Integrand size = 22, antiderivative size = 73 \[ \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=-\frac {a (A b-a B) \left (a+b x^3\right )^{5/3}}{5 b^3}+\frac {(A b-2 a B) \left (a+b x^3\right )^{8/3}}{8 b^3}+\frac {B \left (a+b x^3\right )^{11/3}}{11 b^3} \] Output:
-1/5*a*(A*b-B*a)*(b*x^3+a)^(5/3)/b^3+1/8*(A*b-2*B*a)*(b*x^3+a)^(8/3)/b^3+1 /11*B*(b*x^3+a)^(11/3)/b^3
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {\left (a+b x^3\right )^{5/3} \left (-33 a A b+18 a^2 B+55 A b^2 x^3-30 a b B x^3+40 b^2 B x^6\right )}{440 b^3} \] Input:
Integrate[x^5*(a + b*x^3)^(2/3)*(A + B*x^3),x]
Output:
((a + b*x^3)^(5/3)*(-33*a*A*b + 18*a^2*B + 55*A*b^2*x^3 - 30*a*b*B*x^3 + 4 0*b^2*B*x^6))/(440*b^3)
Time = 0.35 (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.
\(\displaystyle \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int x^3 \left (b x^3+a\right )^{2/3} \left (B x^3+A\right )dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{8/3}}{b^2}+\frac {(A b-2 a B) \left (b x^3+a\right )^{5/3}}{b^2}+\frac {a (a B-A b) \left (b x^3+a\right )^{2/3}}{b^2}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{8/3} (A b-2 a B)}{8 b^3}-\frac {3 a \left (a+b x^3\right )^{5/3} (A b-a B)}{5 b^3}+\frac {3 B \left (a+b x^3\right )^{11/3}}{11 b^3}\right )\) |
Input:
Int[x^5*(a + b*x^3)^(2/3)*(A + B*x^3),x]
Output:
((-3*a*(A*b - a*B)*(a + b*x^3)^(5/3))/(5*b^3) + (3*(A*b - 2*a*B)*(a + b*x^ 3)^(8/3))/(8*b^3) + (3*B*(a + b*x^3)^(11/3))/(11*b^3))/3
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 = 0.76 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {3 \left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-\frac {5 \left (\frac {8 B \,x^{3}}{11}+A \right ) x^{3} b^{2}}{3}+a \left (\frac {10 B \,x^{3}}{11}+A \right ) b -\frac {6 a^{2} B}{11}\right )}{40 b^{3}}\) | \(49\) |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-40 b^{2} B \,x^{6}-55 A \,b^{2} x^{3}+30 B a b \,x^{3}+33 a b A -18 a^{2} B \right )}{440 b^{3}}\) | \(53\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (-40 b^{2} B \,x^{6}-55 A \,b^{2} x^{3}+30 B a b \,x^{3}+33 a b A -18 a^{2} B \right )}{440 b^{3}}\) | \(53\) |
trager | \(-\frac {\left (-40 b^{3} B \,x^{9}-55 A \,b^{3} x^{6}-10 B a \,b^{2} x^{6}-22 a A \,b^{2} x^{3}+12 B \,a^{2} b \,x^{3}+33 a^{2} b A -18 a^{3} B \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{440 b^{3}}\) | \(77\) |
risch | \(-\frac {\left (-40 b^{3} B \,x^{9}-55 A \,b^{3} x^{6}-10 B a \,b^{2} x^{6}-22 a A \,b^{2} x^{3}+12 B \,a^{2} b \,x^{3}+33 a^{2} b A -18 a^{3} B \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{440 b^{3}}\) | \(77\) |
Input:
int(x^5*(b*x^3+a)^(2/3)*(B*x^3+A),x,method=_RETURNVERBOSE)
Output:
-3/40*(b*x^3+a)^(5/3)*(-5/3*(8/11*B*x^3+A)*x^3*b^2+a*(10/11*B*x^3+A)*b-6/1 1*a^2*B)/b^3
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {{\left (40 \, B b^{3} x^{9} + 5 \, {\left (2 \, B a b^{2} + 11 \, A b^{3}\right )} x^{6} + 18 \, B a^{3} - 33 \, A a^{2} b - 2 \, {\left (6 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{440 \, b^{3}} \] Input:
integrate(x^5*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="fricas")
Output:
1/440*(40*B*b^3*x^9 + 5*(2*B*a*b^2 + 11*A*b^3)*x^6 + 18*B*a^3 - 33*A*a^2*b - 2*(6*B*a^2*b - 11*A*a*b^2)*x^3)*(b*x^3 + a)^(2/3)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (63) = 126\).
Time = 0.56 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.22 \[ \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\begin {cases} - \frac {3 A a^{2} \left (a + b x^{3}\right )^{\frac {2}{3}}}{40 b^{2}} + \frac {A a x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{20 b} + \frac {A x^{6} \left (a + b x^{3}\right )^{\frac {2}{3}}}{8} + \frac {9 B a^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{220 b^{3}} - \frac {3 B a^{2} x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{110 b^{2}} + \frac {B a x^{6} \left (a + b x^{3}\right )^{\frac {2}{3}}}{44 b} + \frac {B x^{9} \left (a + b x^{3}\right )^{\frac {2}{3}}}{11} & \text {for}\: b \neq 0 \\a^{\frac {2}{3}} \left (\frac {A x^{6}}{6} + \frac {B x^{9}}{9}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(b*x**3+a)**(2/3)*(B*x**3+A),x)
Output:
Piecewise((-3*A*a**2*(a + b*x**3)**(2/3)/(40*b**2) + A*a*x**3*(a + b*x**3) **(2/3)/(20*b) + A*x**6*(a + b*x**3)**(2/3)/8 + 9*B*a**3*(a + b*x**3)**(2/ 3)/(220*b**3) - 3*B*a**2*x**3*(a + b*x**3)**(2/3)/(110*b**2) + B*a*x**6*(a + b*x**3)**(2/3)/(44*b) + B*x**9*(a + b*x**3)**(2/3)/11, Ne(b, 0)), (a**( 2/3)*(A*x**6/6 + B*x**9/9), True))
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {1}{220} \, B {\left (\frac {20 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}}}{b^{3}} - \frac {55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a}{b^{3}} + \frac {44 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2}}{b^{3}}\right )} + \frac {1}{40} \, A {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}}}{b^{2}} - \frac {8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a}{b^{2}}\right )} \] Input:
integrate(x^5*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="maxima")
Output:
1/220*B*(20*(b*x^3 + a)^(11/3)/b^3 - 55*(b*x^3 + a)^(8/3)*a/b^3 + 44*(b*x^ 3 + a)^(5/3)*a^2/b^3) + 1/40*A*(5*(b*x^3 + a)^(8/3)/b^2 - 8*(b*x^3 + a)^(5 /3)*a/b^2)
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} B - 110 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} B a + 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} B a^{2} + 55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} A b - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} A a b}{440 \, b^{3}} \] Input:
integrate(x^5*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="giac")
Output:
1/440*(40*(b*x^3 + a)^(11/3)*B - 110*(b*x^3 + a)^(8/3)*B*a + 88*(b*x^3 + a )^(5/3)*B*a^2 + 55*(b*x^3 + a)^(8/3)*A*b - 88*(b*x^3 + a)^(5/3)*A*a*b)/b^3
Time = 0.87 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx={\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {B\,x^9}{11}+\frac {18\,B\,a^3-33\,A\,a^2\,b}{440\,b^3}+\frac {x^6\,\left (55\,A\,b^3+10\,B\,a\,b^2\right )}{440\,b^3}+\frac {a\,x^3\,\left (11\,A\,b-6\,B\,a\right )}{220\,b^2}\right ) \] Input:
int(x^5*(A + B*x^3)*(a + b*x^3)^(2/3),x)
Output:
(a + b*x^3)^(2/3)*((B*x^9)/11 + (18*B*a^3 - 33*A*a^2*b)/(440*b^3) + (x^6*( 55*A*b^3 + 10*B*a*b^2))/(440*b^3) + (a*x^3*(11*A*b - 6*B*a))/(220*b^2))
Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63 \[ \int x^5 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (8 b^{3} x^{9}+13 a \,b^{2} x^{6}+2 a^{2} b \,x^{3}-3 a^{3}\right )}{88 b^{2}} \] Input:
int(x^5*(b*x^3+a)^(2/3)*(B*x^3+A),x)
Output:
((a + b*x**3)**(2/3)*( - 3*a**3 + 2*a**2*b*x**3 + 13*a*b**2*x**6 + 8*b**3* x**9))/(88*b**2)