Integrand size = 23, antiderivative size = 109 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{5/2}} \, dx=-\frac {2 a^2 A}{3 x^{3/2}}-\frac {2 a (2 A b+a B)}{\sqrt {x}}+2 \left (2 a b B+A \left (b^2+2 a c\right )\right ) \sqrt {x}+\frac {2}{3} \left (b^2 B+2 A b c+2 a B c\right ) x^{3/2}+\frac {2}{5} c (2 b B+A c) x^{5/2}+\frac {2}{7} B c^2 x^{7/2} \]
-2/3*a^2*A/x^(3/2)+2/3*(2*A*b*c+2*B*a*c+B*b^2)*x^(3/2)+2/5*c*(A*c+2*B*b)*x ^(5/2)+2/7*B*c^2*x^(7/2)-2*a*(2*A*b+B*a)/x^(1/2)+2*(2*a*b*B+A*(2*a*c+b^2)) *x^(1/2)
Time = 0.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{5/2}} \, dx=\frac {2 \left (-35 a^2 (A+3 B x)+70 a x (-3 A (b-c x)+B x (3 b+c x))+x^2 \left (7 A \left (15 b^2+10 b c x+3 c^2 x^2\right )+B x \left (35 b^2+42 b c x+15 c^2 x^2\right )\right )\right )}{105 x^{3/2}} \]
(2*(-35*a^2*(A + 3*B*x) + 70*a*x*(-3*A*(b - c*x) + B*x*(3*b + c*x)) + x^2* (7*A*(15*b^2 + 10*b*c*x + 3*c^2*x^2) + B*x*(35*b^2 + 42*b*c*x + 15*c^2*x^2 ))))/(105*x^(3/2))
Time = 0.24 (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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {a^2 A}{x^{5/2}}+\sqrt {x} \left (2 a B c+2 A b c+b^2 B\right )+\frac {A \left (2 a c+b^2\right )+2 a b B}{\sqrt {x}}+\frac {a (a B+2 A b)}{x^{3/2}}+c x^{3/2} (A c+2 b B)+B c^2 x^{5/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 A}{3 x^{3/2}}+\frac {2}{3} x^{3/2} \left (2 a B c+2 A b c+b^2 B\right )+2 \sqrt {x} \left (A \left (2 a c+b^2\right )+2 a b B\right )-\frac {2 a (a B+2 A b)}{\sqrt {x}}+\frac {2}{5} c x^{5/2} (A c+2 b B)+\frac {2}{7} B c^2 x^{7/2}\) |
(-2*a^2*A)/(3*x^(3/2)) - (2*a*(2*A*b + a*B))/Sqrt[x] + 2*(2*a*b*B + A*(b^2 + 2*a*c))*Sqrt[x] + (2*(b^2*B + 2*A*b*c + 2*a*B*c)*x^(3/2))/3 + (2*c*(2*b *B + A*c)*x^(5/2))/5 + (2*B*c^2*x^(7/2))/7
3.10.96.3.1 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 = 101, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {2 B \,c^{2} x^{\frac {7}{2}}}{7}+\frac {2 A \,c^{2} x^{\frac {5}{2}}}{5}+\frac {4 B b c \,x^{\frac {5}{2}}}{5}+\frac {4 A b c \,x^{\frac {3}{2}}}{3}+\frac {4 a B c \,x^{\frac {3}{2}}}{3}+\frac {2 b^{2} B \,x^{\frac {3}{2}}}{3}+4 a A c \sqrt {x}+2 A \,b^{2} \sqrt {x}+4 a b B \sqrt {x}-\frac {2 a^{2} A}{3 x^{\frac {3}{2}}}-\frac {2 a \left (2 A b +B a \right )}{\sqrt {x}}\) | \(101\) |
default | \(\frac {2 B \,c^{2} x^{\frac {7}{2}}}{7}+\frac {2 A \,c^{2} x^{\frac {5}{2}}}{5}+\frac {4 B b c \,x^{\frac {5}{2}}}{5}+\frac {4 A b c \,x^{\frac {3}{2}}}{3}+\frac {4 a B c \,x^{\frac {3}{2}}}{3}+\frac {2 b^{2} B \,x^{\frac {3}{2}}}{3}+4 a A c \sqrt {x}+2 A \,b^{2} \sqrt {x}+4 a b B \sqrt {x}-\frac {2 a^{2} A}{3 x^{\frac {3}{2}}}-\frac {2 a \left (2 A b +B a \right )}{\sqrt {x}}\) | \(101\) |
gosper | \(-\frac {2 \left (-15 B \,c^{2} x^{5}-21 A \,c^{2} x^{4}-42 x^{4} B b c -70 x^{3} A b c -70 a B c \,x^{3}-35 B \,b^{2} x^{3}-210 a A c \,x^{2}-105 A \,b^{2} x^{2}-210 B a b \,x^{2}+210 a A b x +105 a^{2} B x +35 A \,a^{2}\right )}{105 x^{\frac {3}{2}}}\) | \(102\) |
trager | \(-\frac {2 \left (-15 B \,c^{2} x^{5}-21 A \,c^{2} x^{4}-42 x^{4} B b c -70 x^{3} A b c -70 a B c \,x^{3}-35 B \,b^{2} x^{3}-210 a A c \,x^{2}-105 A \,b^{2} x^{2}-210 B a b \,x^{2}+210 a A b x +105 a^{2} B x +35 A \,a^{2}\right )}{105 x^{\frac {3}{2}}}\) | \(102\) |
risch | \(-\frac {2 \left (-15 B \,c^{2} x^{5}-21 A \,c^{2} x^{4}-42 x^{4} B b c -70 x^{3} A b c -70 a B c \,x^{3}-35 B \,b^{2} x^{3}-210 a A c \,x^{2}-105 A \,b^{2} x^{2}-210 B a b \,x^{2}+210 a A b x +105 a^{2} B x +35 A \,a^{2}\right )}{105 x^{\frac {3}{2}}}\) | \(102\) |
2/7*B*c^2*x^(7/2)+2/5*A*c^2*x^(5/2)+4/5*B*b*c*x^(5/2)+4/3*A*b*c*x^(3/2)+4/ 3*a*B*c*x^(3/2)+2/3*b^2*B*x^(3/2)+4*a*A*c*x^(1/2)+2*A*b^2*x^(1/2)+4*a*b*B* x^(1/2)-2/3*a^2*A/x^(3/2)-2*a*(2*A*b+B*a)/x^(1/2)
Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{5/2}} \, dx=\frac {2 \, {\left (15 \, B c^{2} x^{5} + 21 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 35 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} - 35 \, A a^{2} + 105 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} - 105 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{105 \, x^{\frac {3}{2}}} \]
2/105*(15*B*c^2*x^5 + 21*(2*B*b*c + A*c^2)*x^4 + 35*(B*b^2 + 2*(B*a + A*b) *c)*x^3 - 35*A*a^2 + 105*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 - 105*(B*a^2 + 2* A*a*b)*x)/x^(3/2)
Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.40 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{5/2}} \, dx=- \frac {2 A a^{2}}{3 x^{\frac {3}{2}}} - \frac {4 A a b}{\sqrt {x}} + 4 A a c \sqrt {x} + 2 A b^{2} \sqrt {x} + \frac {4 A b c x^{\frac {3}{2}}}{3} + \frac {2 A c^{2} x^{\frac {5}{2}}}{5} - \frac {2 B a^{2}}{\sqrt {x}} + 4 B a b \sqrt {x} + \frac {4 B a c x^{\frac {3}{2}}}{3} + \frac {2 B b^{2} x^{\frac {3}{2}}}{3} + \frac {4 B b c x^{\frac {5}{2}}}{5} + \frac {2 B c^{2} x^{\frac {7}{2}}}{7} \]
-2*A*a**2/(3*x**(3/2)) - 4*A*a*b/sqrt(x) + 4*A*a*c*sqrt(x) + 2*A*b**2*sqrt (x) + 4*A*b*c*x**(3/2)/3 + 2*A*c**2*x**(5/2)/5 - 2*B*a**2/sqrt(x) + 4*B*a* b*sqrt(x) + 4*B*a*c*x**(3/2)/3 + 2*B*b**2*x**(3/2)/3 + 4*B*b*c*x**(5/2)/5 + 2*B*c**2*x**(7/2)/7
Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{5/2}} \, dx=\frac {2}{7} \, B c^{2} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{\frac {3}{2}} + 2 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} \sqrt {x} - \frac {2 \, {\left (A a^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \]
2/7*B*c^2*x^(7/2) + 2/5*(2*B*b*c + A*c^2)*x^(5/2) + 2/3*(B*b^2 + 2*(B*a + A*b)*c)*x^(3/2) + 2*(2*B*a*b + A*b^2 + 2*A*a*c)*sqrt(x) - 2/3*(A*a^2 + 3*( B*a^2 + 2*A*a*b)*x)/x^(3/2)
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{5/2}} \, dx=\frac {2}{7} \, B c^{2} x^{\frac {7}{2}} + \frac {4}{5} \, B b c x^{\frac {5}{2}} + \frac {2}{5} \, A c^{2} x^{\frac {5}{2}} + \frac {2}{3} \, B b^{2} x^{\frac {3}{2}} + \frac {4}{3} \, B a c x^{\frac {3}{2}} + \frac {4}{3} \, A b c x^{\frac {3}{2}} + 4 \, B a b \sqrt {x} + 2 \, A b^{2} \sqrt {x} + 4 \, A a c \sqrt {x} - \frac {2 \, {\left (3 \, B a^{2} x + 6 \, A a b x + A a^{2}\right )}}{3 \, x^{\frac {3}{2}}} \]
2/7*B*c^2*x^(7/2) + 4/5*B*b*c*x^(5/2) + 2/5*A*c^2*x^(5/2) + 2/3*B*b^2*x^(3 /2) + 4/3*B*a*c*x^(3/2) + 4/3*A*b*c*x^(3/2) + 4*B*a*b*sqrt(x) + 2*A*b^2*sq rt(x) + 4*A*a*c*sqrt(x) - 2/3*(3*B*a^2*x + 6*A*a*b*x + A*a^2)/x^(3/2)
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{5/2}} \, dx=x^{5/2}\,\left (\frac {2\,A\,c^2}{5}+\frac {4\,B\,b\,c}{5}\right )+\sqrt {x}\,\left (2\,A\,b^2+4\,B\,a\,b+4\,A\,a\,c\right )+x^{3/2}\,\left (\frac {2\,B\,b^2}{3}+\frac {4\,A\,c\,b}{3}+\frac {4\,B\,a\,c}{3}\right )-\frac {\frac {2\,A\,a^2}{3}+x\,\left (2\,B\,a^2+4\,A\,b\,a\right )}{x^{3/2}}+\frac {2\,B\,c^2\,x^{7/2}}{7} \]