Integrand size = 22, antiderivative size = 101 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=-\frac {a^2 (A b-a B)}{b^4 \sqrt [3]{a+b x^3}}-\frac {a (2 A b-3 a B) \left (a+b x^3\right )^{2/3}}{2 b^4}+\frac {(A b-3 a B) \left (a+b x^3\right )^{5/3}}{5 b^4}+\frac {B \left (a+b x^3\right )^{8/3}}{8 b^4} \] Output:
-a^2*(A*b-B*a)/b^4/(b*x^3+a)^(1/3)-1/2*a*(2*A*b-3*B*a)*(b*x^3+a)^(2/3)/b^4 +1/5*(A*b-3*B*a)*(b*x^3+a)^(5/3)/b^4+1/8*B*(b*x^3+a)^(8/3)/b^4
Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {81 a^3 B-9 a^2 b \left (8 A-3 B x^3\right )-3 a b^2 x^3 \left (8 A+3 B x^3\right )+b^3 x^6 \left (8 A+5 B x^3\right )}{40 b^4 \sqrt [3]{a+b x^3}} \] Input:
Integrate[(x^8*(A + B*x^3))/(a + b*x^3)^(4/3),x]
Output:
(81*a^3*B - 9*a^2*b*(8*A - 3*B*x^3) - 3*a*b^2*x^3*(8*A + 3*B*x^3) + b^3*x^ 6*(8*A + 5*B*x^3))/(40*b^4*(a + b*x^3)^(1/3))
Time = 0.38 (sec) , antiderivative size = 105, 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.
\(\displaystyle \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6 \left (B x^3+A\right )}{\left (b x^3+a\right )^{4/3}}dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (-\frac {(a B-A b) a^2}{b^3 \left (b x^3+a\right )^{4/3}}+\frac {(3 a B-2 A b) a}{b^3 \sqrt [3]{b x^3+a}}+\frac {B \left (b x^3+a\right )^{5/3}}{b^3}+\frac {(A b-3 a B) \left (b x^3+a\right )^{2/3}}{b^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 a^2 (A b-a B)}{b^4 \sqrt [3]{a+b x^3}}+\frac {3 \left (a+b x^3\right )^{5/3} (A b-3 a B)}{5 b^4}-\frac {3 a \left (a+b x^3\right )^{2/3} (2 A b-3 a B)}{2 b^4}+\frac {3 B \left (a+b x^3\right )^{8/3}}{8 b^4}\right )\) |
Input:
Int[(x^8*(A + B*x^3))/(a + b*x^3)^(4/3),x]
Output:
((-3*a^2*(A*b - a*B))/(b^4*(a + b*x^3)^(1/3)) - (3*a*(2*A*b - 3*a*B)*(a + b*x^3)^(2/3))/(2*b^4) + (3*(A*b - 3*a*B)*(a + b*x^3)^(5/3))/(5*b^4) + (3*B *(a + b*x^3)^(8/3))/(8*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 = 0.79 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {9 \left (-\frac {\left (\frac {5 B \,x^{3}}{8}+A \right ) x^{6} b^{3}}{9}+\frac {\left (\frac {3 B \,x^{3}}{8}+A \right ) a \,x^{3} b^{2}}{3}+a^{2} \left (-\frac {3 B \,x^{3}}{8}+A \right ) b -\frac {9 a^{3} B}{8}\right )}{5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{4}}\) | \(68\) |
gosper | \(-\frac {-5 b^{3} B \,x^{9}-8 A \,b^{3} x^{6}+9 B a \,b^{2} x^{6}+24 a A \,b^{2} x^{3}-27 B \,a^{2} b \,x^{3}+72 a^{2} b A -81 a^{3} B}{40 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{4}}\) | \(77\) |
trager | \(-\frac {-5 b^{3} B \,x^{9}-8 A \,b^{3} x^{6}+9 B a \,b^{2} x^{6}+24 a A \,b^{2} x^{3}-27 B \,a^{2} b \,x^{3}+72 a^{2} b A -81 a^{3} B}{40 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{4}}\) | \(77\) |
orering | \(-\frac {-5 b^{3} B \,x^{9}-8 A \,b^{3} x^{6}+9 B a \,b^{2} x^{6}+24 a A \,b^{2} x^{3}-27 B \,a^{2} b \,x^{3}+72 a^{2} b A -81 a^{3} B}{40 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{4}}\) | \(77\) |
risch | \(-\frac {\left (-5 b^{2} B \,x^{6}-8 A \,b^{2} x^{3}+14 B a b \,x^{3}+32 a b A -41 a^{2} B \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{40 b^{4}}-\frac {a^{2} \left (A b -B a \right )}{b^{4} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}\) | \(79\) |
Input:
int(x^8*(B*x^3+A)/(b*x^3+a)^(4/3),x,method=_RETURNVERBOSE)
Output:
-9/5*(-1/9*(5/8*B*x^3+A)*x^6*b^3+1/3*(3/8*B*x^3+A)*a*x^3*b^2+a^2*(-3/8*B*x ^3+A)*b-9/8*a^3*B)/(b*x^3+a)^(1/3)/b^4
Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {{\left (5 \, B b^{3} x^{9} - {\left (9 \, B a b^{2} - 8 \, A b^{3}\right )} x^{6} + 81 \, B a^{3} - 72 \, A a^{2} b + 3 \, {\left (9 \, B a^{2} b - 8 \, A a b^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{40 \, {\left (b^{5} x^{3} + a b^{4}\right )}} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="fricas")
Output:
1/40*(5*B*b^3*x^9 - (9*B*a*b^2 - 8*A*b^3)*x^6 + 81*B*a^3 - 72*A*a^2*b + 3* (9*B*a^2*b - 8*A*a*b^2)*x^3)*(b*x^3 + a)^(2/3)/(b^5*x^3 + a*b^4)
Time = 0.55 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.70 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\begin {cases} - \frac {9 A a^{2}}{5 b^{3} \sqrt [3]{a + b x^{3}}} - \frac {3 A a x^{3}}{5 b^{2} \sqrt [3]{a + b x^{3}}} + \frac {A x^{6}}{5 b \sqrt [3]{a + b x^{3}}} + \frac {81 B a^{3}}{40 b^{4} \sqrt [3]{a + b x^{3}}} + \frac {27 B a^{2} x^{3}}{40 b^{3} \sqrt [3]{a + b x^{3}}} - \frac {9 B a x^{6}}{40 b^{2} \sqrt [3]{a + b x^{3}}} + \frac {B x^{9}}{8 b \sqrt [3]{a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{9}}{9} + \frac {B x^{12}}{12}}{a^{\frac {4}{3}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**8*(B*x**3+A)/(b*x**3+a)**(4/3),x)
Output:
Piecewise((-9*A*a**2/(5*b**3*(a + b*x**3)**(1/3)) - 3*A*a*x**3/(5*b**2*(a + b*x**3)**(1/3)) + A*x**6/(5*b*(a + b*x**3)**(1/3)) + 81*B*a**3/(40*b**4* (a + b*x**3)**(1/3)) + 27*B*a**2*x**3/(40*b**3*(a + b*x**3)**(1/3)) - 9*B* a*x**6/(40*b**2*(a + b*x**3)**(1/3)) + B*x**9/(8*b*(a + b*x**3)**(1/3)), N e(b, 0)), ((A*x**9/9 + B*x**12/12)/a**(4/3), True))
Time = 0.03 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.16 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {1}{40} \, B {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}}}{b^{4}} - \frac {24 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a}{b^{4}} + \frac {60 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2}}{b^{4}} + \frac {40 \, a^{3}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{4}}\right )} + \frac {1}{5} \, A {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{b^{3}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{b^{3}} - \frac {5 \, a^{2}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{3}}\right )} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="maxima")
Output:
1/40*B*(5*(b*x^3 + a)^(8/3)/b^4 - 24*(b*x^3 + a)^(5/3)*a/b^4 + 60*(b*x^3 + a)^(2/3)*a^2/b^4 + 40*a^3/((b*x^3 + a)^(1/3)*b^4)) + 1/5*A*((b*x^3 + a)^( 5/3)/b^3 - 5*(b*x^3 + a)^(2/3)*a/b^3 - 5*a^2/((b*x^3 + a)^(1/3)*b^3))
Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.12 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {B a^{3} - A a^{2} b}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{4}} + \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} B b^{28} - 24 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} B a b^{28} + 60 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} B a^{2} b^{28} + 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} A b^{29} - 40 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} A a b^{29}}{40 \, b^{32}} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="giac")
Output:
(B*a^3 - A*a^2*b)/((b*x^3 + a)^(1/3)*b^4) + 1/40*(5*(b*x^3 + a)^(8/3)*B*b^ 28 - 24*(b*x^3 + a)^(5/3)*B*a*b^28 + 60*(b*x^3 + a)^(2/3)*B*a^2*b^28 + 8*( b*x^3 + a)^(5/3)*A*b^29 - 40*(b*x^3 + a)^(2/3)*A*a*b^29)/b^32
Time = 1.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {5\,B\,{\left (b\,x^3+a\right )}^3+40\,B\,a^3+8\,A\,b\,{\left (b\,x^3+a\right )}^2-24\,B\,a\,{\left (b\,x^3+a\right )}^2+60\,B\,a^2\,\left (b\,x^3+a\right )-40\,A\,a^2\,b-40\,A\,a\,b\,\left (b\,x^3+a\right )}{40\,b^4\,{\left (b\,x^3+a\right )}^{1/3}} \] Input:
int((x^8*(A + B*x^3))/(a + b*x^3)^(4/3),x)
Output:
(5*B*(a + b*x^3)^3 + 40*B*a^3 + 8*A*b*(a + b*x^3)^2 - 24*B*a*(a + b*x^3)^2 + 60*B*a^2*(a + b*x^3) - 40*A*a^2*b - 40*A*a*b*(a + b*x^3))/(40*b^4*(a + b*x^3)^(1/3))
\[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int \frac {x^{8}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \] Input:
int(x^8*(B*x^3+A)/(b*x^3+a)^(4/3),x)
Output:
int(x**8/(a + b*x**3)**(1/3),x)