Integrand size = 22, antiderivative size = 103 \[ \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 a^2 (A b-a B) \left (a+b x^3\right )^{3/2}}{9 b^4}-\frac {2 a (2 A b-3 a B) \left (a+b x^3\right )^{5/2}}{15 b^4}+\frac {2 (A b-3 a B) \left (a+b x^3\right )^{7/2}}{21 b^4}+\frac {2 B \left (a+b x^3\right )^{9/2}}{27 b^4} \] Output:
2/9*a^2*(A*b-B*a)*(b*x^3+a)^(3/2)/b^4-2/15*a*(2*A*b-3*B*a)*(b*x^3+a)^(5/2) /b^4+2/21*(A*b-3*B*a)*(b*x^3+a)^(7/2)/b^4+2/27*B*(b*x^3+a)^(9/2)/b^4
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.73 \[ \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \left (a+b x^3\right )^{3/2} \left (-16 a^3 B+24 a^2 b \left (A+B x^3\right )-6 a b^2 x^3 \left (6 A+5 B x^3\right )+5 b^3 x^6 \left (9 A+7 B x^3\right )\right )}{945 b^4} \] Input:
Integrate[x^8*Sqrt[a + b*x^3]*(A + B*x^3),x]
Output:
(2*(a + b*x^3)^(3/2)*(-16*a^3*B + 24*a^2*b*(A + B*x^3) - 6*a*b^2*x^3*(6*A + 5*B*x^3) + 5*b^3*x^6*(9*A + 7*B*x^3)))/(945*b^4)
Time = 0.41 (sec) , antiderivative size = 107, 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.
\(\displaystyle \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int x^6 \sqrt {b x^3+a} \left (B x^3+A\right )dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{7/2}}{b^3}+\frac {(A b-3 a B) \left (b x^3+a\right )^{5/2}}{b^3}+\frac {a (3 a B-2 A b) \left (b x^3+a\right )^{3/2}}{b^3}-\frac {a^2 (a B-A b) \sqrt {b x^3+a}}{b^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {2 a^2 \left (a+b x^3\right )^{3/2} (A b-a B)}{3 b^4}+\frac {2 \left (a+b x^3\right )^{7/2} (A b-3 a B)}{7 b^4}-\frac {2 a \left (a+b x^3\right )^{5/2} (2 A b-3 a B)}{5 b^4}+\frac {2 B \left (a+b x^3\right )^{9/2}}{9 b^4}\right )\) |
Input:
Int[x^8*Sqrt[a + b*x^3]*(A + B*x^3),x]
Output:
((2*a^2*(A*b - a*B)*(a + b*x^3)^(3/2))/(3*b^4) - (2*a*(2*A*b - 3*a*B)*(a + b*x^3)^(5/2))/(5*b^4) + (2*(A*b - 3*a*B)*(a + b*x^3)^(7/2))/(7*b^4) + (2* B*(a + b*x^3)^(9/2))/(9*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 = 67, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(\frac {16 \left (\frac {15 \left (\frac {7 B \,x^{3}}{9}+A \right ) x^{6} b^{3}}{8}-\frac {3 a \left (\frac {5 B \,x^{3}}{6}+A \right ) x^{3} b^{2}}{2}+a^{2} \left (B \,x^{3}+A \right ) b -\frac {2 a^{3} B}{3}\right ) \left (b \,x^{3}+a \right )^{\frac {3}{2}}}{315 b^{4}}\) | \(67\) |
gosper | \(\frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}} \left (35 b^{3} B \,x^{9}+45 A \,b^{3} x^{6}-30 B a \,b^{2} x^{6}-36 a A \,b^{2} x^{3}+24 B \,a^{2} b \,x^{3}+24 a^{2} b A -16 a^{3} B \right )}{945 b^{4}}\) | \(77\) |
orering | \(\frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}} \left (35 b^{3} B \,x^{9}+45 A \,b^{3} x^{6}-30 B a \,b^{2} x^{6}-36 a A \,b^{2} x^{3}+24 B \,a^{2} b \,x^{3}+24 a^{2} b A -16 a^{3} B \right )}{945 b^{4}}\) | \(77\) |
trager | \(\frac {2 \left (35 B \,x^{12} b^{4}+45 A \,b^{4} x^{9}+5 B \,x^{9} a \,b^{3}+9 A \,x^{6} a \,b^{3}-6 B \,x^{6} a^{2} b^{2}-12 A \,a^{2} b^{2} x^{3}+8 B \,a^{3} b \,x^{3}+24 A \,a^{3} b -16 B \,a^{4}\right ) \sqrt {b \,x^{3}+a}}{945 b^{4}}\) | \(101\) |
risch | \(\frac {2 \left (35 B \,x^{12} b^{4}+45 A \,b^{4} x^{9}+5 B \,x^{9} a \,b^{3}+9 A \,x^{6} a \,b^{3}-6 B \,x^{6} a^{2} b^{2}-12 A \,a^{2} b^{2} x^{3}+8 B \,a^{3} b \,x^{3}+24 A \,a^{3} b -16 B \,a^{4}\right ) \sqrt {b \,x^{3}+a}}{945 b^{4}}\) | \(101\) |
elliptic | \(\frac {2 B \,x^{12} \sqrt {b \,x^{3}+a}}{27}+\frac {2 \left (A b +\frac {B a}{9}\right ) x^{9} \sqrt {b \,x^{3}+a}}{21 b}+\frac {2 \left (A a -\frac {6 a \left (A b +\frac {B a}{9}\right )}{7 b}\right ) x^{6} \sqrt {b \,x^{3}+a}}{15 b}-\frac {8 a \left (A a -\frac {6 a \left (A b +\frac {B a}{9}\right )}{7 b}\right ) x^{3} \sqrt {b \,x^{3}+a}}{45 b^{2}}+\frac {16 a^{2} \left (A a -\frac {6 a \left (A b +\frac {B a}{9}\right )}{7 b}\right ) \sqrt {b \,x^{3}+a}}{45 b^{3}}\) | \(148\) |
default | \(A \left (\frac {2 x^{9} \sqrt {b \,x^{3}+a}}{21}+\frac {2 a \,x^{6} \sqrt {b \,x^{3}+a}}{105 b}-\frac {8 a^{2} x^{3} \sqrt {b \,x^{3}+a}}{315 b^{2}}+\frac {16 a^{3} \sqrt {b \,x^{3}+a}}{315 b^{3}}\right )+B \left (\frac {2 x^{12} \sqrt {b \,x^{3}+a}}{27}+\frac {2 a \,x^{9} \sqrt {b \,x^{3}+a}}{189 b}-\frac {4 a^{2} x^{6} \sqrt {b \,x^{3}+a}}{315 b^{2}}+\frac {16 a^{3} x^{3} \sqrt {b \,x^{3}+a}}{945 b^{3}}-\frac {32 a^{4} \sqrt {b \,x^{3}+a}}{945 b^{4}}\right )\) | \(166\) |
Input:
int(x^8*(b*x^3+a)^(1/2)*(B*x^3+A),x,method=_RETURNVERBOSE)
Output:
16/315*(15/8*(7/9*B*x^3+A)*x^6*b^3-3/2*a*(5/6*B*x^3+A)*x^3*b^2+a^2*(B*x^3+ A)*b-2/3*a^3*B)*(b*x^3+a)^(3/2)/b^4
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \, {\left (35 \, B b^{4} x^{12} + 5 \, {\left (B a b^{3} + 9 \, A b^{4}\right )} x^{9} - 3 \, {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{6} - 16 \, B a^{4} + 24 \, A a^{3} b + 4 \, {\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{945 \, b^{4}} \] Input:
integrate(x^8*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="fricas")
Output:
2/945*(35*B*b^4*x^12 + 5*(B*a*b^3 + 9*A*b^4)*x^9 - 3*(2*B*a^2*b^2 - 3*A*a* b^3)*x^6 - 16*B*a^4 + 24*A*a^3*b + 4*(2*B*a^3*b - 3*A*a^2*b^2)*x^3)*sqrt(b *x^3 + a)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (100) = 200\).
Time = 0.42 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.13 \[ \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\begin {cases} \frac {16 A a^{3} \sqrt {a + b x^{3}}}{315 b^{3}} - \frac {8 A a^{2} x^{3} \sqrt {a + b x^{3}}}{315 b^{2}} + \frac {2 A a x^{6} \sqrt {a + b x^{3}}}{105 b} + \frac {2 A x^{9} \sqrt {a + b x^{3}}}{21} - \frac {32 B a^{4} \sqrt {a + b x^{3}}}{945 b^{4}} + \frac {16 B a^{3} x^{3} \sqrt {a + b x^{3}}}{945 b^{3}} - \frac {4 B a^{2} x^{6} \sqrt {a + b x^{3}}}{315 b^{2}} + \frac {2 B a x^{9} \sqrt {a + b x^{3}}}{189 b} + \frac {2 B x^{12} \sqrt {a + b x^{3}}}{27} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{9}}{9} + \frac {B x^{12}}{12}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**8*(b*x**3+a)**(1/2)*(B*x**3+A),x)
Output:
Piecewise((16*A*a**3*sqrt(a + b*x**3)/(315*b**3) - 8*A*a**2*x**3*sqrt(a + b*x**3)/(315*b**2) + 2*A*a*x**6*sqrt(a + b*x**3)/(105*b) + 2*A*x**9*sqrt(a + b*x**3)/21 - 32*B*a**4*sqrt(a + b*x**3)/(945*b**4) + 16*B*a**3*x**3*sqr t(a + b*x**3)/(945*b**3) - 4*B*a**2*x**6*sqrt(a + b*x**3)/(315*b**2) + 2*B *a*x**9*sqrt(a + b*x**3)/(189*b) + 2*B*x**12*sqrt(a + b*x**3)/27, Ne(b, 0) ), (sqrt(a)*(A*x**9/9 + B*x**12/12), True))
Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15 \[ \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2}{945} \, B {\left (\frac {35 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}}}{b^{4}} - \frac {135 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} a}{b^{4}} + \frac {189 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{2}}{b^{4}} - \frac {105 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{3}}{b^{4}}\right )} + \frac {2}{315} \, A {\left (\frac {15 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}}}{b^{3}} - \frac {42 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a}{b^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}{b^{3}}\right )} \] Input:
integrate(x^8*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="maxima")
Output:
2/945*B*(35*(b*x^3 + a)^(9/2)/b^4 - 135*(b*x^3 + a)^(7/2)*a/b^4 + 189*(b*x ^3 + a)^(5/2)*a^2/b^4 - 105*(b*x^3 + a)^(3/2)*a^3/b^4) + 2/315*A*(15*(b*x^ 3 + a)^(7/2)/b^3 - 42*(b*x^3 + a)^(5/2)*a/b^3 + 35*(b*x^3 + a)^(3/2)*a^2/b ^3)
Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \, {\left (35 \, {\left (b x^{3} + a\right )}^{\frac {9}{2}} B - 135 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} B a + 189 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B a^{2} - 105 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a^{3} + 45 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} A b - 126 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A a b + 105 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A a^{2} b\right )}}{945 \, b^{4}} \] Input:
integrate(x^8*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="giac")
Output:
2/945*(35*(b*x^3 + a)^(9/2)*B - 135*(b*x^3 + a)^(7/2)*B*a + 189*(b*x^3 + a )^(5/2)*B*a^2 - 105*(b*x^3 + a)^(3/2)*B*a^3 + 45*(b*x^3 + a)^(7/2)*A*b - 1 26*(b*x^3 + a)^(5/2)*A*a*b + 105*(b*x^3 + a)^(3/2)*A*a^2*b)/b^4
Time = 0.79 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.50 \[ \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2\,B\,x^{12}\,\sqrt {b\,x^3+a}}{27}+\frac {x^9\,\sqrt {b\,x^3+a}\,\left (2\,A\,b+\frac {2\,B\,a}{9}\right )}{21\,b}+\frac {8\,a^2\,\left (2\,A\,a-\frac {6\,a\,\left (2\,A\,b+\frac {2\,B\,a}{9}\right )}{7\,b}\right )\,\sqrt {b\,x^3+a}}{45\,b^3}+\frac {x^6\,\left (2\,A\,a-\frac {6\,a\,\left (2\,A\,b+\frac {2\,B\,a}{9}\right )}{7\,b}\right )\,\sqrt {b\,x^3+a}}{15\,b}-\frac {4\,a\,x^3\,\left (2\,A\,a-\frac {6\,a\,\left (2\,A\,b+\frac {2\,B\,a}{9}\right )}{7\,b}\right )\,\sqrt {b\,x^3+a}}{45\,b^2} \] Input:
int(x^8*(A + B*x^3)*(a + b*x^3)^(1/2),x)
Output:
(2*B*x^12*(a + b*x^3)^(1/2))/27 + (x^9*(a + b*x^3)^(1/2)*(2*A*b + (2*B*a)/ 9))/(21*b) + (8*a^2*(2*A*a - (6*a*(2*A*b + (2*B*a)/9))/(7*b))*(a + b*x^3)^ (1/2))/(45*b^3) + (x^6*(2*A*a - (6*a*(2*A*b + (2*B*a)/9))/(7*b))*(a + b*x^ 3)^(1/2))/(15*b) - (4*a*x^3*(2*A*a - (6*a*(2*A*b + (2*B*a)/9))/(7*b))*(a + b*x^3)^(1/2))/(45*b^2)
Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.54 \[ \int x^8 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \sqrt {b \,x^{3}+a}\, \left (35 b^{4} x^{12}+50 a \,b^{3} x^{9}+3 a^{2} b^{2} x^{6}-4 a^{3} b \,x^{3}+8 a^{4}\right )}{945 b^{3}} \] Input:
int(x^8*(b*x^3+a)^(1/2)*(B*x^3+A),x)
Output:
(2*sqrt(a + b*x**3)*(8*a**4 - 4*a**3*b*x**3 + 3*a**2*b**2*x**6 + 50*a*b**3 *x**9 + 35*b**4*x**12))/(945*b**3)