))/(5*b))*(a + b*x^3)^(1/2))/(9*b) + (x^9*(2*A* b^2 + (20*B*a*b)/9)*(a + b*x^3)^(1/2))/(21*b)