Integrand size = 22, antiderivative size = 73 \[ \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 a (A b-a B)}{9 b^3 \left (a+b x^3\right )^{3/2}}-\frac {2 (A b-2 a B)}{3 b^3 \sqrt {a+b x^3}}+\frac {2 B \sqrt {a+b x^3}}{3 b^3} \] Output:
2/9*a*(A*b-B*a)/b^3/(b*x^3+a)^(3/2)-2/3*(A*b-2*B*a)/b^3/(b*x^3+a)^(1/2)+2/ 3*B*(b*x^3+a)^(1/2)/b^3
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 \left (-2 a A b+8 a^2 B-3 A b^2 x^3+12 a b B x^3+3 b^2 B x^6\right )}{9 b^3 \left (a+b x^3\right )^{3/2}} \] Input:
Integrate[(x^5*(A + B*x^3))/(a + b*x^3)^(5/2),x]
Output:
(2*(-2*a*A*b + 8*a^2*B - 3*A*b^2*x^3 + 12*a*b*B*x^3 + 3*b^2*B*x^6))/(9*b^3 *(a + b*x^3)^(3/2))
Time = 0.37 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^3 \left (B x^3+A\right )}{\left (b x^3+a\right )^{5/2}}dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B}{b^2 \sqrt {b x^3+a}}+\frac {A b-2 a B}{b^2 \left (b x^3+a\right )^{3/2}}+\frac {a (a B-A b)}{b^2 \left (b x^3+a\right )^{5/2}}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {2 (A b-2 a B)}{b^3 \sqrt {a+b x^3}}+\frac {2 a (A b-a B)}{3 b^3 \left (a+b x^3\right )^{3/2}}+\frac {2 B \sqrt {a+b x^3}}{b^3}\right )\) |
Input:
Int[(x^5*(A + B*x^3))/(a + b*x^3)^(5/2),x]
Output:
((2*a*(A*b - a*B))/(3*b^3*(a + b*x^3)^(3/2)) - (2*(A*b - 2*a*B))/(b^3*Sqrt [a + b*x^3]) + (2*B*Sqrt[a + b*x^3])/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 = 0.98 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {4 \left (\frac {3 x^{3} \left (-B \,x^{3}+A \right ) b^{2}}{2}+a \left (-6 B \,x^{3}+A \right ) b -4 a^{2} B \right )}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{3}}\) | \(49\) |
gosper | \(-\frac {2 \left (-3 b^{2} B \,x^{6}+3 A \,b^{2} x^{3}-12 B a b \,x^{3}+2 a b A -8 a^{2} B \right )}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{3}}\) | \(53\) |
trager | \(-\frac {2 \left (-3 b^{2} B \,x^{6}+3 A \,b^{2} x^{3}-12 B a b \,x^{3}+2 a b A -8 a^{2} B \right )}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{3}}\) | \(53\) |
orering | \(-\frac {2 \left (-3 b^{2} B \,x^{6}+3 A \,b^{2} x^{3}-12 B a b \,x^{3}+2 a b A -8 a^{2} B \right )}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} b^{3}}\) | \(53\) |
risch | \(\frac {2 B \sqrt {b \,x^{3}+a}}{3 b^{3}}-\frac {2 \left (3 A \,b^{2} x^{3}-6 B a b \,x^{3}+2 a b A -5 a^{2} B \right )}{9 b^{3} \left (b \,x^{3}+a \right )^{\frac {3}{2}}}\) | \(60\) |
elliptic | \(\frac {2 \left (A b -B a \right ) a \sqrt {b \,x^{3}+a}}{9 b^{5} \left (x^{3}+\frac {a}{b}\right )^{2}}-\frac {2 \left (A b -2 B a \right )}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 B \sqrt {b \,x^{3}+a}}{3 b^{3}}\) | \(77\) |
default | \(A \left (\frac {2 a \sqrt {b \,x^{3}+a}}{9 b^{4} \left (x^{3}+\frac {a}{b}\right )^{2}}-\frac {2}{3 b^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}\right )+B \left (-\frac {2 a^{2} \sqrt {b \,x^{3}+a}}{9 b^{5} \left (x^{3}+\frac {a}{b}\right )^{2}}+\frac {4 a}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 \sqrt {b \,x^{3}+a}}{3 b^{3}}\right )\) | \(113\) |
Input:
int(x^5*(B*x^3+A)/(b*x^3+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-4/9*(3/2*x^3*(-B*x^3+A)*b^2+a*(-6*B*x^3+A)*b-4*a^2*B)/(b*x^3+a)^(3/2)/b^3
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b^{2} x^{6} + 3 \, {\left (4 \, B a b - A b^{2}\right )} x^{3} + 8 \, B a^{2} - 2 \, A a b\right )} \sqrt {b x^{3} + a}}{9 \, {\left (b^{5} x^{6} + 2 \, a b^{4} x^{3} + a^{2} b^{3}\right )}} \] Input:
integrate(x^5*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="fricas")
Output:
2/9*(3*B*b^2*x^6 + 3*(4*B*a*b - A*b^2)*x^3 + 8*B*a^2 - 2*A*a*b)*sqrt(b*x^3 + a)/(b^5*x^6 + 2*a*b^4*x^3 + a^2*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (70) = 140\).
Time = 0.54 (sec) , antiderivative size = 240, normalized size of antiderivative = 3.29 \[ \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\begin {cases} - \frac {4 A a b}{9 a b^{3} \sqrt {a + b x^{3}} + 9 b^{4} x^{3} \sqrt {a + b x^{3}}} - \frac {6 A b^{2} x^{3}}{9 a b^{3} \sqrt {a + b x^{3}} + 9 b^{4} x^{3} \sqrt {a + b x^{3}}} + \frac {16 B a^{2}}{9 a b^{3} \sqrt {a + b x^{3}} + 9 b^{4} x^{3} \sqrt {a + b x^{3}}} + \frac {24 B a b x^{3}}{9 a b^{3} \sqrt {a + b x^{3}} + 9 b^{4} x^{3} \sqrt {a + b x^{3}}} + \frac {6 B b^{2} x^{6}}{9 a b^{3} \sqrt {a + b x^{3}} + 9 b^{4} x^{3} \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{6}}{6} + \frac {B x^{9}}{9}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(B*x**3+A)/(b*x**3+a)**(5/2),x)
Output:
Piecewise((-4*A*a*b/(9*a*b**3*sqrt(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x* *3)) - 6*A*b**2*x**3/(9*a*b**3*sqrt(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x **3)) + 16*B*a**2/(9*a*b**3*sqrt(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x**3 )) + 24*B*a*b*x**3/(9*a*b**3*sqrt(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x** 3)) + 6*B*b**2*x**6/(9*a*b**3*sqrt(a + b*x**3) + 9*b**4*x**3*sqrt(a + b*x* *3)), Ne(b, 0)), ((A*x**6/6 + B*x**9/9)/a**(5/2), True))
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2}{9} \, B {\left (\frac {3 \, \sqrt {b x^{3} + a}}{b^{3}} + \frac {6 \, a}{\sqrt {b x^{3} + a} b^{3}} - \frac {a^{2}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3}}\right )} - \frac {2}{9} \, A {\left (\frac {3}{\sqrt {b x^{3} + a} b^{2}} - \frac {a}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2}}\right )} \] Input:
integrate(x^5*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="maxima")
Output:
2/9*B*(3*sqrt(b*x^3 + a)/b^3 + 6*a/(sqrt(b*x^3 + a)*b^3) - a^2/((b*x^3 + a )^(3/2)*b^3)) - 2/9*A*(3/(sqrt(b*x^3 + a)*b^2) - a/((b*x^3 + a)^(3/2)*b^2) )
Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 \, \sqrt {b x^{3} + a} B}{3 \, b^{3}} + \frac {2 \, {\left (6 \, {\left (b x^{3} + a\right )} B a - B a^{2} - 3 \, {\left (b x^{3} + a\right )} A b + A a b\right )}}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3}} \] Input:
integrate(x^5*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="giac")
Output:
2/3*sqrt(b*x^3 + a)*B/b^3 + 2/9*(6*(b*x^3 + a)*B*a - B*a^2 - 3*(b*x^3 + a) *A*b + A*a*b)/((b*x^3 + a)^(3/2)*b^3)
Time = 1.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {6\,B\,{\left (b\,x^3+a\right )}^2-2\,B\,a^2-6\,A\,b\,\left (b\,x^3+a\right )+12\,B\,a\,\left (b\,x^3+a\right )+2\,A\,a\,b}{9\,b^3\,{\left (b\,x^3+a\right )}^{3/2}} \] Input:
int((x^5*(A + B*x^3))/(a + b*x^3)^(5/2),x)
Output:
(6*B*(a + b*x^3)^2 - 2*B*a^2 - 6*A*b*(a + b*x^3) + 12*B*a*(a + b*x^3) + 2* A*a*b)/(9*b^3*(a + b*x^3)^(3/2))
Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.42 \[ \int \frac {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {b \,x^{3}+a}\, \left (b \,x^{3}+2 a \right )}{3 b^{2} \left (b \,x^{3}+a \right )} \] Input:
int(x^5*(B*x^3+A)/(b*x^3+a)^(5/2),x)
Output:
(2*sqrt(a + b*x**3)*(2*a + b*x**3))/(3*b**2*(a + b*x**3))