Integrand size = 18, antiderivative size = 85 \[ \int x^{5/2} (a+b x)^3 (A+B x) \, dx=\frac {2}{7} a^3 A x^{7/2}+\frac {2}{9} a^2 (3 A b+a B) x^{9/2}+\frac {6}{11} a b (A b+a B) x^{11/2}+\frac {2}{13} b^2 (A b+3 a B) x^{13/2}+\frac {2}{15} b^3 B x^{15/2} \] Output:
2/7*a^3*A*x^(7/2)+2/9*a^2*(3*A*b+B*a)*x^(9/2)+6/11*a*b*(A*b+B*a)*x^(11/2)+ 2/13*b^2*(A*b+3*B*a)*x^(13/2)+2/15*b^3*B*x^(15/2)
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int x^{5/2} (a+b x)^3 (A+B x) \, dx=\frac {2 x^{7/2} \left (715 a^3 (9 A+7 B x)+1365 a^2 b x (11 A+9 B x)+945 a b^2 x^2 (13 A+11 B x)+231 b^3 x^3 (15 A+13 B x)\right )}{45045} \] Input:
Integrate[x^(5/2)*(a + b*x)^3*(A + B*x),x]
Output:
(2*x^(7/2)*(715*a^3*(9*A + 7*B*x) + 1365*a^2*b*x*(11*A + 9*B*x) + 945*a*b^ 2*x^2*(13*A + 11*B*x) + 231*b^3*x^3*(15*A + 13*B*x)))/45045
Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {85, 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.
\(\displaystyle \int x^{5/2} (a+b x)^3 (A+B x) \, dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (a^3 A x^{5/2}+a^2 x^{7/2} (a B+3 A b)+b^2 x^{11/2} (3 a B+A b)+3 a b x^{9/2} (a B+A b)+b^3 B x^{13/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{7} a^3 A x^{7/2}+\frac {2}{9} a^2 x^{9/2} (a B+3 A b)+\frac {2}{13} b^2 x^{13/2} (3 a B+A b)+\frac {6}{11} a b x^{11/2} (a B+A b)+\frac {2}{15} b^3 B x^{15/2}\) |
Input:
Int[x^(5/2)*(a + b*x)^3*(A + B*x),x]
Output:
(2*a^3*A*x^(7/2))/7 + (2*a^2*(3*A*b + a*B)*x^(9/2))/9 + (6*a*b*(A*b + a*B) *x^(11/2))/11 + (2*b^2*(A*b + 3*a*B)*x^(13/2))/13 + (2*b^3*B*x^(15/2))/15
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89
method | result | size |
gosper | \(\frac {2 x^{\frac {7}{2}} \left (3003 B \,b^{3} x^{4}+3465 A \,b^{3} x^{3}+10395 B a \,b^{2} x^{3}+12285 a A \,b^{2} x^{2}+12285 B \,a^{2} b \,x^{2}+15015 a^{2} A b x +5005 B \,a^{3} x +6435 a^{3} A \right )}{45045}\) | \(76\) |
derivativedivides | \(\frac {2 b^{3} B \,x^{\frac {15}{2}}}{15}+\frac {2 \left (b^{3} A +3 a \,b^{2} B \right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (3 a^{2} b A +a^{3} B \right ) x^{\frac {9}{2}}}{9}+\frac {2 a^{3} A \,x^{\frac {7}{2}}}{7}\) | \(76\) |
default | \(\frac {2 b^{3} B \,x^{\frac {15}{2}}}{15}+\frac {2 \left (b^{3} A +3 a \,b^{2} B \right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (3 a \,b^{2} A +3 a^{2} b B \right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (3 a^{2} b A +a^{3} B \right ) x^{\frac {9}{2}}}{9}+\frac {2 a^{3} A \,x^{\frac {7}{2}}}{7}\) | \(76\) |
trager | \(\frac {2 x^{\frac {7}{2}} \left (3003 B \,b^{3} x^{4}+3465 A \,b^{3} x^{3}+10395 B a \,b^{2} x^{3}+12285 a A \,b^{2} x^{2}+12285 B \,a^{2} b \,x^{2}+15015 a^{2} A b x +5005 B \,a^{3} x +6435 a^{3} A \right )}{45045}\) | \(76\) |
risch | \(\frac {2 x^{\frac {7}{2}} \left (3003 B \,b^{3} x^{4}+3465 A \,b^{3} x^{3}+10395 B a \,b^{2} x^{3}+12285 a A \,b^{2} x^{2}+12285 B \,a^{2} b \,x^{2}+15015 a^{2} A b x +5005 B \,a^{3} x +6435 a^{3} A \right )}{45045}\) | \(76\) |
orering | \(\frac {2 x^{\frac {7}{2}} \left (3003 B \,b^{3} x^{4}+3465 A \,b^{3} x^{3}+10395 B a \,b^{2} x^{3}+12285 a A \,b^{2} x^{2}+12285 B \,a^{2} b \,x^{2}+15015 a^{2} A b x +5005 B \,a^{3} x +6435 a^{3} A \right )}{45045}\) | \(76\) |
Input:
int(x^(5/2)*(b*x+a)^3*(B*x+A),x,method=_RETURNVERBOSE)
Output:
2/45045*x^(7/2)*(3003*B*b^3*x^4+3465*A*b^3*x^3+10395*B*a*b^2*x^3+12285*A*a *b^2*x^2+12285*B*a^2*b*x^2+15015*A*a^2*b*x+5005*B*a^3*x+6435*A*a^3)
Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int x^{5/2} (a+b x)^3 (A+B x) \, dx=\frac {2}{45045} \, {\left (3003 \, B b^{3} x^{7} + 6435 \, A a^{3} x^{3} + 3465 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 12285 \, {\left (B a^{2} b + A a b^{2}\right )} x^{5} + 5005 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{4}\right )} \sqrt {x} \] Input:
integrate(x^(5/2)*(b*x+a)^3*(B*x+A),x, algorithm="fricas")
Output:
2/45045*(3003*B*b^3*x^7 + 6435*A*a^3*x^3 + 3465*(3*B*a*b^2 + A*b^3)*x^6 + 12285*(B*a^2*b + A*a*b^2)*x^5 + 5005*(B*a^3 + 3*A*a^2*b)*x^4)*sqrt(x)
Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int x^{5/2} (a+b x)^3 (A+B x) \, dx=\frac {2 A a^{3} x^{\frac {7}{2}}}{7} + \frac {2 A a^{2} b x^{\frac {9}{2}}}{3} + \frac {6 A a b^{2} x^{\frac {11}{2}}}{11} + \frac {2 A b^{3} x^{\frac {13}{2}}}{13} + \frac {2 B a^{3} x^{\frac {9}{2}}}{9} + \frac {6 B a^{2} b x^{\frac {11}{2}}}{11} + \frac {6 B a b^{2} x^{\frac {13}{2}}}{13} + \frac {2 B b^{3} x^{\frac {15}{2}}}{15} \] Input:
integrate(x**(5/2)*(b*x+a)**3*(B*x+A),x)
Output:
2*A*a**3*x**(7/2)/7 + 2*A*a**2*b*x**(9/2)/3 + 6*A*a*b**2*x**(11/2)/11 + 2* A*b**3*x**(13/2)/13 + 2*B*a**3*x**(9/2)/9 + 6*B*a**2*b*x**(11/2)/11 + 6*B* a*b**2*x**(13/2)/13 + 2*B*b**3*x**(15/2)/15
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int x^{5/2} (a+b x)^3 (A+B x) \, dx=\frac {2}{15} \, B b^{3} x^{\frac {15}{2}} + \frac {2}{7} \, A a^{3} x^{\frac {7}{2}} + \frac {2}{13} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{\frac {13}{2}} + \frac {6}{11} \, {\left (B a^{2} b + A a b^{2}\right )} x^{\frac {11}{2}} + \frac {2}{9} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac {9}{2}} \] Input:
integrate(x^(5/2)*(b*x+a)^3*(B*x+A),x, algorithm="maxima")
Output:
2/15*B*b^3*x^(15/2) + 2/7*A*a^3*x^(7/2) + 2/13*(3*B*a*b^2 + A*b^3)*x^(13/2 ) + 6/11*(B*a^2*b + A*a*b^2)*x^(11/2) + 2/9*(B*a^3 + 3*A*a^2*b)*x^(9/2)
Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int x^{5/2} (a+b x)^3 (A+B x) \, dx=\frac {2}{15} \, B b^{3} x^{\frac {15}{2}} + \frac {6}{13} \, B a b^{2} x^{\frac {13}{2}} + \frac {2}{13} \, A b^{3} x^{\frac {13}{2}} + \frac {6}{11} \, B a^{2} b x^{\frac {11}{2}} + \frac {6}{11} \, A a b^{2} x^{\frac {11}{2}} + \frac {2}{9} \, B a^{3} x^{\frac {9}{2}} + \frac {2}{3} \, A a^{2} b x^{\frac {9}{2}} + \frac {2}{7} \, A a^{3} x^{\frac {7}{2}} \] Input:
integrate(x^(5/2)*(b*x+a)^3*(B*x+A),x, algorithm="giac")
Output:
2/15*B*b^3*x^(15/2) + 6/13*B*a*b^2*x^(13/2) + 2/13*A*b^3*x^(13/2) + 6/11*B *a^2*b*x^(11/2) + 6/11*A*a*b^2*x^(11/2) + 2/9*B*a^3*x^(9/2) + 2/3*A*a^2*b* x^(9/2) + 2/7*A*a^3*x^(7/2)
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int x^{5/2} (a+b x)^3 (A+B x) \, dx=x^{9/2}\,\left (\frac {2\,B\,a^3}{9}+\frac {2\,A\,b\,a^2}{3}\right )+x^{13/2}\,\left (\frac {2\,A\,b^3}{13}+\frac {6\,B\,a\,b^2}{13}\right )+\frac {2\,A\,a^3\,x^{7/2}}{7}+\frac {2\,B\,b^3\,x^{15/2}}{15}+\frac {6\,a\,b\,x^{11/2}\,\left (A\,b+B\,a\right )}{11} \] Input:
int(x^(5/2)*(A + B*x)*(a + b*x)^3,x)
Output:
x^(9/2)*((2*B*a^3)/9 + (2*A*a^2*b)/3) + x^(13/2)*((2*A*b^3)/13 + (6*B*a*b^ 2)/13) + (2*A*a^3*x^(7/2))/7 + (2*B*b^3*x^(15/2))/15 + (6*a*b*x^(11/2)*(A* b + B*a))/11
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.56 \[ \int x^{5/2} (a+b x)^3 (A+B x) \, dx=\frac {2 \sqrt {x}\, x^{3} \left (3003 b^{4} x^{4}+13860 a \,b^{3} x^{3}+24570 a^{2} b^{2} x^{2}+20020 a^{3} b x +6435 a^{4}\right )}{45045} \] Input:
int(x^(5/2)*(b*x+a)^3*(B*x+A),x)
Output:
(2*sqrt(x)*x**3*(6435*a**4 + 20020*a**3*b*x + 24570*a**2*b**2*x**2 + 13860 *a*b**3*x**3 + 3003*b**4*x**4))/45045