Integrand size = 22, antiderivative size = 100 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {a^2 (A b-a B) \sqrt [3]{a+b x^3}}{b^4}-\frac {a (2 A b-3 a B) \left (a+b x^3\right )^{4/3}}{4 b^4}+\frac {(A b-3 a B) \left (a+b x^3\right )^{7/3}}{7 b^4}+\frac {B \left (a+b x^3\right )^{10/3}}{10 b^4} \] Output:
a^2*(A*b-B*a)*(b*x^3+a)^(1/3)/b^4-1/4*a*(2*A*b-3*B*a)*(b*x^3+a)^(4/3)/b^4+ 1/7*(A*b-3*B*a)*(b*x^3+a)^(7/3)/b^4+1/10*B*(b*x^3+a)^(10/3)/b^4
Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {\sqrt [3]{a+b x^3} \left (90 a^2 A b-81 a^3 B-30 a A b^2 x^3+27 a^2 b B x^3+20 A b^3 x^6-18 a b^2 B x^6+14 b^3 B x^9\right )}{140 b^4} \] Input:
Integrate[(x^8*(A + B*x^3))/(a + b*x^3)^(2/3),x]
Output:
((a + b*x^3)^(1/3)*(90*a^2*A*b - 81*a^3*B - 30*a*A*b^2*x^3 + 27*a^2*b*B*x^ 3 + 20*A*b^3*x^6 - 18*a*b^2*B*x^6 + 14*b^3*B*x^9))/(140*b^4)
Time = 0.40 (sec) , antiderivative size = 105, 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.
\(\displaystyle \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6 \left (B x^3+A\right )}{\left (b x^3+a\right )^{2/3}}dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{7/3}}{b^3}+\frac {(A b-3 a B) \left (b x^3+a\right )^{4/3}}{b^3}+\frac {a (3 a B-2 A b) \sqrt [3]{b x^3+a}}{b^3}-\frac {a^2 (a B-A b)}{b^3 \left (b x^3+a\right )^{2/3}}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {3 a^2 \sqrt [3]{a+b x^3} (A b-a B)}{b^4}+\frac {3 \left (a+b x^3\right )^{7/3} (A b-3 a B)}{7 b^4}-\frac {3 a \left (a+b x^3\right )^{4/3} (2 A b-3 a B)}{4 b^4}+\frac {3 B \left (a+b x^3\right )^{10/3}}{10 b^4}\right )\) |
Input:
Int[(x^8*(A + B*x^3))/(a + b*x^3)^(2/3),x]
Output:
((3*a^2*(A*b - a*B)*(a + b*x^3)^(1/3))/b^4 - (3*a*(2*A*b - 3*a*B)*(a + b*x ^3)^(4/3))/(4*b^4) + (3*(A*b - 3*a*B)*(a + b*x^3)^(7/3))/(7*b^4) + (3*B*(a + b*x^3)^(10/3))/(10*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 = 0.88 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {9 \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (\frac {2 x^{6} \left (\frac {7 B \,x^{3}}{10}+A \right ) b^{3}}{9}-\frac {\left (\frac {3 B \,x^{3}}{5}+A \right ) a \,x^{3} b^{2}}{3}+a^{2} \left (\frac {3 B \,x^{3}}{10}+A \right ) b -\frac {9 a^{3} B}{10}\right )}{14 b^{4}}\) | \(68\) |
gosper | \(\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (14 b^{3} B \,x^{9}+20 A \,b^{3} x^{6}-18 B a \,b^{2} x^{6}-30 a A \,b^{2} x^{3}+27 B \,a^{2} b \,x^{3}+90 a^{2} b A -81 a^{3} B \right )}{140 b^{4}}\) | \(77\) |
trager | \(\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (14 b^{3} B \,x^{9}+20 A \,b^{3} x^{6}-18 B a \,b^{2} x^{6}-30 a A \,b^{2} x^{3}+27 B \,a^{2} b \,x^{3}+90 a^{2} b A -81 a^{3} B \right )}{140 b^{4}}\) | \(77\) |
risch | \(\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (14 b^{3} B \,x^{9}+20 A \,b^{3} x^{6}-18 B a \,b^{2} x^{6}-30 a A \,b^{2} x^{3}+27 B \,a^{2} b \,x^{3}+90 a^{2} b A -81 a^{3} B \right )}{140 b^{4}}\) | \(77\) |
orering | \(\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (14 b^{3} B \,x^{9}+20 A \,b^{3} x^{6}-18 B a \,b^{2} x^{6}-30 a A \,b^{2} x^{3}+27 B \,a^{2} b \,x^{3}+90 a^{2} b A -81 a^{3} B \right )}{140 b^{4}}\) | \(77\) |
Input:
int(x^8*(B*x^3+A)/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)
Output:
9/14*(b*x^3+a)^(1/3)*(2/9*x^6*(7/10*B*x^3+A)*b^3-1/3*(3/5*B*x^3+A)*a*x^3*b ^2+a^2*(3/10*B*x^3+A)*b-9/10*a^3*B)/b^4
Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {{\left (14 \, B b^{3} x^{9} - 2 \, {\left (9 \, B a b^{2} - 10 \, A b^{3}\right )} x^{6} - 81 \, B a^{3} + 90 \, A a^{2} b + 3 \, {\left (9 \, B a^{2} b - 10 \, A a b^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{140 \, b^{4}} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="fricas")
Output:
1/140*(14*B*b^3*x^9 - 2*(9*B*a*b^2 - 10*A*b^3)*x^6 - 81*B*a^3 + 90*A*a^2*b + 3*(9*B*a^2*b - 10*A*a*b^2)*x^3)*(b*x^3 + a)^(1/3)/b^4
Time = 0.53 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.72 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\begin {cases} \frac {9 A a^{2} \sqrt [3]{a + b x^{3}}}{14 b^{3}} - \frac {3 A a x^{3} \sqrt [3]{a + b x^{3}}}{14 b^{2}} + \frac {A x^{6} \sqrt [3]{a + b x^{3}}}{7 b} - \frac {81 B a^{3} \sqrt [3]{a + b x^{3}}}{140 b^{4}} + \frac {27 B a^{2} x^{3} \sqrt [3]{a + b x^{3}}}{140 b^{3}} - \frac {9 B a x^{6} \sqrt [3]{a + b x^{3}}}{70 b^{2}} + \frac {B x^{9} \sqrt [3]{a + b x^{3}}}{10 b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{9}}{9} + \frac {B x^{12}}{12}}{a^{\frac {2}{3}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**8*(B*x**3+A)/(b*x**3+a)**(2/3),x)
Output:
Piecewise((9*A*a**2*(a + b*x**3)**(1/3)/(14*b**3) - 3*A*a*x**3*(a + b*x**3 )**(1/3)/(14*b**2) + A*x**6*(a + b*x**3)**(1/3)/(7*b) - 81*B*a**3*(a + b*x **3)**(1/3)/(140*b**4) + 27*B*a**2*x**3*(a + b*x**3)**(1/3)/(140*b**3) - 9 *B*a*x**6*(a + b*x**3)**(1/3)/(70*b**2) + B*x**9*(a + b*x**3)**(1/3)/(10*b ), Ne(b, 0)), ((A*x**9/9 + B*x**12/12)/a**(2/3), True))
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {1}{140} \, B {\left (\frac {14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}}}{b^{4}} - \frac {60 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a}{b^{4}} + \frac {105 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2}}{b^{4}} - \frac {140 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{3}}{b^{4}}\right )} + \frac {1}{14} \, A {\left (\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}}}{b^{3}} - \frac {7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a}{b^{3}} + \frac {14 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2}}{b^{3}}\right )} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="maxima")
Output:
1/140*B*(14*(b*x^3 + a)^(10/3)/b^4 - 60*(b*x^3 + a)^(7/3)*a/b^4 + 105*(b*x ^3 + a)^(4/3)*a^2/b^4 - 140*(b*x^3 + a)^(1/3)*a^3/b^4) + 1/14*A*(2*(b*x^3 + a)^(7/3)/b^3 - 7*(b*x^3 + a)^(4/3)*a/b^3 + 14*(b*x^3 + a)^(1/3)*a^2/b^3)
Time = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=-\frac {{\left (B a^{3} - A a^{2} b\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{b^{4}} + \frac {14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} B - 60 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} B a + 105 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} B a^{2} + 20 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} A b - 70 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} A a b}{140 \, b^{4}} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="giac")
Output:
-(B*a^3 - A*a^2*b)*(b*x^3 + a)^(1/3)/b^4 + 1/140*(14*(b*x^3 + a)^(10/3)*B - 60*(b*x^3 + a)^(7/3)*B*a + 105*(b*x^3 + a)^(4/3)*B*a^2 + 20*(b*x^3 + a)^ (7/3)*A*b - 70*(b*x^3 + a)^(4/3)*A*a*b)/b^4
Time = 0.87 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=-{\left (b\,x^3+a\right )}^{1/3}\,\left (\frac {81\,B\,a^3-90\,A\,a^2\,b}{140\,b^4}-\frac {B\,x^9}{10\,b}-\frac {x^6\,\left (20\,A\,b^3-18\,B\,a\,b^2\right )}{140\,b^4}+\frac {3\,a\,x^3\,\left (10\,A\,b-9\,B\,a\right )}{140\,b^3}\right ) \] Input:
int((x^8*(A + B*x^3))/(a + b*x^3)^(2/3),x)
Output:
-(a + b*x^3)^(1/3)*((81*B*a^3 - 90*A*a^2*b)/(140*b^4) - (B*x^9)/(10*b) - ( x^6*(20*A*b^3 - 18*B*a*b^2))/(140*b^4) + (3*a*x^3*(10*A*b - 9*B*a))/(140*b ^3))
\[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\left (\int \frac {x^{11}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) b +\left (\int \frac {x^{8}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a \] Input:
int(x^8*(B*x^3+A)/(b*x^3+a)^(2/3),x)
Output:
int(x**11/(a + b*x**3)**(2/3),x)*b + int(x**8/(a + b*x**3)**(2/3),x)*a