3.11.6 \(\int \frac {(A+B x) (a+b x+c x^2)^3}{x^{7/2}} \, dx\) [1006]

3.11.6.1 Optimal result
3.11.6.2 Mathematica [A] (verified)
3.11.6.3 Rubi [A] (verified)
3.11.6.4 Maple [A] (verified)
3.11.6.5 Fricas [A] (verification not implemented)
3.11.6.6 Sympy [A] (verification not implemented)
3.11.6.7 Maxima [A] (verification not implemented)
3.11.6.8 Giac [A] (verification not implemented)
3.11.6.9 Mupad [B] (verification not implemented)

3.11.6.1 Optimal result

Integrand size = 23, antiderivative size = 178 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{7/2}} \, dx=-\frac {2 a^3 A}{5 x^{5/2}}-\frac {2 a^2 (3 A b+a B)}{3 x^{3/2}}-\frac {6 a \left (a b B+A \left (b^2+a c\right )\right )}{\sqrt {x}}+2 \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) \sqrt {x}+\frac {2}{3} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^{3/2}+\frac {6}{5} c \left (b^2 B+A b c+a B c\right ) x^{5/2}+\frac {2}{7} c^2 (3 b B+A c) x^{7/2}+\frac {2}{9} B c^3 x^{9/2} \]

output
-2/5*a^3*A/x^(5/2)-2/3*a^2*(3*A*b+B*a)/x^(3/2)+2/3*(3*A*a*c^2+3*A*b^2*c+6* 
B*a*b*c+B*b^3)*x^(3/2)+6/5*c*(A*b*c+B*a*c+B*b^2)*x^(5/2)+2/7*c^2*(A*c+3*B* 
b)*x^(7/2)+2/9*B*c^3*x^(9/2)-6*a*(a*b*B+A*(a*c+b^2))/x^(1/2)+2*(3*a*B*(a*c 
+b^2)+A*(6*a*b*c+b^3))*x^(1/2)
 
3.11.6.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{7/2}} \, dx=\frac {2 \left (-21 a^3 (3 A+5 B x)-315 a^2 x (3 B x (b-c x)+A (b+3 c x))+63 a x^2 \left (5 A \left (-3 b^2+6 b c x+c^2 x^2\right )+B x \left (15 b^2+10 b c x+3 c^2 x^2\right )\right )+x^3 \left (9 A \left (35 b^3+35 b^2 c x+21 b c^2 x^2+5 c^3 x^3\right )+B x \left (105 b^3+189 b^2 c x+135 b c^2 x^2+35 c^3 x^3\right )\right )\right )}{315 x^{5/2}} \]

input
Integrate[((A + B*x)*(a + b*x + c*x^2)^3)/x^(7/2),x]
 
output
(2*(-21*a^3*(3*A + 5*B*x) - 315*a^2*x*(3*B*x*(b - c*x) + A*(b + 3*c*x)) + 
63*a*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)) + x^3*(9*A*(35*b^3 + 35*b^2*c*x + 21*b*c^2*x^2 + 5*c^3*x^3) + B*x* 
(105*b^3 + 189*b^2*c*x + 135*b*c^2*x^2 + 35*c^3*x^3))))/(315*x^(5/2))
 
3.11.6.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 178, 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.

\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{7/2}} \, dx\)

\(\Big \downarrow \) 1195

\(\displaystyle \int \left (\frac {a^3 A}{x^{7/2}}+\frac {a^2 (a B+3 A b)}{x^{5/2}}+3 c x^{3/2} \left (a B c+A b c+b^2 B\right )+\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x^{3/2}}+\sqrt {x} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{\sqrt {x}}+c^2 x^{5/2} (A c+3 b B)+B c^3 x^{7/2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^3 A}{5 x^{5/2}}-\frac {2 a^2 (a B+3 A b)}{3 x^{3/2}}+\frac {6}{5} c x^{5/2} \left (a B c+A b c+b^2 B\right )-\frac {6 a \left (A \left (a c+b^2\right )+a b B\right )}{\sqrt {x}}+\frac {2}{3} x^{3/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+2 \sqrt {x} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {2}{7} c^2 x^{7/2} (A c+3 b B)+\frac {2}{9} B c^3 x^{9/2}\)

input
Int[((A + B*x)*(a + b*x + c*x^2)^3)/x^(7/2),x]
 
output
(-2*a^3*A)/(5*x^(5/2)) - (2*a^2*(3*A*b + a*B))/(3*x^(3/2)) - (6*a*(a*b*B + 
 A*(b^2 + a*c)))/Sqrt[x] + 2*(3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*Sqrt[ 
x] + (2*(b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^(3/2))/3 + (6*c*(b^2 
*B + A*b*c + a*B*c)*x^(5/2))/5 + (2*c^2*(3*b*B + A*c)*x^(7/2))/7 + (2*B*c^ 
3*x^(9/2))/9
 

3.11.6.3.1 Defintions of rubi rules used

rule 1195
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]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
3.11.6.4 Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.02

method result size
derivativedivides \(\frac {2 B \,c^{3} x^{\frac {9}{2}}}{9}+\frac {2 A \,c^{3} x^{\frac {7}{2}}}{7}+\frac {6 B b \,c^{2} x^{\frac {7}{2}}}{7}+\frac {6 A b \,c^{2} x^{\frac {5}{2}}}{5}+\frac {6 a B \,c^{2} x^{\frac {5}{2}}}{5}+\frac {6 B \,b^{2} c \,x^{\frac {5}{2}}}{5}+2 a A \,c^{2} x^{\frac {3}{2}}+2 A \,b^{2} c \,x^{\frac {3}{2}}+4 B a b c \,x^{\frac {3}{2}}+\frac {2 B \,b^{3} x^{\frac {3}{2}}}{3}+12 A a b c \sqrt {x}+2 A \,b^{3} \sqrt {x}+6 a^{2} B c \sqrt {x}+6 B a \,b^{2} \sqrt {x}-\frac {2 a^{3} A}{5 x^{\frac {5}{2}}}-\frac {2 a^{2} \left (3 A b +B a \right )}{3 x^{\frac {3}{2}}}-\frac {6 a \left (A a c +A \,b^{2}+a b B \right )}{\sqrt {x}}\) \(181\)
default \(\frac {2 B \,c^{3} x^{\frac {9}{2}}}{9}+\frac {2 A \,c^{3} x^{\frac {7}{2}}}{7}+\frac {6 B b \,c^{2} x^{\frac {7}{2}}}{7}+\frac {6 A b \,c^{2} x^{\frac {5}{2}}}{5}+\frac {6 a B \,c^{2} x^{\frac {5}{2}}}{5}+\frac {6 B \,b^{2} c \,x^{\frac {5}{2}}}{5}+2 a A \,c^{2} x^{\frac {3}{2}}+2 A \,b^{2} c \,x^{\frac {3}{2}}+4 B a b c \,x^{\frac {3}{2}}+\frac {2 B \,b^{3} x^{\frac {3}{2}}}{3}+12 A a b c \sqrt {x}+2 A \,b^{3} \sqrt {x}+6 a^{2} B c \sqrt {x}+6 B a \,b^{2} \sqrt {x}-\frac {2 a^{3} A}{5 x^{\frac {5}{2}}}-\frac {2 a^{2} \left (3 A b +B a \right )}{3 x^{\frac {3}{2}}}-\frac {6 a \left (A a c +A \,b^{2}+a b B \right )}{\sqrt {x}}\) \(181\)
gosper \(-\frac {2 \left (-35 B \,c^{3} x^{7}-45 A \,c^{3} x^{6}-135 B b \,c^{2} x^{6}-189 A b \,c^{2} x^{5}-189 a B \,c^{2} x^{5}-189 B \,b^{2} c \,x^{5}-315 a A \,c^{2} x^{4}-315 A \,b^{2} c \,x^{4}-630 B a b c \,x^{4}-105 x^{4} B \,b^{3}-1890 A a b c \,x^{3}-315 A \,b^{3} x^{3}-945 a^{2} B c \,x^{3}-945 B a \,b^{2} x^{3}+945 a^{2} A c \,x^{2}+945 A a \,b^{2} x^{2}+945 B \,a^{2} b \,x^{2}+315 A \,a^{2} b x +105 a^{3} B x +63 A \,a^{3}\right )}{315 x^{\frac {5}{2}}}\) \(192\)
trager \(-\frac {2 \left (-35 B \,c^{3} x^{7}-45 A \,c^{3} x^{6}-135 B b \,c^{2} x^{6}-189 A b \,c^{2} x^{5}-189 a B \,c^{2} x^{5}-189 B \,b^{2} c \,x^{5}-315 a A \,c^{2} x^{4}-315 A \,b^{2} c \,x^{4}-630 B a b c \,x^{4}-105 x^{4} B \,b^{3}-1890 A a b c \,x^{3}-315 A \,b^{3} x^{3}-945 a^{2} B c \,x^{3}-945 B a \,b^{2} x^{3}+945 a^{2} A c \,x^{2}+945 A a \,b^{2} x^{2}+945 B \,a^{2} b \,x^{2}+315 A \,a^{2} b x +105 a^{3} B x +63 A \,a^{3}\right )}{315 x^{\frac {5}{2}}}\) \(192\)
risch \(-\frac {2 \left (-35 B \,c^{3} x^{7}-45 A \,c^{3} x^{6}-135 B b \,c^{2} x^{6}-189 A b \,c^{2} x^{5}-189 a B \,c^{2} x^{5}-189 B \,b^{2} c \,x^{5}-315 a A \,c^{2} x^{4}-315 A \,b^{2} c \,x^{4}-630 B a b c \,x^{4}-105 x^{4} B \,b^{3}-1890 A a b c \,x^{3}-315 A \,b^{3} x^{3}-945 a^{2} B c \,x^{3}-945 B a \,b^{2} x^{3}+945 a^{2} A c \,x^{2}+945 A a \,b^{2} x^{2}+945 B \,a^{2} b \,x^{2}+315 A \,a^{2} b x +105 a^{3} B x +63 A \,a^{3}\right )}{315 x^{\frac {5}{2}}}\) \(192\)

input
int((B*x+A)*(c*x^2+b*x+a)^3/x^(7/2),x,method=_RETURNVERBOSE)
 
output
2/9*B*c^3*x^(9/2)+2/7*A*c^3*x^(7/2)+6/7*B*b*c^2*x^(7/2)+6/5*A*b*c^2*x^(5/2 
)+6/5*a*B*c^2*x^(5/2)+6/5*B*b^2*c*x^(5/2)+2*a*A*c^2*x^(3/2)+2*A*b^2*c*x^(3 
/2)+4*B*a*b*c*x^(3/2)+2/3*B*b^3*x^(3/2)+12*A*a*b*c*x^(1/2)+2*A*b^3*x^(1/2) 
+6*a^2*B*c*x^(1/2)+6*B*a*b^2*x^(1/2)-2/5*a^3*A/x^(5/2)-2/3*a^2*(3*A*b+B*a) 
/x^(3/2)-6*a*(A*a*c+A*b^2+B*a*b)/x^(1/2)
 
3.11.6.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{7/2}} \, dx=\frac {2 \, {\left (35 \, B c^{3} x^{7} + 45 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 189 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 105 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} - 63 \, A a^{3} + 315 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} - 945 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 105 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{315 \, x^{\frac {5}{2}}} \]

input
integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(7/2),x, algorithm="fricas")
 
output
2/315*(35*B*c^3*x^7 + 45*(3*B*b*c^2 + A*c^3)*x^6 + 189*(B*b^2*c + (B*a + A 
*b)*c^2)*x^5 + 105*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 - 63*A* 
a^3 + 315*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 - 945*(B*a^2*b + 
 A*a*b^2 + A*a^2*c)*x^2 - 105*(B*a^3 + 3*A*a^2*b)*x)/x^(5/2)
 
3.11.6.6 Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.54 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{7/2}} \, dx=- \frac {2 A a^{3}}{5 x^{\frac {5}{2}}} - \frac {2 A a^{2} b}{x^{\frac {3}{2}}} - \frac {6 A a^{2} c}{\sqrt {x}} - \frac {6 A a b^{2}}{\sqrt {x}} + 12 A a b c \sqrt {x} + 2 A a c^{2} x^{\frac {3}{2}} + 2 A b^{3} \sqrt {x} + 2 A b^{2} c x^{\frac {3}{2}} + \frac {6 A b c^{2} x^{\frac {5}{2}}}{5} + \frac {2 A c^{3} x^{\frac {7}{2}}}{7} - \frac {2 B a^{3}}{3 x^{\frac {3}{2}}} - \frac {6 B a^{2} b}{\sqrt {x}} + 6 B a^{2} c \sqrt {x} + 6 B a b^{2} \sqrt {x} + 4 B a b c x^{\frac {3}{2}} + \frac {6 B a c^{2} x^{\frac {5}{2}}}{5} + \frac {2 B b^{3} x^{\frac {3}{2}}}{3} + \frac {6 B b^{2} c x^{\frac {5}{2}}}{5} + \frac {6 B b c^{2} x^{\frac {7}{2}}}{7} + \frac {2 B c^{3} x^{\frac {9}{2}}}{9} \]

input
integrate((B*x+A)*(c*x**2+b*x+a)**3/x**(7/2),x)
 
output
-2*A*a**3/(5*x**(5/2)) - 2*A*a**2*b/x**(3/2) - 6*A*a**2*c/sqrt(x) - 6*A*a* 
b**2/sqrt(x) + 12*A*a*b*c*sqrt(x) + 2*A*a*c**2*x**(3/2) + 2*A*b**3*sqrt(x) 
 + 2*A*b**2*c*x**(3/2) + 6*A*b*c**2*x**(5/2)/5 + 2*A*c**3*x**(7/2)/7 - 2*B 
*a**3/(3*x**(3/2)) - 6*B*a**2*b/sqrt(x) + 6*B*a**2*c*sqrt(x) + 6*B*a*b**2* 
sqrt(x) + 4*B*a*b*c*x**(3/2) + 6*B*a*c**2*x**(5/2)/5 + 2*B*b**3*x**(3/2)/3 
 + 6*B*b**2*c*x**(5/2)/5 + 6*B*b*c**2*x**(7/2)/7 + 2*B*c**3*x**(9/2)/9
 
3.11.6.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{7/2}} \, dx=\frac {2}{9} \, B c^{3} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {7}{2}} + \frac {6}{5} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{\frac {3}{2}} + 2 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} \sqrt {x} - \frac {2 \, {\left (3 \, A a^{3} + 45 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

input
integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(7/2),x, algorithm="maxima")
 
output
2/9*B*c^3*x^(9/2) + 2/7*(3*B*b*c^2 + A*c^3)*x^(7/2) + 6/5*(B*b^2*c + (B*a 
+ A*b)*c^2)*x^(5/2) + 2/3*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^(3 
/2) + 2*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*sqrt(x) - 2/15*(3*A*a^ 
3 + 45*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 5*(B*a^3 + 3*A*a^2*b)*x)/x^(5/2 
)
 
3.11.6.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{7/2}} \, dx=\frac {2}{9} \, B c^{3} x^{\frac {9}{2}} + \frac {6}{7} \, B b c^{2} x^{\frac {7}{2}} + \frac {2}{7} \, A c^{3} x^{\frac {7}{2}} + \frac {6}{5} \, B b^{2} c x^{\frac {5}{2}} + \frac {6}{5} \, B a c^{2} x^{\frac {5}{2}} + \frac {6}{5} \, A b c^{2} x^{\frac {5}{2}} + \frac {2}{3} \, B b^{3} x^{\frac {3}{2}} + 4 \, B a b c x^{\frac {3}{2}} + 2 \, A b^{2} c x^{\frac {3}{2}} + 2 \, A a c^{2} x^{\frac {3}{2}} + 6 \, B a b^{2} \sqrt {x} + 2 \, A b^{3} \sqrt {x} + 6 \, B a^{2} c \sqrt {x} + 12 \, A a b c \sqrt {x} - \frac {2 \, {\left (45 \, B a^{2} b x^{2} + 45 \, A a b^{2} x^{2} + 45 \, A a^{2} c x^{2} + 5 \, B a^{3} x + 15 \, A a^{2} b x + 3 \, A a^{3}\right )}}{15 \, x^{\frac {5}{2}}} \]

input
integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(7/2),x, algorithm="giac")
 
output
2/9*B*c^3*x^(9/2) + 6/7*B*b*c^2*x^(7/2) + 2/7*A*c^3*x^(7/2) + 6/5*B*b^2*c* 
x^(5/2) + 6/5*B*a*c^2*x^(5/2) + 6/5*A*b*c^2*x^(5/2) + 2/3*B*b^3*x^(3/2) + 
4*B*a*b*c*x^(3/2) + 2*A*b^2*c*x^(3/2) + 2*A*a*c^2*x^(3/2) + 6*B*a*b^2*sqrt 
(x) + 2*A*b^3*sqrt(x) + 6*B*a^2*c*sqrt(x) + 12*A*a*b*c*sqrt(x) - 2/15*(45* 
B*a^2*b*x^2 + 45*A*a*b^2*x^2 + 45*A*a^2*c*x^2 + 5*B*a^3*x + 15*A*a^2*b*x + 
 3*A*a^3)/x^(5/2)
 
3.11.6.9 Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^{7/2}} \, dx=\sqrt {x}\,\left (6\,B\,c\,a^2+6\,B\,a\,b^2+12\,A\,c\,a\,b+2\,A\,b^3\right )-\frac {x\,\left (\frac {2\,B\,a^3}{3}+2\,A\,b\,a^2\right )+\frac {2\,A\,a^3}{5}+x^2\,\left (6\,B\,a^2\,b+6\,A\,c\,a^2+6\,A\,a\,b^2\right )}{x^{5/2}}+x^{3/2}\,\left (\frac {2\,B\,b^3}{3}+2\,A\,b^2\,c+4\,B\,a\,b\,c+2\,A\,a\,c^2\right )+x^{7/2}\,\left (\frac {2\,A\,c^3}{7}+\frac {6\,B\,b\,c^2}{7}\right )+x^{5/2}\,\left (\frac {6\,B\,b^2\,c}{5}+\frac {6\,A\,b\,c^2}{5}+\frac {6\,B\,a\,c^2}{5}\right )+\frac {2\,B\,c^3\,x^{9/2}}{9} \]

input
int(((A + B*x)*(a + b*x + c*x^2)^3)/x^(7/2),x)
 
output
x^(1/2)*(2*A*b^3 + 6*B*a*b^2 + 6*B*a^2*c + 12*A*a*b*c) - (x*((2*B*a^3)/3 + 
 2*A*a^2*b) + (2*A*a^3)/5 + x^2*(6*A*a*b^2 + 6*A*a^2*c + 6*B*a^2*b))/x^(5/ 
2) + x^(3/2)*((2*B*b^3)/3 + 2*A*a*c^2 + 2*A*b^2*c + 4*B*a*b*c) + x^(7/2)*( 
(2*A*c^3)/7 + (6*B*b*c^2)/7) + x^(5/2)*((6*A*b*c^2)/5 + (6*B*a*c^2)/5 + (6 
*B*b^2*c)/5) + (2*B*c^3*x^(9/2))/9