Integrand size = 22, antiderivative size = 103 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 a^2 (A b-a B)}{3 b^4 \sqrt {a+b x^3}}-\frac {2 a (2 A b-3 a B) \sqrt {a+b x^3}}{3 b^4}+\frac {2 (A b-3 a B) \left (a+b x^3\right )^{3/2}}{9 b^4}+\frac {2 B \left (a+b x^3\right )^{5/2}}{15 b^4} \] Output:
-2/3*a^2*(A*b-B*a)/b^4/(b*x^3+a)^(1/2)-2/3*a*(2*A*b-3*B*a)*(b*x^3+a)^(1/2) /b^4+2/9*(A*b-3*B*a)*(b*x^3+a)^(3/2)/b^4+2/15*B*(b*x^3+a)^(5/2)/b^4
Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 \left (48 a^3 B-8 a^2 b \left (5 A-3 B x^3\right )+b^3 x^6 \left (5 A+3 B x^3\right )-2 a b^2 x^3 \left (10 A+3 B x^3\right )\right )}{45 b^4 \sqrt {a+b x^3}} \] Input:
Integrate[(x^8*(A + B*x^3))/(a + b*x^3)^(3/2),x]
Output:
(2*(48*a^3*B - 8*a^2*b*(5*A - 3*B*x^3) + b^3*x^6*(5*A + 3*B*x^3) - 2*a*b^2 *x^3*(10*A + 3*B*x^3)))/(45*b^4*Sqrt[a + b*x^3])
Time = 0.42 (sec) , antiderivative size = 103, 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.
\(\displaystyle \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6 \left (B x^3+A\right )}{\left (b x^3+a\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (-\frac {(a B-A b) a^2}{b^3 \left (b x^3+a\right )^{3/2}}+\frac {(3 a B-2 A b) a}{b^3 \sqrt {b x^3+a}}+\frac {B \left (b x^3+a\right )^{3/2}}{b^3}+\frac {(A b-3 a B) \sqrt {b x^3+a}}{b^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {2 a^2 (A b-a B)}{b^4 \sqrt {a+b x^3}}+\frac {2 \left (a+b x^3\right )^{3/2} (A b-3 a B)}{3 b^4}-\frac {2 a \sqrt {a+b x^3} (2 A b-3 a B)}{b^4}+\frac {2 B \left (a+b x^3\right )^{5/2}}{5 b^4}\right )\) |
Input:
Int[(x^8*(A + B*x^3))/(a + b*x^3)^(3/2),x]
Output:
((-2*a^2*(A*b - a*B))/(b^4*Sqrt[a + b*x^3]) - (2*a*(2*A*b - 3*a*B)*Sqrt[a + b*x^3])/b^4 + (2*(A*b - 3*a*B)*(a + b*x^3)^(3/2))/(3*b^4) + (2*B*(a + b* x^3)^(5/2))/(5*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 = 0.82 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(-\frac {16 \left (-\frac {\left (\frac {3 B \,x^{3}}{5}+A \right ) x^{6} b^{3}}{8}+\frac {a \,x^{3} \left (\frac {3 B \,x^{3}}{10}+A \right ) b^{2}}{2}+a^{2} \left (-\frac {3 B \,x^{3}}{5}+A \right ) b -\frac {6 a^{3} B}{5}\right )}{9 \sqrt {b \,x^{3}+a}\, b^{4}}\) | \(68\) |
gosper | \(-\frac {2 \left (-3 b^{3} B \,x^{9}-5 A \,b^{3} x^{6}+6 B a \,b^{2} x^{6}+20 a A \,b^{2} x^{3}-24 B \,a^{2} b \,x^{3}+40 a^{2} b A -48 a^{3} B \right )}{45 \sqrt {b \,x^{3}+a}\, b^{4}}\) | \(77\) |
trager | \(-\frac {2 \left (-3 b^{3} B \,x^{9}-5 A \,b^{3} x^{6}+6 B a \,b^{2} x^{6}+20 a A \,b^{2} x^{3}-24 B \,a^{2} b \,x^{3}+40 a^{2} b A -48 a^{3} B \right )}{45 \sqrt {b \,x^{3}+a}\, b^{4}}\) | \(77\) |
orering | \(-\frac {2 \left (-3 b^{3} B \,x^{9}-5 A \,b^{3} x^{6}+6 B a \,b^{2} x^{6}+20 a A \,b^{2} x^{3}-24 B \,a^{2} b \,x^{3}+40 a^{2} b A -48 a^{3} B \right )}{45 \sqrt {b \,x^{3}+a}\, b^{4}}\) | \(77\) |
risch | \(-\frac {2 \left (-3 b^{2} B \,x^{6}-5 A \,b^{2} x^{3}+9 B a b \,x^{3}+25 a b A -33 a^{2} B \right ) \sqrt {b \,x^{3}+a}}{45 b^{4}}-\frac {2 a^{2} \left (A b -B a \right )}{3 b^{4} \sqrt {b \,x^{3}+a}}\) | \(79\) |
default | \(A \left (-\frac {2 a^{2}}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{3} \sqrt {b \,x^{3}+a}}{9 b^{2}}-\frac {10 a \sqrt {b \,x^{3}+a}}{9 b^{3}}\right )+B \left (\frac {2 a^{3}}{3 b^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{6} \sqrt {b \,x^{3}+a}}{15 b^{2}}-\frac {2 a \,x^{3} \sqrt {b \,x^{3}+a}}{5 b^{3}}+\frac {22 a^{2} \sqrt {b \,x^{3}+a}}{15 b^{4}}\right )\) | \(134\) |
elliptic | \(-\frac {2 a^{2} \left (A b -B a \right )}{3 b^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 B \,x^{6} \sqrt {b \,x^{3}+a}}{15 b^{2}}+\frac {2 \left (\frac {A b -B a}{b^{2}}-\frac {4 B a}{5 b^{2}}\right ) x^{3} \sqrt {b \,x^{3}+a}}{9 b}+\frac {2 \left (-\frac {a \left (A b -B a \right )}{b^{3}}-\frac {2 \left (\frac {A b -B a}{b^{2}}-\frac {4 B a}{5 b^{2}}\right ) a}{3 b}\right ) \sqrt {b \,x^{3}+a}}{3 b}\) | \(141\) |
Input:
int(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-16/9*(-1/8*(3/5*B*x^3+A)*x^6*b^3+1/2*a*x^3*(3/10*B*x^3+A)*b^2+a^2*(-3/5*B *x^3+A)*b-6/5*a^3*B)/(b*x^3+a)^(1/2)/b^4
Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.85 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (3 \, B b^{3} x^{9} - {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 48 \, B a^{3} - 40 \, A a^{2} b + 4 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{45 \, {\left (b^{5} x^{3} + a b^{4}\right )}} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="fricas")
Output:
2/45*(3*B*b^3*x^9 - (6*B*a*b^2 - 5*A*b^3)*x^6 + 48*B*a^3 - 40*A*a^2*b + 4* (6*B*a^2*b - 5*A*a*b^2)*x^3)*sqrt(b*x^3 + a)/(b^5*x^3 + a*b^4)
Time = 0.46 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.70 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\begin {cases} - \frac {16 A a^{2}}{9 b^{3} \sqrt {a + b x^{3}}} - \frac {8 A a x^{3}}{9 b^{2} \sqrt {a + b x^{3}}} + \frac {2 A x^{6}}{9 b \sqrt {a + b x^{3}}} + \frac {32 B a^{3}}{15 b^{4} \sqrt {a + b x^{3}}} + \frac {16 B a^{2} x^{3}}{15 b^{3} \sqrt {a + b x^{3}}} - \frac {4 B a x^{6}}{15 b^{2} \sqrt {a + b x^{3}}} + \frac {2 B x^{9}}{15 b \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{9}}{9} + \frac {B x^{12}}{12}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(x**8*(B*x**3+A)/(b*x**3+a)**(3/2),x)
Output:
Piecewise((-16*A*a**2/(9*b**3*sqrt(a + b*x**3)) - 8*A*a*x**3/(9*b**2*sqrt( a + b*x**3)) + 2*A*x**6/(9*b*sqrt(a + b*x**3)) + 32*B*a**3/(15*b**4*sqrt(a + b*x**3)) + 16*B*a**2*x**3/(15*b**3*sqrt(a + b*x**3)) - 4*B*a*x**6/(15*b **2*sqrt(a + b*x**3)) + 2*B*x**9/(15*b*sqrt(a + b*x**3)), Ne(b, 0)), ((A*x **9/9 + B*x**12/12)/a**(3/2), True))
Time = 0.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2}{15} \, B {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {5}{2}}}{b^{4}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a}{b^{4}} + \frac {15 \, \sqrt {b x^{3} + a} a^{2}}{b^{4}} + \frac {5 \, a^{3}}{\sqrt {b x^{3} + a} b^{4}}\right )} + \frac {2}{9} \, A {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}}}{b^{3}} - \frac {6 \, \sqrt {b x^{3} + a} a}{b^{3}} - \frac {3 \, a^{2}}{\sqrt {b x^{3} + a} b^{3}}\right )} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="maxima")
Output:
2/15*B*((b*x^3 + a)^(5/2)/b^4 - 5*(b*x^3 + a)^(3/2)*a/b^4 + 15*sqrt(b*x^3 + a)*a^2/b^4 + 5*a^3/(sqrt(b*x^3 + a)*b^4)) + 2/9*A*((b*x^3 + a)^(3/2)/b^3 - 6*sqrt(b*x^3 + a)*a/b^3 - 3*a^2/(sqrt(b*x^3 + a)*b^3))
Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.11 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (B a^{3} - A a^{2} b\right )}}{3 \, \sqrt {b x^{3} + a} b^{4}} + \frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B b^{16} - 15 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a b^{16} + 45 \, \sqrt {b x^{3} + a} B a^{2} b^{16} + 5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A b^{17} - 30 \, \sqrt {b x^{3} + a} A a b^{17}\right )}}{45 \, b^{20}} \] Input:
integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="giac")
Output:
2/3*(B*a^3 - A*a^2*b)/(sqrt(b*x^3 + a)*b^4) + 2/45*(3*(b*x^3 + a)^(5/2)*B* b^16 - 15*(b*x^3 + a)^(3/2)*B*a*b^16 + 45*sqrt(b*x^3 + a)*B*a^2*b^16 + 5*( b*x^3 + a)^(3/2)*A*b^17 - 30*sqrt(b*x^3 + a)*A*a*b^17)/b^20
Time = 1.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.48 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {b\,x^3+a}\,\left (\frac {2\,\left (B\,a^2-A\,a\,b\right )}{b^3}-\frac {2\,a\,\left (\frac {2\,\left (A\,b^2-B\,a\,b\right )}{b^3}-\frac {8\,B\,a}{5\,b^2}\right )}{3\,b}\right )}{3\,b}+\frac {x^3\,\sqrt {b\,x^3+a}\,\left (\frac {2\,\left (A\,b^2-B\,a\,b\right )}{b^3}-\frac {8\,B\,a}{5\,b^2}\right )}{9\,b}-\frac {a^2\,\left (\frac {2\,A}{3\,b}-\frac {2\,B\,a}{3\,b^2}\right )}{b^2\,\sqrt {b\,x^3+a}}+\frac {2\,B\,x^6\,\sqrt {b\,x^3+a}}{15\,b^2} \] Input:
int((x^8*(A + B*x^3))/(a + b*x^3)^(3/2),x)
Output:
((a + b*x^3)^(1/2)*((2*(B*a^2 - A*a*b))/b^3 - (2*a*((2*(A*b^2 - B*a*b))/b^ 3 - (8*B*a)/(5*b^2)))/(3*b)))/(3*b) + (x^3*(a + b*x^3)^(1/2)*((2*(A*b^2 - B*a*b))/b^3 - (8*B*a)/(5*b^2)))/(9*b) - (a^2*((2*A)/(3*b) - (2*B*a)/(3*b^2 )))/(b^2*(a + b*x^3)^(1/2)) + (2*B*x^6*(a + b*x^3)^(1/2))/(15*b^2)
Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.33 \[ \int \frac {x^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 \sqrt {b \,x^{3}+a}\, \left (3 b^{2} x^{6}-4 a b \,x^{3}+8 a^{2}\right )}{45 b^{3}} \] Input:
int(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x)
Output:
(2*sqrt(a + b*x**3)*(8*a**2 - 4*a*b*x**3 + 3*b**2*x**6))/(45*b**3)