Integrand size = 22, antiderivative size = 103 \[ \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {a^2 (A b-a B) \left (a+b x^3\right )^{4/3}}{4 b^4}-\frac {a (2 A b-3 a B) \left (a+b x^3\right )^{7/3}}{7 b^4}+\frac {(A b-3 a B) \left (a+b x^3\right )^{10/3}}{10 b^4}+\frac {B \left (a+b x^3\right )^{13/3}}{13 b^4} \] Output:
1/4*a^2*(A*b-B*a)*(b*x^3+a)^(4/3)/b^4-1/7*a*(2*A*b-3*B*a)*(b*x^3+a)^(7/3)/ b^4+1/10*(A*b-3*B*a)*(b*x^3+a)^(10/3)/b^4+1/13*B*(b*x^3+a)^(13/3)/b^4
Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {\left (a+b x^3\right )^{4/3} \left (117 a^2 A b-81 a^3 B-156 a A b^2 x^3+108 a^2 b B x^3+182 A b^3 x^6-126 a b^2 B x^6+140 b^3 B x^9\right )}{1820 b^4} \] Input:
Integrate[x^8*(a + b*x^3)^(1/3)*(A + B*x^3),x]
Output:
((a + b*x^3)^(4/3)*(117*a^2*A*b - 81*a^3*B - 156*a*A*b^2*x^3 + 108*a^2*b*B *x^3 + 182*A*b^3*x^6 - 126*a*b^2*B*x^6 + 140*b^3*B*x^9))/(1820*b^4)
Time = 0.42 (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.
\(\displaystyle \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int x^6 \sqrt [3]{b x^3+a} \left (B x^3+A\right )dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{10/3}}{b^3}+\frac {(A b-3 a B) \left (b x^3+a\right )^{7/3}}{b^3}+\frac {a (3 a B-2 A b) \left (b x^3+a\right )^{4/3}}{b^3}-\frac {a^2 (a B-A b) \sqrt [3]{b x^3+a}}{b^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {3 a^2 \left (a+b x^3\right )^{4/3} (A b-a B)}{4 b^4}+\frac {3 \left (a+b x^3\right )^{10/3} (A b-3 a B)}{10 b^4}-\frac {3 a \left (a+b x^3\right )^{7/3} (2 A b-3 a B)}{7 b^4}+\frac {3 B \left (a+b x^3\right )^{13/3}}{13 b^4}\right )\) |
Input:
Int[x^8*(a + b*x^3)^(1/3)*(A + B*x^3),x]
Output:
((3*a^2*(A*b - a*B)*(a + b*x^3)^(4/3))/(4*b^4) - (3*a*(2*A*b - 3*a*B)*(a + b*x^3)^(7/3))/(7*b^4) + (3*(A*b - 3*a*B)*(a + b*x^3)^(10/3))/(10*b^4) + ( 3*B*(a + b*x^3)^(13/3))/(13*b^4))/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 = 68, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {9 \left (\frac {14 \left (\frac {10 B \,x^{3}}{13}+A \right ) x^{6} b^{3}}{9}-\frac {4 a \left (\frac {21 B \,x^{3}}{26}+A \right ) x^{3} b^{2}}{3}+a^{2} \left (\frac {12 B \,x^{3}}{13}+A \right ) b -\frac {9 a^{3} B}{13}\right ) \left (b \,x^{3}+a \right )^{\frac {4}{3}}}{140 b^{4}}\) | \(68\) |
gosper | \(\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (140 b^{3} B \,x^{9}+182 A \,b^{3} x^{6}-126 B a \,b^{2} x^{6}-156 a A \,b^{2} x^{3}+108 B \,a^{2} b \,x^{3}+117 a^{2} b A -81 a^{3} B \right )}{1820 b^{4}}\) | \(77\) |
orering | \(\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (140 b^{3} B \,x^{9}+182 A \,b^{3} x^{6}-126 B a \,b^{2} x^{6}-156 a A \,b^{2} x^{3}+108 B \,a^{2} b \,x^{3}+117 a^{2} b A -81 a^{3} B \right )}{1820 b^{4}}\) | \(77\) |
trager | \(\frac {\left (140 B \,b^{4} x^{12}+182 A \,b^{4} x^{9}+14 B \,x^{9} a \,b^{3}+26 A \,x^{6} a \,b^{3}-18 B \,x^{6} a^{2} b^{2}-39 A \,a^{2} b^{2} x^{3}+27 B \,a^{3} b \,x^{3}+117 A \,a^{3} b -81 B \,a^{4}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{1820 b^{4}}\) | \(101\) |
risch | \(\frac {\left (140 B \,b^{4} x^{12}+182 A \,b^{4} x^{9}+14 B \,x^{9} a \,b^{3}+26 A \,x^{6} a \,b^{3}-18 B \,x^{6} a^{2} b^{2}-39 A \,a^{2} b^{2} x^{3}+27 B \,a^{3} b \,x^{3}+117 A \,a^{3} b -81 B \,a^{4}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{1820 b^{4}}\) | \(101\) |
Input:
int(x^8*(b*x^3+a)^(1/3)*(B*x^3+A),x,method=_RETURNVERBOSE)
Output:
9/140*(14/9*(10/13*B*x^3+A)*x^6*b^3-4/3*a*(21/26*B*x^3+A)*x^3*b^2+a^2*(12/ 13*B*x^3+A)*b-9/13*a^3*B)*(b*x^3+a)^(4/3)/b^4
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {{\left (140 \, B b^{4} x^{12} + 14 \, {\left (B a b^{3} + 13 \, A b^{4}\right )} x^{9} - 2 \, {\left (9 \, B a^{2} b^{2} - 13 \, A a b^{3}\right )} x^{6} - 81 \, B a^{4} + 117 \, A a^{3} b + 3 \, {\left (9 \, B a^{3} b - 13 \, A a^{2} b^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{1820 \, b^{4}} \] Input:
integrate(x^8*(b*x^3+a)^(1/3)*(B*x^3+A),x, algorithm="fricas")
Output:
1/1820*(140*B*b^4*x^12 + 14*(B*a*b^3 + 13*A*b^4)*x^9 - 2*(9*B*a^2*b^2 - 13 *A*a*b^3)*x^6 - 81*B*a^4 + 117*A*a^3*b + 3*(9*B*a^3*b - 13*A*a^2*b^2)*x^3) *(b*x^3 + a)^(1/3)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (94) = 188\).
Time = 0.51 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.06 \[ \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\begin {cases} \frac {9 A a^{3} \sqrt [3]{a + b x^{3}}}{140 b^{3}} - \frac {3 A a^{2} x^{3} \sqrt [3]{a + b x^{3}}}{140 b^{2}} + \frac {A a x^{6} \sqrt [3]{a + b x^{3}}}{70 b} + \frac {A x^{9} \sqrt [3]{a + b x^{3}}}{10} - \frac {81 B a^{4} \sqrt [3]{a + b x^{3}}}{1820 b^{4}} + \frac {27 B a^{3} x^{3} \sqrt [3]{a + b x^{3}}}{1820 b^{3}} - \frac {9 B a^{2} x^{6} \sqrt [3]{a + b x^{3}}}{910 b^{2}} + \frac {B a x^{9} \sqrt [3]{a + b x^{3}}}{130 b} + \frac {B x^{12} \sqrt [3]{a + b x^{3}}}{13} & \text {for}\: b \neq 0 \\\sqrt [3]{a} \left (\frac {A x^{9}}{9} + \frac {B x^{12}}{12}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**8*(b*x**3+a)**(1/3)*(B*x**3+A),x)
Output:
Piecewise((9*A*a**3*(a + b*x**3)**(1/3)/(140*b**3) - 3*A*a**2*x**3*(a + b* x**3)**(1/3)/(140*b**2) + A*a*x**6*(a + b*x**3)**(1/3)/(70*b) + A*x**9*(a + b*x**3)**(1/3)/10 - 81*B*a**4*(a + b*x**3)**(1/3)/(1820*b**4) + 27*B*a** 3*x**3*(a + b*x**3)**(1/3)/(1820*b**3) - 9*B*a**2*x**6*(a + b*x**3)**(1/3) /(910*b**2) + B*a*x**9*(a + b*x**3)**(1/3)/(130*b) + B*x**12*(a + b*x**3)* *(1/3)/13, Ne(b, 0)), (a**(1/3)*(A*x**9/9 + B*x**12/12), True))
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15 \[ \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {1}{1820} \, B {\left (\frac {140 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}}}{b^{4}} - \frac {546 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a}{b^{4}} + \frac {780 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{2}}{b^{4}} - \frac {455 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{3}}{b^{4}}\right )} + \frac {1}{140} \, A {\left (\frac {14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}}}{b^{3}} - \frac {40 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a}{b^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2}}{b^{3}}\right )} \] Input:
integrate(x^8*(b*x^3+a)^(1/3)*(B*x^3+A),x, algorithm="maxima")
Output:
1/1820*B*(140*(b*x^3 + a)^(13/3)/b^4 - 546*(b*x^3 + a)^(10/3)*a/b^4 + 780* (b*x^3 + a)^(7/3)*a^2/b^4 - 455*(b*x^3 + a)^(4/3)*a^3/b^4) + 1/140*A*(14*( b*x^3 + a)^(10/3)/b^3 - 40*(b*x^3 + a)^(7/3)*a/b^3 + 35*(b*x^3 + a)^(4/3)* a^2/b^3)
Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {140 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} B - 546 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} B a + 780 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} B a^{2} - 455 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} B a^{3} + 182 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} A b - 520 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} A a b + 455 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} A a^{2} b}{1820 \, b^{4}} \] Input:
integrate(x^8*(b*x^3+a)^(1/3)*(B*x^3+A),x, algorithm="giac")
Output:
1/1820*(140*(b*x^3 + a)^(13/3)*B - 546*(b*x^3 + a)^(10/3)*B*a + 780*(b*x^3 + a)^(7/3)*B*a^2 - 455*(b*x^3 + a)^(4/3)*B*a^3 + 182*(b*x^3 + a)^(10/3)*A *b - 520*(b*x^3 + a)^(7/3)*A*a*b + 455*(b*x^3 + a)^(4/3)*A*a^2*b)/b^4
Time = 0.85 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.93 \[ \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx={\left (b\,x^3+a\right )}^{1/3}\,\left (\frac {B\,x^{12}}{13}-\frac {81\,B\,a^4-117\,A\,a^3\,b}{1820\,b^4}+\frac {x^9\,\left (182\,A\,b^4+14\,B\,a\,b^3\right )}{1820\,b^4}-\frac {3\,a^2\,x^3\,\left (13\,A\,b-9\,B\,a\right )}{1820\,b^3}+\frac {a\,x^6\,\left (13\,A\,b-9\,B\,a\right )}{910\,b^2}\right ) \] Input:
int(x^8*(A + B*x^3)*(a + b*x^3)^(1/3),x)
Output:
(a + b*x^3)^(1/3)*((B*x^12)/13 - (81*B*a^4 - 117*A*a^3*b)/(1820*b^4) + (x^ 9*(182*A*b^4 + 14*B*a*b^3))/(1820*b^4) - (3*a^2*x^3*(13*A*b - 9*B*a))/(182 0*b^3) + (a*x^6*(13*A*b - 9*B*a))/(910*b^2))
Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.55 \[ \int x^8 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (35 b^{4} x^{12}+49 a \,b^{3} x^{9}+2 a^{2} b^{2} x^{6}-3 a^{3} b \,x^{3}+9 a^{4}\right )}{455 b^{3}} \] Input:
int(x^8*(b*x^3+a)^(1/3)*(B*x^3+A),x)
Output:
((a + b*x**3)**(1/3)*(9*a**4 - 3*a**3*b*x**3 + 2*a**2*b**2*x**6 + 49*a*b** 3*x**9 + 35*b**4*x**12))/(455*b**3)