Integrand size = 23, antiderivative size = 109 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx=-\frac {2 a^2 A}{5 x^{5/2}}-\frac {2 a (2 A b+a B)}{3 x^{3/2}}-\frac {2 \left (2 a b B+A \left (b^2+2 a c\right )\right )}{\sqrt {x}}+2 \left (b^2 B+2 A b c+2 a B c\right ) \sqrt {x}+\frac {2}{3} c (2 b B+A c) x^{3/2}+\frac {2}{5} B c^2 x^{5/2} \] Output:
-2/5*a^2*A/x^(5/2)-2/3*a*(2*A*b+B*a)/x^(3/2)-2*(2*a*b*B+A*(2*a*c+b^2))/x^( 1/2)+2*(2*A*b*c+2*B*a*c+B*b^2)*x^(1/2)+2/3*c*(A*c+2*B*b)*x^(3/2)+2/5*B*c^2 *x^(5/2)
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx=\frac {-2 a^2 (3 A+5 B x)-20 a x (3 B x (b-c x)+A (b+3 c x))+2 x^2 \left (5 A \left (-3 b^2+6 b c x+c^2 x^2\right )+B x \left (15 b^2+10 b c x+3 c^2 x^2\right )\right )}{15 x^{5/2}} \] Input:
Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x^(7/2),x]
Output:
(-2*a^2*(3*A + 5*B*x) - 20*a*x*(3*B*x*(b - c*x) + A*(b + 3*c*x)) + 2*x^2*( 5*A*(-3*b^2 + 6*b*c*x + c^2*x^2) + B*x*(15*b^2 + 10*b*c*x + 3*c^2*x^2)))/( 15*x^(5/2))
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.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {a^2 A}{x^{7/2}}+\frac {A \left (2 a c+b^2\right )+2 a b B}{x^{3/2}}+\frac {2 a B c+2 A b c+b^2 B}{\sqrt {x}}+\frac {a (a B+2 A b)}{x^{5/2}}+c \sqrt {x} (A c+2 b B)+B c^2 x^{3/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 A}{5 x^{5/2}}+2 \sqrt {x} \left (2 a B c+2 A b c+b^2 B\right )-\frac {2 \left (A \left (2 a c+b^2\right )+2 a b B\right )}{\sqrt {x}}-\frac {2 a (a B+2 A b)}{3 x^{3/2}}+\frac {2}{3} c x^{3/2} (A c+2 b B)+\frac {2}{5} B c^2 x^{5/2}\) |
Input:
Int[((A + B*x)*(a + b*x + c*x^2)^2)/x^(7/2),x]
Output:
(-2*a^2*A)/(5*x^(5/2)) - (2*a*(2*A*b + a*B))/(3*x^(3/2)) - (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)*Sqrt[x] + (2*c*(2 *b*B + A*c)*x^(3/2))/3 + (2*B*c^2*x^(5/2))/5
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 = 1.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {2 B \,c^{2} x^{\frac {5}{2}}}{5}+\frac {2 A \,c^{2} x^{\frac {3}{2}}}{3}+\frac {4 B b c \,x^{\frac {3}{2}}}{3}+4 A b c \sqrt {x}+4 B a c \sqrt {x}+2 B \,b^{2} \sqrt {x}-\frac {2 a \left (2 A b +B a \right )}{3 x^{\frac {3}{2}}}-\frac {2 a^{2} A}{5 x^{\frac {5}{2}}}-\frac {2 \left (2 A a c +b^{2} A +2 a b B \right )}{\sqrt {x}}\) | \(97\) |
default | \(\frac {2 B \,c^{2} x^{\frac {5}{2}}}{5}+\frac {2 A \,c^{2} x^{\frac {3}{2}}}{3}+\frac {4 B b c \,x^{\frac {3}{2}}}{3}+4 A b c \sqrt {x}+4 B a c \sqrt {x}+2 B \,b^{2} \sqrt {x}-\frac {2 a \left (2 A b +B a \right )}{3 x^{\frac {3}{2}}}-\frac {2 a^{2} A}{5 x^{\frac {5}{2}}}-\frac {2 \left (2 A a c +b^{2} A +2 a b B \right )}{\sqrt {x}}\) | \(97\) |
gosper | \(-\frac {2 \left (-3 B \,c^{2} x^{5}-5 x^{4} A \,c^{2}-10 x^{4} B b c -30 x^{3} A b c -30 B a c \,x^{3}-15 x^{3} B \,b^{2}+30 A a c \,x^{2}+15 x^{2} b^{2} A +30 B a \,x^{2} b +10 a b A x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}}}\) | \(102\) |
trager | \(-\frac {2 \left (-3 B \,c^{2} x^{5}-5 x^{4} A \,c^{2}-10 x^{4} B b c -30 x^{3} A b c -30 B a c \,x^{3}-15 x^{3} B \,b^{2}+30 A a c \,x^{2}+15 x^{2} b^{2} A +30 B a \,x^{2} b +10 a b A x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}}}\) | \(102\) |
risch | \(-\frac {2 \left (-3 B \,c^{2} x^{5}-5 x^{4} A \,c^{2}-10 x^{4} B b c -30 x^{3} A b c -30 B a c \,x^{3}-15 x^{3} B \,b^{2}+30 A a c \,x^{2}+15 x^{2} b^{2} A +30 B a \,x^{2} b +10 a b A x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}}}\) | \(102\) |
orering | \(-\frac {2 \left (-3 B \,c^{2} x^{5}-5 x^{4} A \,c^{2}-10 x^{4} B b c -30 x^{3} A b c -30 B a c \,x^{3}-15 x^{3} B \,b^{2}+30 A a c \,x^{2}+15 x^{2} b^{2} A +30 B a \,x^{2} b +10 a b A x +5 a^{2} B x +3 a^{2} A \right )}{15 x^{\frac {5}{2}}}\) | \(102\) |
Input:
int((B*x+A)*(c*x^2+b*x+a)^2/x^(7/2),x,method=_RETURNVERBOSE)
Output:
2/5*B*c^2*x^(5/2)+2/3*A*c^2*x^(3/2)+4/3*B*b*c*x^(3/2)+4*A*b*c*x^(1/2)+4*B* a*c*x^(1/2)+2*B*b^2*x^(1/2)-2/3*a*(2*A*b+B*a)/x^(3/2)-2/5*a^2*A/x^(5/2)-2* (2*A*a*c+A*b^2+2*B*a*b)/x^(1/2)
Time = 0.10 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx=\frac {2 \, {\left (3 \, B c^{2} x^{5} + 5 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 15 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} - 3 \, A a^{2} - 15 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} - 5 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x^(7/2),x, algorithm="fricas")
Output:
2/15*(3*B*c^2*x^5 + 5*(2*B*b*c + A*c^2)*x^4 + 15*(B*b^2 + 2*(B*a + A*b)*c) *x^3 - 3*A*a^2 - 15*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 - 5*(B*a^2 + 2*A*a*b)* x)/x^(5/2)
Time = 0.37 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx=- \frac {2 A a^{2}}{5 x^{\frac {5}{2}}} - \frac {4 A a b}{3 x^{\frac {3}{2}}} - \frac {4 A a c}{\sqrt {x}} - \frac {2 A b^{2}}{\sqrt {x}} + 4 A b c \sqrt {x} + \frac {2 A c^{2} x^{\frac {3}{2}}}{3} - \frac {2 B a^{2}}{3 x^{\frac {3}{2}}} - \frac {4 B a b}{\sqrt {x}} + 4 B a c \sqrt {x} + 2 B b^{2} \sqrt {x} + \frac {4 B b c x^{\frac {3}{2}}}{3} + \frac {2 B c^{2} x^{\frac {5}{2}}}{5} \] Input:
integrate((B*x+A)*(c*x**2+b*x+a)**2/x**(7/2),x)
Output:
-2*A*a**2/(5*x**(5/2)) - 4*A*a*b/(3*x**(3/2)) - 4*A*a*c/sqrt(x) - 2*A*b**2 /sqrt(x) + 4*A*b*c*sqrt(x) + 2*A*c**2*x**(3/2)/3 - 2*B*a**2/(3*x**(3/2)) - 4*B*a*b/sqrt(x) + 4*B*a*c*sqrt(x) + 2*B*b**2*sqrt(x) + 4*B*b*c*x**(3/2)/3 + 2*B*c**2*x**(5/2)/5
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx=\frac {2}{5} \, B c^{2} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {3}{2}} + 2 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} \sqrt {x} - \frac {2 \, {\left (3 \, A a^{2} + 15 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 5 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x^(7/2),x, algorithm="maxima")
Output:
2/5*B*c^2*x^(5/2) + 2/3*(2*B*b*c + A*c^2)*x^(3/2) + 2*(B*b^2 + 2*(B*a + A* b)*c)*sqrt(x) - 2/15*(3*A*a^2 + 15*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 + 5*(B* a^2 + 2*A*a*b)*x)/x^(5/2)
Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx=\frac {2}{5} \, B c^{2} x^{\frac {5}{2}} + \frac {4}{3} \, B b c x^{\frac {3}{2}} + \frac {2}{3} \, A c^{2} x^{\frac {3}{2}} + 2 \, B b^{2} \sqrt {x} + 4 \, B a c \sqrt {x} + 4 \, A b c \sqrt {x} - \frac {2 \, {\left (30 \, B a b x^{2} + 15 \, A b^{2} x^{2} + 30 \, A a c x^{2} + 5 \, B a^{2} x + 10 \, A a b x + 3 \, A a^{2}\right )}}{15 \, x^{\frac {5}{2}}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^2/x^(7/2),x, algorithm="giac")
Output:
2/5*B*c^2*x^(5/2) + 4/3*B*b*c*x^(3/2) + 2/3*A*c^2*x^(3/2) + 2*B*b^2*sqrt(x ) + 4*B*a*c*sqrt(x) + 4*A*b*c*sqrt(x) - 2/15*(30*B*a*b*x^2 + 15*A*b^2*x^2 + 30*A*a*c*x^2 + 5*B*a^2*x + 10*A*a*b*x + 3*A*a^2)/x^(5/2)
Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx=x^{3/2}\,\left (\frac {2\,A\,c^2}{3}+\frac {4\,B\,b\,c}{3}\right )-\frac {\frac {2\,A\,a^2}{5}+x^2\,\left (2\,A\,b^2+4\,B\,a\,b+4\,A\,a\,c\right )+x\,\left (\frac {2\,B\,a^2}{3}+\frac {4\,A\,b\,a}{3}\right )}{x^{5/2}}+\sqrt {x}\,\left (2\,B\,b^2+4\,A\,c\,b+4\,B\,a\,c\right )+\frac {2\,B\,c^2\,x^{5/2}}{5} \] Input:
int(((A + B*x)*(a + b*x + c*x^2)^2)/x^(7/2),x)
Output:
x^(3/2)*((2*A*c^2)/3 + (4*B*b*c)/3) - ((2*A*a^2)/5 + x^2*(2*A*b^2 + 4*A*a* c + 4*B*a*b) + x*((2*B*a^2)/3 + (4*A*a*b)/3))/x^(5/2) + x^(1/2)*(2*B*b^2 + 4*A*b*c + 4*B*a*c) + (2*B*c^2*x^(5/2))/5
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^{7/2}} \, dx=\frac {\frac {2}{5} b \,c^{2} x^{5}+\frac {2}{3} a \,c^{2} x^{4}+\frac {4}{3} b^{2} c \,x^{4}+8 a b c \,x^{3}+2 b^{3} x^{3}-4 a^{2} c \,x^{2}-6 a \,b^{2} x^{2}-2 a^{2} b x -\frac {2}{5} a^{3}}{\sqrt {x}\, x^{2}} \] Input:
int((B*x+A)*(c*x^2+b*x+a)^2/x^(7/2),x)
Output:
(2*( - 3*a**3 - 15*a**2*b*x - 30*a**2*c*x**2 - 45*a*b**2*x**2 + 60*a*b*c*x **3 + 5*a*c**2*x**4 + 15*b**3*x**3 + 10*b**2*c*x**4 + 3*b*c**2*x**5))/(15* sqrt(x)*x**2)