Integrand size = 29, antiderivative size = 111 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{5} a^4 A x^{5/2}+\frac {2}{7} a^3 (4 A b+a B) x^{7/2}+\frac {4}{9} a^2 b (3 A b+2 a B) x^{9/2}+\frac {4}{11} a b^2 (2 A b+3 a B) x^{11/2}+\frac {2}{13} b^3 (A b+4 a B) x^{13/2}+\frac {2}{15} b^4 B x^{15/2} \] Output:
2/5*a^4*A*x^(5/2)+2/7*a^3*(4*A*b+B*a)*x^(7/2)+4/9*a^2*b*(3*A*b+2*B*a)*x^(9 /2)+4/11*a*b^2*(2*A*b+3*B*a)*x^(11/2)+2/13*b^3*(A*b+4*B*a)*x^(13/2)+2/15*b ^4*B*x^(15/2)
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2 x^{5/2} \left (1287 a^4 (7 A+5 B x)+2860 a^3 b x (9 A+7 B x)+2730 a^2 b^2 x^2 (11 A+9 B x)+1260 a b^3 x^3 (13 A+11 B x)+231 b^4 x^4 (15 A+13 B x)\right )}{45045} \] Input:
Integrate[x^(3/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(2*x^(5/2)*(1287*a^4*(7*A + 5*B*x) + 2860*a^3*b*x*(9*A + 7*B*x) + 2730*a^2 *b^2*x^2*(11*A + 9*B*x) + 1260*a*b^3*x^3*(13*A + 11*B*x) + 231*b^4*x^4*(15 *A + 13*B*x)))/45045
Time = 0.42 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 27, 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.
\(\displaystyle \int x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x) \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int b^4 x^{3/2} (a+b x)^4 (A+B x)dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int x^{3/2} (a+b x)^4 (A+B x)dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (a^4 A x^{3/2}+a^3 x^{5/2} (a B+4 A b)+2 a^2 b x^{7/2} (2 a B+3 A b)+b^3 x^{11/2} (4 a B+A b)+2 a b^2 x^{9/2} (3 a B+2 A b)+b^4 B x^{13/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} a^4 A x^{5/2}+\frac {2}{7} a^3 x^{7/2} (a B+4 A b)+\frac {4}{9} a^2 b x^{9/2} (2 a B+3 A b)+\frac {2}{13} b^3 x^{13/2} (4 a B+A b)+\frac {4}{11} a b^2 x^{11/2} (3 a B+2 A b)+\frac {2}{15} b^4 B x^{15/2}\) |
Input:
Int[x^(3/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(2*a^4*A*x^(5/2))/5 + (2*a^3*(4*A*b + a*B)*x^(7/2))/7 + (4*a^2*b*(3*A*b + 2*a*B)*x^(9/2))/9 + (4*a*b^2*(2*A*b + 3*a*B)*x^(11/2))/11 + (2*b^3*(A*b + 4*a*B)*x^(13/2))/13 + (2*b^4*B*x^(15/2))/15
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.96 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.90
method | result | size |
gosper | \(\frac {2 x^{\frac {5}{2}} \left (3003 b^{4} B \,x^{5}+3465 A \,b^{4} x^{4}+13860 B a \,b^{3} x^{4}+16380 A a \,b^{3} x^{3}+24570 B \,a^{2} b^{2} x^{3}+30030 A \,a^{2} b^{2} x^{2}+20020 B \,a^{3} b \,x^{2}+25740 A \,a^{3} b x +6435 a^{4} B x +9009 a^{4} A \right )}{45045}\) | \(100\) |
derivativedivides | \(\frac {2 b^{4} B \,x^{\frac {15}{2}}}{15}+\frac {2 \left (A \,b^{4}+4 B a \,b^{3}\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (6 a^{2} A \,b^{2}+4 B \,a^{3} b \right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (4 A \,a^{3} b +a^{4} B \right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{4} A \,x^{\frac {5}{2}}}{5}\) | \(100\) |
default | \(\frac {2 b^{4} B \,x^{\frac {15}{2}}}{15}+\frac {2 \left (A \,b^{4}+4 B a \,b^{3}\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (6 a^{2} A \,b^{2}+4 B \,a^{3} b \right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (4 A \,a^{3} b +a^{4} B \right ) x^{\frac {7}{2}}}{7}+\frac {2 a^{4} A \,x^{\frac {5}{2}}}{5}\) | \(100\) |
trager | \(\frac {2 x^{\frac {5}{2}} \left (3003 b^{4} B \,x^{5}+3465 A \,b^{4} x^{4}+13860 B a \,b^{3} x^{4}+16380 A a \,b^{3} x^{3}+24570 B \,a^{2} b^{2} x^{3}+30030 A \,a^{2} b^{2} x^{2}+20020 B \,a^{3} b \,x^{2}+25740 A \,a^{3} b x +6435 a^{4} B x +9009 a^{4} A \right )}{45045}\) | \(100\) |
risch | \(\frac {2 x^{\frac {5}{2}} \left (3003 b^{4} B \,x^{5}+3465 A \,b^{4} x^{4}+13860 B a \,b^{3} x^{4}+16380 A a \,b^{3} x^{3}+24570 B \,a^{2} b^{2} x^{3}+30030 A \,a^{2} b^{2} x^{2}+20020 B \,a^{3} b \,x^{2}+25740 A \,a^{3} b x +6435 a^{4} B x +9009 a^{4} A \right )}{45045}\) | \(100\) |
orering | \(\frac {2 \left (3003 b^{4} B \,x^{5}+3465 A \,b^{4} x^{4}+13860 B a \,b^{3} x^{4}+16380 A a \,b^{3} x^{3}+24570 B \,a^{2} b^{2} x^{3}+30030 A \,a^{2} b^{2} x^{2}+20020 B \,a^{3} b \,x^{2}+25740 A \,a^{3} b x +6435 a^{4} B x +9009 a^{4} A \right ) x^{\frac {5}{2}} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}{45045 \left (b x +a \right )^{4}}\) | \(125\) |
Input:
int(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
2/45045*x^(5/2)*(3003*B*b^4*x^5+3465*A*b^4*x^4+13860*B*a*b^3*x^4+16380*A*a *b^3*x^3+24570*B*a^2*b^2*x^3+30030*A*a^2*b^2*x^2+20020*B*a^3*b*x^2+25740*A *a^3*b*x+6435*B*a^4*x+9009*A*a^4)
Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.94 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{45045} \, {\left (3003 \, B b^{4} x^{7} + 9009 \, A a^{4} x^{2} + 3465 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{6} + 8190 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{5} + 10010 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} + 6435 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{3}\right )} \sqrt {x} \] Input:
integrate(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
2/45045*(3003*B*b^4*x^7 + 9009*A*a^4*x^2 + 3465*(4*B*a*b^3 + A*b^4)*x^6 + 8190*(3*B*a^2*b^2 + 2*A*a*b^3)*x^5 + 10010*(2*B*a^3*b + 3*A*a^2*b^2)*x^4 + 6435*(B*a^4 + 4*A*a^3*b)*x^3)*sqrt(x)
Time = 0.32 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.33 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2 A a^{4} x^{\frac {5}{2}}}{5} + \frac {8 A a^{3} b x^{\frac {7}{2}}}{7} + \frac {4 A a^{2} b^{2} x^{\frac {9}{2}}}{3} + \frac {8 A a b^{3} x^{\frac {11}{2}}}{11} + \frac {2 A b^{4} x^{\frac {13}{2}}}{13} + \frac {2 B a^{4} x^{\frac {7}{2}}}{7} + \frac {8 B a^{3} b x^{\frac {9}{2}}}{9} + \frac {12 B a^{2} b^{2} x^{\frac {11}{2}}}{11} + \frac {8 B a b^{3} x^{\frac {13}{2}}}{13} + \frac {2 B b^{4} x^{\frac {15}{2}}}{15} \] Input:
integrate(x**(3/2)*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
2*A*a**4*x**(5/2)/5 + 8*A*a**3*b*x**(7/2)/7 + 4*A*a**2*b**2*x**(9/2)/3 + 8 *A*a*b**3*x**(11/2)/11 + 2*A*b**4*x**(13/2)/13 + 2*B*a**4*x**(7/2)/7 + 8*B *a**3*b*x**(9/2)/9 + 12*B*a**2*b**2*x**(11/2)/11 + 8*B*a*b**3*x**(13/2)/13 + 2*B*b**4*x**(15/2)/15
Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.89 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{15} \, B b^{4} x^{\frac {15}{2}} + \frac {2}{5} \, A a^{4} x^{\frac {5}{2}} + \frac {2}{13} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{\frac {13}{2}} + \frac {4}{11} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{\frac {11}{2}} + \frac {4}{9} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{\frac {7}{2}} \] Input:
integrate(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
2/15*B*b^4*x^(15/2) + 2/5*A*a^4*x^(5/2) + 2/13*(4*B*a*b^3 + A*b^4)*x^(13/2 ) + 4/11*(3*B*a^2*b^2 + 2*A*a*b^3)*x^(11/2) + 4/9*(2*B*a^3*b + 3*A*a^2*b^2 )*x^(9/2) + 2/7*(B*a^4 + 4*A*a^3*b)*x^(7/2)
Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.91 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2}{15} \, B b^{4} x^{\frac {15}{2}} + \frac {8}{13} \, B a b^{3} x^{\frac {13}{2}} + \frac {2}{13} \, A b^{4} x^{\frac {13}{2}} + \frac {12}{11} \, B a^{2} b^{2} x^{\frac {11}{2}} + \frac {8}{11} \, A a b^{3} x^{\frac {11}{2}} + \frac {8}{9} \, B a^{3} b x^{\frac {9}{2}} + \frac {4}{3} \, A a^{2} b^{2} x^{\frac {9}{2}} + \frac {2}{7} \, B a^{4} x^{\frac {7}{2}} + \frac {8}{7} \, A a^{3} b x^{\frac {7}{2}} + \frac {2}{5} \, A a^{4} x^{\frac {5}{2}} \] Input:
integrate(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
2/15*B*b^4*x^(15/2) + 8/13*B*a*b^3*x^(13/2) + 2/13*A*b^4*x^(13/2) + 12/11* B*a^2*b^2*x^(11/2) + 8/11*A*a*b^3*x^(11/2) + 8/9*B*a^3*b*x^(9/2) + 4/3*A*a ^2*b^2*x^(9/2) + 2/7*B*a^4*x^(7/2) + 8/7*A*a^3*b*x^(7/2) + 2/5*A*a^4*x^(5/ 2)
Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=x^{7/2}\,\left (\frac {2\,B\,a^4}{7}+\frac {8\,A\,b\,a^3}{7}\right )+x^{13/2}\,\left (\frac {2\,A\,b^4}{13}+\frac {8\,B\,a\,b^3}{13}\right )+\frac {2\,A\,a^4\,x^{5/2}}{5}+\frac {2\,B\,b^4\,x^{15/2}}{15}+\frac {4\,a^2\,b\,x^{9/2}\,\left (3\,A\,b+2\,B\,a\right )}{9}+\frac {4\,a\,b^2\,x^{11/2}\,\left (2\,A\,b+3\,B\,a\right )}{11} \] Input:
int(x^(3/2)*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
x^(7/2)*((2*B*a^4)/7 + (8*A*a^3*b)/7) + x^(13/2)*((2*A*b^4)/13 + (8*B*a*b^ 3)/13) + (2*A*a^4*x^(5/2))/5 + (2*B*b^4*x^(15/2))/15 + (4*a^2*b*x^(9/2)*(3 *A*b + 2*B*a))/9 + (4*a*b^2*x^(11/2)*(2*A*b + 3*B*a))/11
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int x^{3/2} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {2 \sqrt {x}\, x^{2} \left (3003 b^{5} x^{5}+17325 a \,b^{4} x^{4}+40950 a^{2} b^{3} x^{3}+50050 a^{3} b^{2} x^{2}+32175 a^{4} b x +9009 a^{5}\right )}{45045} \] Input:
int(x^(3/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
(2*sqrt(x)*x**2*(9009*a**5 + 32175*a**4*b*x + 50050*a**3*b**2*x**2 + 40950 *a**2*b**3*x**3 + 17325*a*b**4*x**4 + 3003*b**5*x**5))/45045