Integrand size = 23, antiderivative size = 109 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{3/2}} \, dx=-\frac {2 a^2 A}{\sqrt {x}}+2 a (2 A b+a B) \sqrt {x}+\frac {2}{3} \left (2 a b B+A \left (b^2+2 a c\right )\right ) x^{3/2}+\frac {2}{5} \left (b^2 B+2 A b c+2 a B c\right ) x^{5/2}+\frac {2}{7} c (2 b B+A c) x^{7/2}+\frac {2}{9} B c^2 x^{9/2} \]
2/3*(2*a*b*B+A*(2*a*c+b^2))*x^(3/2)+2/5*(2*A*b*c+2*B*a*c+B*b^2)*x^(5/2)+2/ 7*c*(A*c+2*B*b)*x^(7/2)+2/9*B*c^2*x^(9/2)-2*a^2*A/x^(1/2)+2*a*(2*A*b+B*a)* x^(1/2)
Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{3/2}} \, dx=\frac {-630 a^2 (A-B x)+84 a x (5 A (3 b+c x)+B x (5 b+3 c x))+2 x^2 \left (3 A \left (35 b^2+42 b c x+15 c^2 x^2\right )+B x \left (63 b^2+90 b c x+35 c^2 x^2\right )\right )}{315 \sqrt {x}} \]
(-630*a^2*(A - B*x) + 84*a*x*(5*A*(3*b + c*x) + B*x*(5*b + 3*c*x)) + 2*x^2 *(3*A*(35*b^2 + 42*b*c*x + 15*c^2*x^2) + B*x*(63*b^2 + 90*b*c*x + 35*c^2*x ^2)))/(315*Sqrt[x])
Time = 0.25 (sec) , antiderivative size = 109, 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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {a^2 A}{x^{3/2}}+x^{3/2} \left (2 a B c+2 A b c+b^2 B\right )+\sqrt {x} \left (A \left (2 a c+b^2\right )+2 a b B\right )+\frac {a (a B+2 A b)}{\sqrt {x}}+c x^{5/2} (A c+2 b B)+B c^2 x^{7/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 A}{\sqrt {x}}+\frac {2}{5} x^{5/2} \left (2 a B c+2 A b c+b^2 B\right )+\frac {2}{3} x^{3/2} \left (A \left (2 a c+b^2\right )+2 a b B\right )+2 a \sqrt {x} (a B+2 A b)+\frac {2}{7} c x^{7/2} (A c+2 b B)+\frac {2}{9} B c^2 x^{9/2}\) |
(-2*a^2*A)/Sqrt[x] + 2*a*(2*A*b + a*B)*Sqrt[x] + (2*(2*a*b*B + A*(b^2 + 2* a*c))*x^(3/2))/3 + (2*(b^2*B + 2*A*b*c + 2*a*B*c)*x^(5/2))/5 + (2*c*(2*b*B + A*c)*x^(7/2))/7 + (2*B*c^2*x^(9/2))/9
3.10.95.3.1 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 = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94
method | result | size |
gosper | \(-\frac {2 \left (-35 B \,c^{2} x^{5}-45 A \,c^{2} x^{4}-90 x^{4} B b c -126 x^{3} A b c -126 a B c \,x^{3}-63 B \,b^{2} x^{3}-210 a A c \,x^{2}-105 A \,b^{2} x^{2}-210 B a b \,x^{2}-630 a A b x -315 a^{2} B x +315 A \,a^{2}\right )}{315 \sqrt {x}}\) | \(102\) |
trager | \(-\frac {2 \left (-35 B \,c^{2} x^{5}-45 A \,c^{2} x^{4}-90 x^{4} B b c -126 x^{3} A b c -126 a B c \,x^{3}-63 B \,b^{2} x^{3}-210 a A c \,x^{2}-105 A \,b^{2} x^{2}-210 B a b \,x^{2}-630 a A b x -315 a^{2} B x +315 A \,a^{2}\right )}{315 \sqrt {x}}\) | \(102\) |
risch | \(-\frac {2 \left (-35 B \,c^{2} x^{5}-45 A \,c^{2} x^{4}-90 x^{4} B b c -126 x^{3} A b c -126 a B c \,x^{3}-63 B \,b^{2} x^{3}-210 a A c \,x^{2}-105 A \,b^{2} x^{2}-210 B a b \,x^{2}-630 a A b x -315 a^{2} B x +315 A \,a^{2}\right )}{315 \sqrt {x}}\) | \(102\) |
derivativedivides | \(\frac {2 B \,c^{2} x^{\frac {9}{2}}}{9}+\frac {2 A \,c^{2} x^{\frac {7}{2}}}{7}+\frac {4 B b c \,x^{\frac {7}{2}}}{7}+\frac {4 A b c \,x^{\frac {5}{2}}}{5}+\frac {4 a B c \,x^{\frac {5}{2}}}{5}+\frac {2 b^{2} B \,x^{\frac {5}{2}}}{5}+\frac {4 a A c \,x^{\frac {3}{2}}}{3}+\frac {2 A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {4 B a b \,x^{\frac {3}{2}}}{3}+4 A b a \sqrt {x}+2 a^{2} B \sqrt {x}-\frac {2 a^{2} A}{\sqrt {x}}\) | \(104\) |
default | \(\frac {2 B \,c^{2} x^{\frac {9}{2}}}{9}+\frac {2 A \,c^{2} x^{\frac {7}{2}}}{7}+\frac {4 B b c \,x^{\frac {7}{2}}}{7}+\frac {4 A b c \,x^{\frac {5}{2}}}{5}+\frac {4 a B c \,x^{\frac {5}{2}}}{5}+\frac {2 b^{2} B \,x^{\frac {5}{2}}}{5}+\frac {4 a A c \,x^{\frac {3}{2}}}{3}+\frac {2 A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {4 B a b \,x^{\frac {3}{2}}}{3}+4 A b a \sqrt {x}+2 a^{2} B \sqrt {x}-\frac {2 a^{2} A}{\sqrt {x}}\) | \(104\) |
-2/315*(-35*B*c^2*x^5-45*A*c^2*x^4-90*B*b*c*x^4-126*A*b*c*x^3-126*B*a*c*x^ 3-63*B*b^2*x^3-210*A*a*c*x^2-105*A*b^2*x^2-210*B*a*b*x^2-630*A*a*b*x-315*B *a^2*x+315*A*a^2)/x^(1/2)
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{3/2}} \, dx=\frac {2 \, {\left (35 \, B c^{2} x^{5} + 45 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 63 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} - 315 \, A a^{2} + 105 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 315 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{315 \, \sqrt {x}} \]
2/315*(35*B*c^2*x^5 + 45*(2*B*b*c + A*c^2)*x^4 + 63*(B*b^2 + 2*(B*a + A*b) *c)*x^3 - 315*A*a^2 + 105*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 + 315*(B*a^2 + 2 *A*a*b)*x)/sqrt(x)
Time = 0.27 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{3/2}} \, dx=- \frac {2 A a^{2}}{\sqrt {x}} + 4 A a b \sqrt {x} + \frac {4 A a c x^{\frac {3}{2}}}{3} + \frac {2 A b^{2} x^{\frac {3}{2}}}{3} + \frac {4 A b c x^{\frac {5}{2}}}{5} + \frac {2 A c^{2} x^{\frac {7}{2}}}{7} + 2 B a^{2} \sqrt {x} + \frac {4 B a b x^{\frac {3}{2}}}{3} + \frac {4 B a c x^{\frac {5}{2}}}{5} + \frac {2 B b^{2} x^{\frac {5}{2}}}{5} + \frac {4 B b c x^{\frac {7}{2}}}{7} + \frac {2 B c^{2} x^{\frac {9}{2}}}{9} \]
-2*A*a**2/sqrt(x) + 4*A*a*b*sqrt(x) + 4*A*a*c*x**(3/2)/3 + 2*A*b**2*x**(3/ 2)/3 + 4*A*b*c*x**(5/2)/5 + 2*A*c**2*x**(7/2)/7 + 2*B*a**2*sqrt(x) + 4*B*a *b*x**(3/2)/3 + 4*B*a*c*x**(5/2)/5 + 2*B*b**2*x**(5/2)/5 + 4*B*b*c*x**(7/2 )/7 + 2*B*c**2*x**(9/2)/9
Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{3/2}} \, dx=\frac {2}{9} \, B c^{2} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{\frac {5}{2}} - \frac {2 \, A a^{2}}{\sqrt {x}} + \frac {2}{3} \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{\frac {3}{2}} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \sqrt {x} \]
2/9*B*c^2*x^(9/2) + 2/7*(2*B*b*c + A*c^2)*x^(7/2) + 2/5*(B*b^2 + 2*(B*a + A*b)*c)*x^(5/2) - 2*A*a^2/sqrt(x) + 2/3*(2*B*a*b + A*b^2 + 2*A*a*c)*x^(3/2 ) + 2*(B*a^2 + 2*A*a*b)*sqrt(x)
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{3/2}} \, dx=\frac {2}{9} \, B c^{2} x^{\frac {9}{2}} + \frac {4}{7} \, B b c x^{\frac {7}{2}} + \frac {2}{7} \, A c^{2} x^{\frac {7}{2}} + \frac {2}{5} \, B b^{2} x^{\frac {5}{2}} + \frac {4}{5} \, B a c x^{\frac {5}{2}} + \frac {4}{5} \, A b c x^{\frac {5}{2}} + \frac {4}{3} \, B a b x^{\frac {3}{2}} + \frac {2}{3} \, A b^{2} x^{\frac {3}{2}} + \frac {4}{3} \, A a c x^{\frac {3}{2}} + 2 \, B a^{2} \sqrt {x} + 4 \, A a b \sqrt {x} - \frac {2 \, A a^{2}}{\sqrt {x}} \]
2/9*B*c^2*x^(9/2) + 4/7*B*b*c*x^(7/2) + 2/7*A*c^2*x^(7/2) + 2/5*B*b^2*x^(5 /2) + 4/5*B*a*c*x^(5/2) + 4/5*A*b*c*x^(5/2) + 4/3*B*a*b*x^(3/2) + 2/3*A*b^ 2*x^(3/2) + 4/3*A*a*c*x^(3/2) + 2*B*a^2*sqrt(x) + 4*A*a*b*sqrt(x) - 2*A*a^ 2/sqrt(x)
Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{3/2}} \, dx=\sqrt {x}\,\left (2\,B\,a^2+4\,A\,b\,a\right )+x^{7/2}\,\left (\frac {2\,A\,c^2}{7}+\frac {4\,B\,b\,c}{7}\right )+x^{3/2}\,\left (\frac {2\,A\,b^2}{3}+\frac {4\,B\,a\,b}{3}+\frac {4\,A\,a\,c}{3}\right )+x^{5/2}\,\left (\frac {2\,B\,b^2}{5}+\frac {4\,A\,c\,b}{5}+\frac {4\,B\,a\,c}{5}\right )-\frac {2\,A\,a^2}{\sqrt {x}}+\frac {2\,B\,c^2\,x^{9/2}}{9} \]