Integrand size = 22, antiderivative size = 73 \[ \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=-\frac {2 a (A b-a B) \left (a+b x^3\right )^{3/2}}{9 b^3}+\frac {2 (A b-2 a B) \left (a+b x^3\right )^{5/2}}{15 b^3}+\frac {2 B \left (a+b x^3\right )^{7/2}}{21 b^3} \] Output:
-2/9*a*(A*b-B*a)*(b*x^3+a)^(3/2)/b^3+2/15*(A*b-2*B*a)*(b*x^3+a)^(5/2)/b^3+ 2/21*B*(b*x^3+a)^(7/2)/b^3
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \left (a+b x^3\right )^{3/2} \left (-14 a A b+8 a^2 B+21 A b^2 x^3-12 a b B x^3+15 b^2 B x^6\right )}{315 b^3} \] Input:
Integrate[x^5*Sqrt[a + b*x^3]*(A + B*x^3),x]
Output:
(2*(a + b*x^3)^(3/2)*(-14*a*A*b + 8*a^2*B + 21*A*b^2*x^3 - 12*a*b*B*x^3 + 15*b^2*B*x^6))/(315*b^3)
Time = 0.37 (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.
\(\displaystyle \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int x^3 \sqrt {b x^3+a} \left (B x^3+A\right )dx^3\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{3} \int \left (\frac {B \left (b x^3+a\right )^{5/2}}{b^2}+\frac {(A b-2 a B) \left (b x^3+a\right )^{3/2}}{b^2}+\frac {a (a B-A b) \sqrt {b x^3+a}}{b^2}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {2 \left (a+b x^3\right )^{5/2} (A b-2 a B)}{5 b^3}-\frac {2 a \left (a+b x^3\right )^{3/2} (A b-a B)}{3 b^3}+\frac {2 B \left (a+b x^3\right )^{7/2}}{7 b^3}\right )\) |
Input:
Int[x^5*Sqrt[a + b*x^3]*(A + B*x^3),x]
Output:
((-2*a*(A*b - a*B)*(a + b*x^3)^(3/2))/(3*b^3) + (2*(A*b - 2*a*B)*(a + b*x^ 3)^(5/2))/(5*b^3) + (2*B*(a + b*x^3)^(7/2))/(7*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 = 0.80 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {4 \left (-\frac {3 \left (\frac {5 B \,x^{3}}{7}+A \right ) x^{3} b^{2}}{2}+a \left (\frac {6 B \,x^{3}}{7}+A \right ) b -\frac {4 a^{2} B}{7}\right ) \left (b \,x^{3}+a \right )^{\frac {3}{2}}}{45 b^{3}}\) | \(49\) |
gosper | \(-\frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}} \left (-15 b^{2} B \,x^{6}-21 A \,b^{2} x^{3}+12 B a b \,x^{3}+14 a b A -8 a^{2} B \right )}{315 b^{3}}\) | \(53\) |
orering | \(-\frac {2 \left (b \,x^{3}+a \right )^{\frac {3}{2}} \left (-15 b^{2} B \,x^{6}-21 A \,b^{2} x^{3}+12 B a b \,x^{3}+14 a b A -8 a^{2} B \right )}{315 b^{3}}\) | \(53\) |
trager | \(-\frac {2 \left (-15 b^{3} B \,x^{9}-21 A \,b^{3} x^{6}-3 B a \,b^{2} x^{6}-7 a A \,b^{2} x^{3}+4 B \,a^{2} b \,x^{3}+14 a^{2} b A -8 a^{3} B \right ) \sqrt {b \,x^{3}+a}}{315 b^{3}}\) | \(77\) |
risch | \(-\frac {2 \left (-15 b^{3} B \,x^{9}-21 A \,b^{3} x^{6}-3 B a \,b^{2} x^{6}-7 a A \,b^{2} x^{3}+4 B \,a^{2} b \,x^{3}+14 a^{2} b A -8 a^{3} B \right ) \sqrt {b \,x^{3}+a}}{315 b^{3}}\) | \(77\) |
elliptic | \(\frac {2 B \,x^{9} \sqrt {b \,x^{3}+a}}{21}+\frac {2 \left (A b +\frac {B a}{7}\right ) x^{6} \sqrt {b \,x^{3}+a}}{15 b}+\frac {2 \left (A a -\frac {4 a \left (A b +\frac {B a}{7}\right )}{5 b}\right ) x^{3} \sqrt {b \,x^{3}+a}}{9 b}-\frac {4 a \left (A a -\frac {4 a \left (A b +\frac {B a}{7}\right )}{5 b}\right ) \sqrt {b \,x^{3}+a}}{9 b^{2}}\) | \(110\) |
default | \(A \left (\frac {2 x^{6} \sqrt {b \,x^{3}+a}}{15}+\frac {2 a \,x^{3} \sqrt {b \,x^{3}+a}}{45 b}-\frac {4 a^{2} \sqrt {b \,x^{3}+a}}{45 b^{2}}\right )+B \left (\frac {2 x^{9} \sqrt {b \,x^{3}+a}}{21}+\frac {2 a \,x^{6} \sqrt {b \,x^{3}+a}}{105 b}-\frac {8 a^{2} x^{3} \sqrt {b \,x^{3}+a}}{315 b^{2}}+\frac {16 a^{3} \sqrt {b \,x^{3}+a}}{315 b^{3}}\right )\) | \(126\) |
Input:
int(x^5*(b*x^3+a)^(1/2)*(B*x^3+A),x,method=_RETURNVERBOSE)
Output:
-4/45*(-3/2*(5/7*B*x^3+A)*x^3*b^2+a*(6/7*B*x^3+A)*b-4/7*a^2*B)*(b*x^3+a)^( 3/2)/b^3
Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \, {\left (15 \, B b^{3} x^{9} + 3 \, {\left (B a b^{2} + 7 \, A b^{3}\right )} x^{6} + 8 \, B a^{3} - 14 \, A a^{2} b - {\left (4 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{315 \, b^{3}} \] Input:
integrate(x^5*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="fricas")
Output:
2/315*(15*B*b^3*x^9 + 3*(B*a*b^2 + 7*A*b^3)*x^6 + 8*B*a^3 - 14*A*a^2*b - ( 4*B*a^2*b - 7*A*a*b^2)*x^3)*sqrt(b*x^3 + a)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (70) = 140\).
Time = 0.30 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.30 \[ \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\begin {cases} - \frac {4 A a^{2} \sqrt {a + b x^{3}}}{45 b^{2}} + \frac {2 A a x^{3} \sqrt {a + b x^{3}}}{45 b} + \frac {2 A x^{6} \sqrt {a + b x^{3}}}{15} + \frac {16 B a^{3} \sqrt {a + b x^{3}}}{315 b^{3}} - \frac {8 B a^{2} x^{3} \sqrt {a + b x^{3}}}{315 b^{2}} + \frac {2 B a x^{6} \sqrt {a + b x^{3}}}{105 b} + \frac {2 B x^{9} \sqrt {a + b x^{3}}}{21} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{6}}{6} + \frac {B x^{9}}{9}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**5*(b*x**3+a)**(1/2)*(B*x**3+A),x)
Output:
Piecewise((-4*A*a**2*sqrt(a + b*x**3)/(45*b**2) + 2*A*a*x**3*sqrt(a + b*x* *3)/(45*b) + 2*A*x**6*sqrt(a + b*x**3)/15 + 16*B*a**3*sqrt(a + b*x**3)/(31 5*b**3) - 8*B*a**2*x**3*sqrt(a + b*x**3)/(315*b**2) + 2*B*a*x**6*sqrt(a + b*x**3)/(105*b) + 2*B*x**9*sqrt(a + b*x**3)/21, Ne(b, 0)), (sqrt(a)*(A*x** 6/6 + B*x**9/9), True))
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2}{315} \, B {\left (\frac {15 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}}}{b^{3}} - \frac {42 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} a}{b^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}{b^{3}}\right )} + \frac {2}{45} \, A {\left (\frac {3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a}{b^{2}}\right )} \] Input:
integrate(x^5*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="maxima")
Output:
2/315*B*(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) + 2/45*A*(3*(b*x^3 + a)^(5/2)/b^2 - 5*(b*x^3 + a)^(3/ 2)*a/b^2)
Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \, {\left (15 \, {\left (b x^{3} + a\right )}^{\frac {7}{2}} B - 42 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B a + 35 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a^{2} + 21 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} A b - 35 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A a b\right )}}{315 \, b^{3}} \] Input:
integrate(x^5*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="giac")
Output:
2/315*(15*(b*x^3 + a)^(7/2)*B - 42*(b*x^3 + a)^(5/2)*B*a + 35*(b*x^3 + a)^ (3/2)*B*a^2 + 21*(b*x^3 + a)^(5/2)*A*b - 35*(b*x^3 + a)^(3/2)*A*a*b)/b^3
Time = 0.77 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.56 \[ \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2\,B\,x^9\,\sqrt {b\,x^3+a}}{21}+\frac {x^6\,\sqrt {b\,x^3+a}\,\left (2\,A\,b+\frac {2\,B\,a}{7}\right )}{15\,b}-\frac {2\,a\,\left (2\,A\,a-\frac {4\,a\,\left (2\,A\,b+\frac {2\,B\,a}{7}\right )}{5\,b}\right )\,\sqrt {b\,x^3+a}}{9\,b^2}+\frac {x^3\,\left (2\,A\,a-\frac {4\,a\,\left (2\,A\,b+\frac {2\,B\,a}{7}\right )}{5\,b}\right )\,\sqrt {b\,x^3+a}}{9\,b} \] Input:
int(x^5*(A + B*x^3)*(a + b*x^3)^(1/2),x)
Output:
(2*B*x^9*(a + b*x^3)^(1/2))/21 + (x^6*(a + b*x^3)^(1/2)*(2*A*b + (2*B*a)/7 ))/(15*b) - (2*a*(2*A*a - (4*a*(2*A*b + (2*B*a)/7))/(5*b))*(a + b*x^3)^(1/ 2))/(9*b^2) + (x^3*(2*A*a - (4*a*(2*A*b + (2*B*a)/7))/(5*b))*(a + b*x^3)^( 1/2))/(9*b)
Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.60 \[ \int x^5 \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \sqrt {b \,x^{3}+a}\, \left (5 b^{3} x^{9}+8 a \,b^{2} x^{6}+a^{2} b \,x^{3}-2 a^{3}\right )}{105 b^{2}} \] Input:
int(x^5*(b*x^3+a)^(1/2)*(B*x^3+A),x)
Output:
(2*sqrt(a + b*x**3)*( - 2*a**3 + a**2*b*x**3 + 8*a*b**2*x**6 + 5*b**3*x**9 ))/(105*b**2)