Integrand size = 23, antiderivative size = 174 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{9/2}} \, dx=-\frac {2 a^3 A}{7 x^{7/2}}-\frac {2 a^2 (3 A b+a B)}{5 x^{5/2}}-\frac {2 a \left (a b B+A \left (b^2+a c\right )\right )}{x^{3/2}}-\frac {2 \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right )}{\sqrt {x}}+2 \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) \sqrt {x}+2 c \left (b^2 B+A b c+a B c\right ) x^{3/2}+\frac {2}{5} c^2 (3 b B+A c) x^{5/2}+\frac {2}{7} B c^3 x^{7/2} \]
-2/7*a^3*A/x^(7/2)-2/5*a^2*(3*A*b+B*a)/x^(5/2)-2*a*(a*b*B+A*(a*c+b^2))/x^( 3/2)+2*c*(A*b*c+B*a*c+B*b^2)*x^(3/2)+2/5*c^2*(A*c+3*B*b)*x^(5/2)+2/7*B*c^3 *x^(7/2)-2*(3*a*B*(a*c+b^2)+A*(6*a*b*c+b^3))/x^(1/2)+2*(3*A*a*c^2+3*A*b^2* c+6*B*a*b*c+B*b^3)*x^(1/2)
Time = 0.18 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{9/2}} \, dx=\frac {2 \left (-a^3 (5 A+7 B x)-7 a^2 x (5 B x (b+3 c x)+A (3 b+5 c x))-35 a x^2 \left (A \left (b^2+6 b c x-3 c^2 x^2\right )-B x \left (-3 b^2+6 b c x+c^2 x^2\right )\right )+x^3 \left (7 A \left (-5 b^3+15 b^2 c x+5 b c^2 x^2+c^3 x^3\right )+B x \left (35 b^3+35 b^2 c x+21 b c^2 x^2+5 c^3 x^3\right )\right )\right )}{35 x^{7/2}} \]
(2*(-(a^3*(5*A + 7*B*x)) - 7*a^2*x*(5*B*x*(b + 3*c*x) + A*(3*b + 5*c*x)) - 35*a*x^2*(A*(b^2 + 6*b*c*x - 3*c^2*x^2) - B*x*(-3*b^2 + 6*b*c*x + c^2*x^2 )) + x^3*(7*A*(-5*b^3 + 15*b^2*c*x + 5*b*c^2*x^2 + c^3*x^3) + B*x*(35*b^3 + 35*b^2*c*x + 21*b*c^2*x^2 + 5*c^3*x^3))))/(35*x^(7/2))
Time = 0.32 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1195, 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.
| \(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{9/2}} \, dx\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \int \left (\frac {a^3 A}{x^{9/2}}+\frac {a^2 (a B+3 A b)}{x^{7/2}}+\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x^{5/2}}+3 c \sqrt {x} \left (a B c+A b c+b^2 B\right )+\frac {3 a A c^2+6 a b B c+3 A b^2 c+b^3 B}{\sqrt {x}}+\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{x^{3/2}}+c^2 x^{3/2} (A c+3 b B)+B c^3 x^{5/2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^3 A}{7 x^{7/2}}-\frac {2 a^2 (a B+3 A b)}{5 x^{5/2}}+2 c x^{3/2} \left (a B c+A b c+b^2 B\right )-\frac {2 a \left (A \left (a c+b^2\right )+a b B\right )}{x^{3/2}}+2 \sqrt {x} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )-\frac {2 \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )}{\sqrt {x}}+\frac {2}{5} c^2 x^{5/2} (A c+3 b B)+\frac {2}{7} B c^3 x^{7/2}\) |
(-2*a^3*A)/(7*x^(7/2)) - (2*a^2*(3*A*b + a*B))/(5*x^(5/2)) - (2*a*(a*b*B + A*(b^2 + a*c)))/x^(3/2) - (2*(3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c)))/Sqr t[x] + 2*(b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*Sqrt[x] + 2*c*(b^2*B + A*b*c + a*B*c)*x^(3/2) + (2*c^2*(3*b*B + A*c)*x^(5/2))/5 + (2*B*c^3*x^(7 /2))/7
3.11.7.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00
| method | result | size |
| derivativedivides | \(\frac {2 B \,c^{3} x^{\frac {7}{2}}}{7}+\frac {2 A \,c^{3} x^{\frac {5}{2}}}{5}+\frac {6 x^{\frac {5}{2}} B b \,c^{2}}{5}+2 x^{\frac {3}{2}} A b \,c^{2}+2 a B \,c^{2} x^{\frac {3}{2}}+2 x^{\frac {3}{2}} B \,b^{2} c +6 a A \,c^{2} \sqrt {x}+6 \sqrt {x}\, A \,b^{2} c +12 B a b c \sqrt {x}+2 \sqrt {x}\, B \,b^{3}-\frac {2 a^{2} \left (3 A b +B a \right )}{5 x^{\frac {5}{2}}}-\frac {2 a^{3} A}{7 x^{\frac {7}{2}}}-\frac {2 a \left (A a c +A \,b^{2}+a b B \right )}{x^{\frac {3}{2}}}-\frac {2 \left (6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}\right )}{\sqrt {x}}\) | \(174\) |
| default | \(\frac {2 B \,c^{3} x^{\frac {7}{2}}}{7}+\frac {2 A \,c^{3} x^{\frac {5}{2}}}{5}+\frac {6 x^{\frac {5}{2}} B b \,c^{2}}{5}+2 x^{\frac {3}{2}} A b \,c^{2}+2 a B \,c^{2} x^{\frac {3}{2}}+2 x^{\frac {3}{2}} B \,b^{2} c +6 a A \,c^{2} \sqrt {x}+6 \sqrt {x}\, A \,b^{2} c +12 B a b c \sqrt {x}+2 \sqrt {x}\, B \,b^{3}-\frac {2 a^{2} \left (3 A b +B a \right )}{5 x^{\frac {5}{2}}}-\frac {2 a^{3} A}{7 x^{\frac {7}{2}}}-\frac {2 a \left (A a c +A \,b^{2}+a b B \right )}{x^{\frac {3}{2}}}-\frac {2 \left (6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}\right )}{\sqrt {x}}\) | \(174\) |
| gosper | \(-\frac {2 \left (-5 B \,c^{3} x^{7}-7 A \,c^{3} x^{6}-21 B b \,c^{2} x^{6}-35 A b \,c^{2} x^{5}-35 a B \,c^{2} x^{5}-35 B \,b^{2} c \,x^{5}-105 a A \,c^{2} x^{4}-105 A \,b^{2} c \,x^{4}-210 B a b c \,x^{4}-35 x^{4} B \,b^{3}+210 A a b c \,x^{3}+35 A \,b^{3} x^{3}+105 a^{2} B c \,x^{3}+105 B a \,b^{2} x^{3}+35 a^{2} A c \,x^{2}+35 A a \,b^{2} x^{2}+35 B \,a^{2} b \,x^{2}+21 A \,a^{2} b x +7 a^{3} B x +5 A \,a^{3}\right )}{35 x^{\frac {7}{2}}}\) | \(192\) |
| trager | \(-\frac {2 \left (-5 B \,c^{3} x^{7}-7 A \,c^{3} x^{6}-21 B b \,c^{2} x^{6}-35 A b \,c^{2} x^{5}-35 a B \,c^{2} x^{5}-35 B \,b^{2} c \,x^{5}-105 a A \,c^{2} x^{4}-105 A \,b^{2} c \,x^{4}-210 B a b c \,x^{4}-35 x^{4} B \,b^{3}+210 A a b c \,x^{3}+35 A \,b^{3} x^{3}+105 a^{2} B c \,x^{3}+105 B a \,b^{2} x^{3}+35 a^{2} A c \,x^{2}+35 A a \,b^{2} x^{2}+35 B \,a^{2} b \,x^{2}+21 A \,a^{2} b x +7 a^{3} B x +5 A \,a^{3}\right )}{35 x^{\frac {7}{2}}}\) | \(192\) |
| risch | \(-\frac {2 \left (-5 B \,c^{3} x^{7}-7 A \,c^{3} x^{6}-21 B b \,c^{2} x^{6}-35 A b \,c^{2} x^{5}-35 a B \,c^{2} x^{5}-35 B \,b^{2} c \,x^{5}-105 a A \,c^{2} x^{4}-105 A \,b^{2} c \,x^{4}-210 B a b c \,x^{4}-35 x^{4} B \,b^{3}+210 A a b c \,x^{3}+35 A \,b^{3} x^{3}+105 a^{2} B c \,x^{3}+105 B a \,b^{2} x^{3}+35 a^{2} A c \,x^{2}+35 A a \,b^{2} x^{2}+35 B \,a^{2} b \,x^{2}+21 A \,a^{2} b x +7 a^{3} B x +5 A \,a^{3}\right )}{35 x^{\frac {7}{2}}}\) | \(192\) |
2/7*B*c^3*x^(7/2)+2/5*A*c^3*x^(5/2)+6/5*x^(5/2)*B*b*c^2+2*x^(3/2)*A*b*c^2+ 2*a*B*c^2*x^(3/2)+2*x^(3/2)*B*b^2*c+6*a*A*c^2*x^(1/2)+6*x^(1/2)*A*b^2*c+12 *B*a*b*c*x^(1/2)+2*x^(1/2)*B*b^3-2/5*a^2*(3*A*b+B*a)/x^(5/2)-2/7*a^3*A/x^( 7/2)-2*a*(A*a*c+A*b^2+B*a*b)/x^(3/2)-2*(6*A*a*b*c+A*b^3+3*B*a^2*c+3*B*a*b^ 2)/x^(1/2)
Time = 0.32 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{9/2}} \, dx=\frac {2 \, {\left (5 \, B c^{3} x^{7} + 7 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 35 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 35 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} - 5 \, A a^{3} - 35 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} - 35 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 7 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{35 \, x^{\frac {7}{2}}} \]
2/35*(5*B*c^3*x^7 + 7*(3*B*b*c^2 + A*c^3)*x^6 + 35*(B*b^2*c + (B*a + A*b)* c^2)*x^5 + 35*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 - 5*A*a^3 - 35*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 - 35*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 - 7*(B*a^3 + 3*A*a^2*b)*x)/x^(7/2)
Time = 0.65 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.55 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{9/2}} \, dx=- \frac {2 A a^{3}}{7 x^{\frac {7}{2}}} - \frac {6 A a^{2} b}{5 x^{\frac {5}{2}}} - \frac {2 A a^{2} c}{x^{\frac {3}{2}}} - \frac {2 A a b^{2}}{x^{\frac {3}{2}}} - \frac {12 A a b c}{\sqrt {x}} + 6 A a c^{2} \sqrt {x} - \frac {2 A b^{3}}{\sqrt {x}} + 6 A b^{2} c \sqrt {x} + 2 A b c^{2} x^{\frac {3}{2}} + \frac {2 A c^{3} x^{\frac {5}{2}}}{5} - \frac {2 B a^{3}}{5 x^{\frac {5}{2}}} - \frac {2 B a^{2} b}{x^{\frac {3}{2}}} - \frac {6 B a^{2} c}{\sqrt {x}} - \frac {6 B a b^{2}}{\sqrt {x}} + 12 B a b c \sqrt {x} + 2 B a c^{2} x^{\frac {3}{2}} + 2 B b^{3} \sqrt {x} + 2 B b^{2} c x^{\frac {3}{2}} + \frac {6 B b c^{2} x^{\frac {5}{2}}}{5} + \frac {2 B c^{3} x^{\frac {7}{2}}}{7} \]
-2*A*a**3/(7*x**(7/2)) - 6*A*a**2*b/(5*x**(5/2)) - 2*A*a**2*c/x**(3/2) - 2 *A*a*b**2/x**(3/2) - 12*A*a*b*c/sqrt(x) + 6*A*a*c**2*sqrt(x) - 2*A*b**3/sq rt(x) + 6*A*b**2*c*sqrt(x) + 2*A*b*c**2*x**(3/2) + 2*A*c**3*x**(5/2)/5 - 2 *B*a**3/(5*x**(5/2)) - 2*B*a**2*b/x**(3/2) - 6*B*a**2*c/sqrt(x) - 6*B*a*b* *2/sqrt(x) + 12*B*a*b*c*sqrt(x) + 2*B*a*c**2*x**(3/2) + 2*B*b**3*sqrt(x) + 2*B*b**2*c*x**(3/2) + 6*B*b*c**2*x**(5/2)/5 + 2*B*c**3*x**(7/2)/7
Time = 0.20 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{9/2}} \, dx=\frac {2}{7} \, B c^{3} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {5}{2}} + 2 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{\frac {3}{2}} + 2 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {x} - \frac {2 \, {\left (5 \, A a^{3} + 35 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 35 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 7 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{35 \, x^{\frac {7}{2}}} \]
2/7*B*c^3*x^(7/2) + 2/5*(3*B*b*c^2 + A*c^3)*x^(5/2) + 2*(B*b^2*c + (B*a + A*b)*c^2)*x^(3/2) + 2*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*sqrt(x) - 2/35*(5*A*a^3 + 35*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 + 35* (B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 7*(B*a^3 + 3*A*a^2*b)*x)/x^(7/2)
Time = 0.29 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{9/2}} \, dx=\frac {2}{7} \, B c^{3} x^{\frac {7}{2}} + \frac {6}{5} \, B b c^{2} x^{\frac {5}{2}} + \frac {2}{5} \, A c^{3} x^{\frac {5}{2}} + 2 \, B b^{2} c x^{\frac {3}{2}} + 2 \, B a c^{2} x^{\frac {3}{2}} + 2 \, A b c^{2} x^{\frac {3}{2}} + 2 \, B b^{3} \sqrt {x} + 12 \, B a b c \sqrt {x} + 6 \, A b^{2} c \sqrt {x} + 6 \, A a c^{2} \sqrt {x} - \frac {2 \, {\left (105 \, B a b^{2} x^{3} + 35 \, A b^{3} x^{3} + 105 \, B a^{2} c x^{3} + 210 \, A a b c x^{3} + 35 \, B a^{2} b x^{2} + 35 \, A a b^{2} x^{2} + 35 \, A a^{2} c x^{2} + 7 \, B a^{3} x + 21 \, A a^{2} b x + 5 \, A a^{3}\right )}}{35 \, x^{\frac {7}{2}}} \]
2/7*B*c^3*x^(7/2) + 6/5*B*b*c^2*x^(5/2) + 2/5*A*c^3*x^(5/2) + 2*B*b^2*c*x^ (3/2) + 2*B*a*c^2*x^(3/2) + 2*A*b*c^2*x^(3/2) + 2*B*b^3*sqrt(x) + 12*B*a*b *c*sqrt(x) + 6*A*b^2*c*sqrt(x) + 6*A*a*c^2*sqrt(x) - 2/35*(105*B*a*b^2*x^3 + 35*A*b^3*x^3 + 105*B*a^2*c*x^3 + 210*A*a*b*c*x^3 + 35*B*a^2*b*x^2 + 35* A*a*b^2*x^2 + 35*A*a^2*c*x^2 + 7*B*a^3*x + 21*A*a^2*b*x + 5*A*a^3)/x^(7/2)
Time = 0.06 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{9/2}} \, dx=\sqrt {x}\,\left (2\,B\,b^3+6\,A\,b^2\,c+12\,B\,a\,b\,c+6\,A\,a\,c^2\right )-\frac {x^3\,\left (6\,B\,c\,a^2+6\,B\,a\,b^2+12\,A\,c\,a\,b+2\,A\,b^3\right )+x\,\left (\frac {2\,B\,a^3}{5}+\frac {6\,A\,b\,a^2}{5}\right )+\frac {2\,A\,a^3}{7}+x^2\,\left (2\,B\,a^2\,b+2\,A\,c\,a^2+2\,A\,a\,b^2\right )}{x^{7/2}}+x^{5/2}\,\left (\frac {2\,A\,c^3}{5}+\frac {6\,B\,b\,c^2}{5}\right )+x^{3/2}\,\left (2\,B\,b^2\,c+2\,A\,b\,c^2+2\,B\,a\,c^2\right )+\frac {2\,B\,c^3\,x^{7/2}}{7} \]
x^(1/2)*(2*B*b^3 + 6*A*a*c^2 + 6*A*b^2*c + 12*B*a*b*c) - (x^3*(2*A*b^3 + 6 *B*a*b^2 + 6*B*a^2*c + 12*A*a*b*c) + x*((2*B*a^3)/5 + (6*A*a^2*b)/5) + (2* A*a^3)/7 + x^2*(2*A*a*b^2 + 2*A*a^2*c + 2*B*a^2*b))/x^(7/2) + x^(5/2)*((2* A*c^3)/5 + (6*B*b*c^2)/5) + x^(3/2)*(2*A*b*c^2 + 2*B*a*c^2 + 2*B*b^2*c) + (2*B*c^3*x^(7/2))/7