Integrand size = 22, antiderivative size = 73 \[ \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=-\frac {a (A b-a B) \left (a+b x^3\right )^{2/3}}{2 b^3}+\frac {(A b-2 a B) \left (a+b x^3\right )^{5/3}}{5 b^3}+\frac {B \left (a+b x^3\right )^{8/3}}{8 b^3} \] Output:
-1/2*a*(A*b-B*a)*(b*x^3+a)^(2/3)/b^3+1/5*(A*b-2*B*a)*(b*x^3+a)^(5/3)/b^3+1 /8*B*(b*x^3+a)^(8/3)/b^3
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-12 a A b+9 a^2 B+8 A b^2 x^3-6 a b B x^3+5 b^2 B x^6\right )}{40 b^3} \] Input:
Integrate[(x^5*(A + B*x^3))/(a + b*x^3)^(1/3),x]
Output:
((a + b*x^3)^(2/3)*(-12*a*A*b + 9*a^2*B + 8*A*b^2*x^3 - 6*a*b*B*x^3 + 5*b^ 2*B*x^6))/(40*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.
\(\displaystyle \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^3 \left (B x^3+A\right )}{\sqrt [3]{b x^3+a}}dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{5/3}}{b^2}+\frac {(A b-2 a B) \left (b x^3+a\right )^{2/3}}{b^2}+\frac {a (a B-A b)}{b^2 \sqrt [3]{b x^3+a}}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{5/3} (A b-2 a B)}{5 b^3}-\frac {3 a \left (a+b x^3\right )^{2/3} (A b-a B)}{2 b^3}+\frac {3 B \left (a+b x^3\right )^{8/3}}{8 b^3}\right )\) |
Input:
Int[(x^5*(A + B*x^3))/(a + b*x^3)^(1/3),x]
Output:
((-3*a*(A*b - a*B)*(a + b*x^3)^(2/3))/(2*b^3) + (3*(A*b - 2*a*B)*(a + b*x^ 3)^(5/3))/(5*b^3) + (3*B*(a + b*x^3)^(8/3))/(8*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 = 0.78 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {3 \left (-\frac {2 \left (\frac {5 B \,x^{3}}{8}+A \right ) x^{3} b^{2}}{3}+a \left (\frac {B \,x^{3}}{2}+A \right ) b -\frac {3 a^{2} B}{4}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{10 b^{3}}\) | \(49\) |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-5 b^{2} B \,x^{6}-8 A \,b^{2} x^{3}+6 B a b \,x^{3}+12 a b A -9 a^{2} B \right )}{40 b^{3}}\) | \(53\) |
trager | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-5 b^{2} B \,x^{6}-8 A \,b^{2} x^{3}+6 B a b \,x^{3}+12 a b A -9 a^{2} B \right )}{40 b^{3}}\) | \(53\) |
risch | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-5 b^{2} B \,x^{6}-8 A \,b^{2} x^{3}+6 B a b \,x^{3}+12 a b A -9 a^{2} B \right )}{40 b^{3}}\) | \(53\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (-5 b^{2} B \,x^{6}-8 A \,b^{2} x^{3}+6 B a b \,x^{3}+12 a b A -9 a^{2} B \right )}{40 b^{3}}\) | \(53\) |
Input:
int(x^5*(B*x^3+A)/(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)
Output:
-3/10*(-2/3*(5/8*B*x^3+A)*x^3*b^2+a*(1/2*B*x^3+A)*b-3/4*a^2*B)*(b*x^3+a)^( 2/3)/b^3
Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {{\left (5 \, B b^{2} x^{6} - 2 \, {\left (3 \, B a b - 4 \, A b^{2}\right )} x^{3} + 9 \, B a^{2} - 12 \, A a b\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{40 \, b^{3}} \] Input:
integrate(x^5*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="fricas")
Output:
1/40*(5*B*b^2*x^6 - 2*(3*B*a*b - 4*A*b^2)*x^3 + 9*B*a^2 - 12*A*a*b)*(b*x^3 + a)^(2/3)/b^3
Time = 0.36 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.66 \[ \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\begin {cases} - \frac {3 A a \left (a + b x^{3}\right )^{\frac {2}{3}}}{10 b^{2}} + \frac {A x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{5 b} + \frac {9 B a^{2} \left (a + b x^{3}\right )^{\frac {2}{3}}}{40 b^{3}} - \frac {3 B a x^{3} \left (a + b x^{3}\right )^{\frac {2}{3}}}{20 b^{2}} + \frac {B x^{6} \left (a + b x^{3}\right )^{\frac {2}{3}}}{8 b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{6}}{6} + \frac {B x^{9}}{9}}{\sqrt [3]{a}} & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(B*x**3+A)/(b*x**3+a)**(1/3),x)
Output:
Piecewise((-3*A*a*(a + b*x**3)**(2/3)/(10*b**2) + A*x**3*(a + b*x**3)**(2/ 3)/(5*b) + 9*B*a**2*(a + b*x**3)**(2/3)/(40*b**3) - 3*B*a*x**3*(a + b*x**3 )**(2/3)/(20*b**2) + B*x**6*(a + b*x**3)**(2/3)/(8*b), Ne(b, 0)), ((A*x**6 /6 + B*x**9/9)/a**(1/3), True))
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {1}{40} \, B {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}}}{b^{3}} - \frac {16 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a}{b^{3}} + \frac {20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2}}{b^{3}}\right )} + \frac {1}{10} \, A {\left (\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}}}{b^{2}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{b^{2}}\right )} \] Input:
integrate(x^5*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="maxima")
Output:
1/40*B*(5*(b*x^3 + a)^(8/3)/b^3 - 16*(b*x^3 + a)^(5/3)*a/b^3 + 20*(b*x^3 + a)^(2/3)*a^2/b^3) + 1/10*A*(2*(b*x^3 + a)^(5/3)/b^2 - 5*(b*x^3 + a)^(2/3) *a/b^2)
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08 \[ \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} B - 16 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} B a + 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} A b}{40 \, b^{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} B a^{2} - {\left (b x^{3} + a\right )}^{\frac {2}{3}} A a b}{2 \, b^{3}} \] Input:
integrate(x^5*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="giac")
Output:
1/40*(5*(b*x^3 + a)^(8/3)*B - 16*(b*x^3 + a)^(5/3)*B*a + 8*(b*x^3 + a)^(5/ 3)*A*b)/b^3 + 1/2*((b*x^3 + a)^(2/3)*B*a^2 - (b*x^3 + a)^(2/3)*A*a*b)/b^3
Time = 0.84 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx={\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {9\,B\,a^2-12\,A\,a\,b}{40\,b^3}+\frac {x^3\,\left (8\,A\,b^2-6\,B\,a\,b\right )}{40\,b^3}+\frac {B\,x^6}{8\,b}\right ) \] Input:
int((x^5*(A + B*x^3))/(a + b*x^3)^(1/3),x)
Output:
(a + b*x^3)^(2/3)*((9*B*a^2 - 12*A*a*b)/(40*b^3) + (x^3*(8*A*b^2 - 6*B*a*b ))/(40*b^3) + (B*x^6)/(8*b))
\[ \int \frac {x^5 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\left (\int \frac {x^{8}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {x^{5}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a \] Input:
int(x^5*(B*x^3+A)/(b*x^3+a)^(1/3),x)
Output:
int(x**8/(a + b*x**3)**(1/3),x)*b + int(x**5/(a + b*x**3)**(1/3),x)*a