\(\int x^4 (A+B x) (a^2+2 a b x+b^2 x^2)^3 \, dx\) [169]

Optimal result

Integrand size = 27, antiderivative size = 139 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {a^4 (A b-a B) (a+b x)^7}{7 b^6}-\frac {a^3 (4 A b-5 a B) (a+b x)^8}{8 b^6}+\frac {2 a^2 (3 A b-5 a B) (a+b x)^9}{9 b^6}-\frac {a (2 A b-5 a B) (a+b x)^{10}}{5 b^6}+\frac {(A b-5 a B) (a+b x)^{11}}{11 b^6}+\frac {B (a+b x)^{12}}{12 b^6} \] Output:

1/7*a^4*(A*b-B*a)*(b*x+a)^7/b^6-1/8*a^3*(4*A*b-5*B*a)*(b*x+a)^8/b^6+2/9*a^ 
2*(3*A*b-5*B*a)*(b*x+a)^9/b^6-1/5*a*(2*A*b-5*B*a)*(b*x+a)^10/b^6+1/11*(A*b 
-5*B*a)*(b*x+a)^11/b^6+1/12*B*(b*x+a)^12/b^6
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {1}{5} a^6 A x^5+\frac {1}{6} a^5 (6 A b+a B) x^6+\frac {3}{7} a^4 b (5 A b+2 a B) x^7+\frac {5}{8} a^3 b^2 (4 A b+3 a B) x^8+\frac {5}{9} a^2 b^3 (3 A b+4 a B) x^9+\frac {3}{10} a b^4 (2 A b+5 a B) x^{10}+\frac {1}{11} b^5 (A b+6 a B) x^{11}+\frac {1}{12} b^6 B x^{12} \] Input:

Integrate[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3,x]
 

Output:

(a^6*A*x^5)/5 + (a^5*(6*A*b + a*B)*x^6)/6 + (3*a^4*b*(5*A*b + 2*a*B)*x^7)/ 
7 + (5*a^3*b^2*(4*A*b + 3*a*B)*x^8)/8 + (5*a^2*b^3*(3*A*b + 4*a*B)*x^9)/9 
+ (3*a*b^4*(2*A*b + 5*a*B)*x^10)/10 + (b^5*(A*b + 6*a*B)*x^11)/11 + (b^6*B 
*x^12)/12
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1184, 27, 85, 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.

\(\displaystyle \int x^4 \left (a^2+2 a b x+b^2 x^2\right )^3 (A+B x) \, dx\)

\(\Big \downarrow \) 1184

\(\displaystyle \frac {\int b^6 x^4 (a+b x)^6 (A+B x)dx}{b^6}\)

\(\Big \downarrow \) 27

\(\displaystyle \int x^4 (a+b x)^6 (A+B x)dx\)

\(\Big \downarrow \) 85

\(\displaystyle \int \left (-\frac {a^4 (a+b x)^6 (a B-A b)}{b^5}+\frac {a^3 (a+b x)^7 (5 a B-4 A b)}{b^5}-\frac {2 a^2 (a+b x)^8 (5 a B-3 A b)}{b^5}+\frac {(a+b x)^{10} (A b-5 a B)}{b^5}+\frac {2 a (a+b x)^9 (5 a B-2 A b)}{b^5}+\frac {B (a+b x)^{11}}{b^5}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^4 (a+b x)^7 (A b-a B)}{7 b^6}-\frac {a^3 (a+b x)^8 (4 A b-5 a B)}{8 b^6}+\frac {2 a^2 (a+b x)^9 (3 A b-5 a B)}{9 b^6}+\frac {(a+b x)^{11} (A b-5 a B)}{11 b^6}-\frac {a (a+b x)^{10} (2 A b-5 a B)}{5 b^6}+\frac {B (a+b x)^{12}}{12 b^6}\)

Input:

Int[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3,x]
 

Output:

(a^4*(A*b - a*B)*(a + b*x)^7)/(7*b^6) - (a^3*(4*A*b - 5*a*B)*(a + b*x)^8)/ 
(8*b^6) + (2*a^2*(3*A*b - 5*a*B)*(a + b*x)^9)/(9*b^6) - (a*(2*A*b - 5*a*B) 
*(a + b*x)^10)/(5*b^6) + ((A*b - 5*a*B)*(a + b*x)^11)/(11*b^6) + (B*(a + b 
*x)^12)/(12*b^6)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 85

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : 
> Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, 
 d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* 
f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n 
 + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 
1])
 

rule 1184

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(f + g*x 
)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03

method	result	size
norman	\(\frac {B \,b^{6} x^{12}}{12}+\left (\frac {1}{11} A \,b^{6}+\frac {6}{11} B a \,b^{5}\right ) x^{11}+\left (\frac {3}{5} A a \,b^{5}+\frac {3}{2} B \,a^{2} b^{4}\right ) x^{10}+\left (\frac {5}{3} A \,a^{2} b^{4}+\frac {20}{9} B \,a^{3} b^{3}\right ) x^{9}+\left (\frac {5}{2} A \,a^{3} b^{3}+\frac {15}{8} B \,a^{4} b^{2}\right ) x^{8}+\left (\frac {15}{7} A \,a^{4} b^{2}+\frac {6}{7} B \,a^{5} b \right ) x^{7}+\left (A \,a^{5} b +\frac {1}{6} B \,a^{6}\right ) x^{6}+\frac {A \,a^{6} x^{5}}{5}\)	\(143\)
gosper	\(\frac {x^{5} \left (2310 b^{6} B \,x^{7}+2520 A \,b^{6} x^{6}+15120 B a \,b^{5} x^{6}+16632 A a \,b^{5} x^{5}+41580 B \,a^{2} b^{4} x^{5}+46200 A \,a^{2} b^{4} x^{4}+61600 B \,a^{3} b^{3} x^{4}+69300 A \,a^{3} b^{3} x^{3}+51975 B \,a^{4} b^{2} x^{3}+59400 A \,a^{4} b^{2} x^{2}+23760 B \,a^{5} b \,x^{2}+27720 A \,a^{5} b x +4620 B \,a^{6} x +5544 A \,a^{6}\right )}{27720}\)	\(148\)
default	\(\frac {B \,b^{6} x^{12}}{12}+\frac {\left (A \,b^{6}+6 B a \,b^{5}\right ) x^{11}}{11}+\frac {\left (6 A a \,b^{5}+15 B \,a^{2} b^{4}\right ) x^{10}}{10}+\frac {\left (15 A \,a^{2} b^{4}+20 B \,a^{3} b^{3}\right ) x^{9}}{9}+\frac {\left (20 A \,a^{3} b^{3}+15 B \,a^{4} b^{2}\right ) x^{8}}{8}+\frac {\left (15 A \,a^{4} b^{2}+6 B \,a^{5} b \right ) x^{7}}{7}+\frac {\left (6 A \,a^{5} b +B \,a^{6}\right ) x^{6}}{6}+\frac {A \,a^{6} x^{5}}{5}\)	\(148\)
risch	\(\frac {1}{12} B \,b^{6} x^{12}+\frac {1}{11} x^{11} A \,b^{6}+\frac {6}{11} x^{11} B a \,b^{5}+\frac {3}{5} x^{10} A a \,b^{5}+\frac {3}{2} x^{10} B \,a^{2} b^{4}+\frac {5}{3} x^{9} A \,a^{2} b^{4}+\frac {20}{9} x^{9} B \,a^{3} b^{3}+\frac {5}{2} x^{8} A \,a^{3} b^{3}+\frac {15}{8} x^{8} B \,a^{4} b^{2}+\frac {15}{7} x^{7} A \,a^{4} b^{2}+\frac {6}{7} x^{7} B \,a^{5} b +x^{6} A \,a^{5} b +\frac {1}{6} x^{6} B \,a^{6}+\frac {1}{5} A \,a^{6} x^{5}\)	\(149\)
parallelrisch	\(\frac {1}{12} B \,b^{6} x^{12}+\frac {1}{11} x^{11} A \,b^{6}+\frac {6}{11} x^{11} B a \,b^{5}+\frac {3}{5} x^{10} A a \,b^{5}+\frac {3}{2} x^{10} B \,a^{2} b^{4}+\frac {5}{3} x^{9} A \,a^{2} b^{4}+\frac {20}{9} x^{9} B \,a^{3} b^{3}+\frac {5}{2} x^{8} A \,a^{3} b^{3}+\frac {15}{8} x^{8} B \,a^{4} b^{2}+\frac {15}{7} x^{7} A \,a^{4} b^{2}+\frac {6}{7} x^{7} B \,a^{5} b +x^{6} A \,a^{5} b +\frac {1}{6} x^{6} B \,a^{6}+\frac {1}{5} A \,a^{6} x^{5}\)	\(149\)
orering	\(\frac {x^{5} \left (2310 b^{6} B \,x^{7}+2520 A \,b^{6} x^{6}+15120 B a \,b^{5} x^{6}+16632 A a \,b^{5} x^{5}+41580 B \,a^{2} b^{4} x^{5}+46200 A \,a^{2} b^{4} x^{4}+61600 B \,a^{3} b^{3} x^{4}+69300 A \,a^{3} b^{3} x^{3}+51975 B \,a^{4} b^{2} x^{3}+59400 A \,a^{4} b^{2} x^{2}+23760 B \,a^{5} b \,x^{2}+27720 A \,a^{5} b x +4620 B \,a^{6} x +5544 A \,a^{6}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{3}}{27720 \left (b x +a \right )^{6}}\)	\(173\)

Input:

int(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/12*B*b^6*x^12+(1/11*A*b^6+6/11*B*a*b^5)*x^11+(3/5*A*a*b^5+3/2*B*a^2*b^4) 
*x^10+(5/3*A*a^2*b^4+20/9*B*a^3*b^3)*x^9+(5/2*A*a^3*b^3+15/8*B*a^4*b^2)*x^ 
8+(15/7*A*a^4*b^2+6/7*B*a^5*b)*x^7+(A*a^5*b+1/6*B*a^6)*x^6+1/5*A*a^6*x^5
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.06 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {1}{12} \, B b^{6} x^{12} + \frac {1}{5} \, A a^{6} x^{5} + \frac {1}{11} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{11} + \frac {3}{10} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{10} + \frac {5}{9} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{9} + \frac {5}{8} \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{8} + \frac {3}{7} \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x^{6} \] Input:

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")
 

Output:

1/12*B*b^6*x^12 + 1/5*A*a^6*x^5 + 1/11*(6*B*a*b^5 + A*b^6)*x^11 + 3/10*(5* 
B*a^2*b^4 + 2*A*a*b^5)*x^10 + 5/9*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^9 + 5/8*(3 
*B*a^4*b^2 + 4*A*a^3*b^3)*x^8 + 3/7*(2*B*a^5*b + 5*A*a^4*b^2)*x^7 + 1/6*(B 
*a^6 + 6*A*a^5*b)*x^6
 

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.17 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {A a^{6} x^{5}}{5} + \frac {B b^{6} x^{12}}{12} + x^{11} \left (\frac {A b^{6}}{11} + \frac {6 B a b^{5}}{11}\right ) + x^{10} \cdot \left (\frac {3 A a b^{5}}{5} + \frac {3 B a^{2} b^{4}}{2}\right ) + x^{9} \cdot \left (\frac {5 A a^{2} b^{4}}{3} + \frac {20 B a^{3} b^{3}}{9}\right ) + x^{8} \cdot \left (\frac {5 A a^{3} b^{3}}{2} + \frac {15 B a^{4} b^{2}}{8}\right ) + x^{7} \cdot \left (\frac {15 A a^{4} b^{2}}{7} + \frac {6 B a^{5} b}{7}\right ) + x^{6} \left (A a^{5} b + \frac {B a^{6}}{6}\right ) \] Input:

integrate(x**4*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**3,x)
 

Output:

A*a**6*x**5/5 + B*b**6*x**12/12 + x**11*(A*b**6/11 + 6*B*a*b**5/11) + x**1 
0*(3*A*a*b**5/5 + 3*B*a**2*b**4/2) + x**9*(5*A*a**2*b**4/3 + 20*B*a**3*b** 
3/9) + x**8*(5*A*a**3*b**3/2 + 15*B*a**4*b**2/8) + x**7*(15*A*a**4*b**2/7 
+ 6*B*a**5*b/7) + x**6*(A*a**5*b + B*a**6/6)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.06 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {1}{12} \, B b^{6} x^{12} + \frac {1}{5} \, A a^{6} x^{5} + \frac {1}{11} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{11} + \frac {3}{10} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{10} + \frac {5}{9} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{9} + \frac {5}{8} \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{8} + \frac {3}{7} \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x^{6} \] Input:

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")
 

Output:

1/12*B*b^6*x^12 + 1/5*A*a^6*x^5 + 1/11*(6*B*a*b^5 + A*b^6)*x^11 + 3/10*(5* 
B*a^2*b^4 + 2*A*a*b^5)*x^10 + 5/9*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^9 + 5/8*(3 
*B*a^4*b^2 + 4*A*a^3*b^3)*x^8 + 3/7*(2*B*a^5*b + 5*A*a^4*b^2)*x^7 + 1/6*(B 
*a^6 + 6*A*a^5*b)*x^6
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {1}{12} \, B b^{6} x^{12} + \frac {6}{11} \, B a b^{5} x^{11} + \frac {1}{11} \, A b^{6} x^{11} + \frac {3}{2} \, B a^{2} b^{4} x^{10} + \frac {3}{5} \, A a b^{5} x^{10} + \frac {20}{9} \, B a^{3} b^{3} x^{9} + \frac {5}{3} \, A a^{2} b^{4} x^{9} + \frac {15}{8} \, B a^{4} b^{2} x^{8} + \frac {5}{2} \, A a^{3} b^{3} x^{8} + \frac {6}{7} \, B a^{5} b x^{7} + \frac {15}{7} \, A a^{4} b^{2} x^{7} + \frac {1}{6} \, B a^{6} x^{6} + A a^{5} b x^{6} + \frac {1}{5} \, A a^{6} x^{5} \] Input:

integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")
 

Output:

1/12*B*b^6*x^12 + 6/11*B*a*b^5*x^11 + 1/11*A*b^6*x^11 + 3/2*B*a^2*b^4*x^10 
 + 3/5*A*a*b^5*x^10 + 20/9*B*a^3*b^3*x^9 + 5/3*A*a^2*b^4*x^9 + 15/8*B*a^4* 
b^2*x^8 + 5/2*A*a^3*b^3*x^8 + 6/7*B*a^5*b*x^7 + 15/7*A*a^4*b^2*x^7 + 1/6*B 
*a^6*x^6 + A*a^5*b*x^6 + 1/5*A*a^6*x^5
 

Mupad [B] (verification not implemented)

Time = 10.52 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.94 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=x^6\,\left (\frac {B\,a^6}{6}+A\,b\,a^5\right )+x^{11}\,\left (\frac {A\,b^6}{11}+\frac {6\,B\,a\,b^5}{11}\right )+\frac {A\,a^6\,x^5}{5}+\frac {B\,b^6\,x^{12}}{12}+\frac {5\,a^3\,b^2\,x^8\,\left (4\,A\,b+3\,B\,a\right )}{8}+\frac {5\,a^2\,b^3\,x^9\,\left (3\,A\,b+4\,B\,a\right )}{9}+\frac {3\,a^4\,b\,x^7\,\left (5\,A\,b+2\,B\,a\right )}{7}+\frac {3\,a\,b^4\,x^{10}\,\left (2\,A\,b+5\,B\,a\right )}{10} \] Input:

int(x^4*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^3,x)
 

Output:

x^6*((B*a^6)/6 + A*a^5*b) + x^11*((A*b^6)/11 + (6*B*a*b^5)/11) + (A*a^6*x^ 
5)/5 + (B*b^6*x^12)/12 + (5*a^3*b^2*x^8*(4*A*b + 3*B*a))/8 + (5*a^2*b^3*x^ 
9*(3*A*b + 4*B*a))/9 + (3*a^4*b*x^7*(5*A*b + 2*B*a))/7 + (3*a*b^4*x^10*(2* 
A*b + 5*B*a))/10
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.57 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {x^{5} \left (330 b^{7} x^{7}+2520 a \,b^{6} x^{6}+8316 a^{2} b^{5} x^{5}+15400 a^{3} b^{4} x^{4}+17325 a^{4} b^{3} x^{3}+11880 a^{5} b^{2} x^{2}+4620 a^{6} b x +792 a^{7}\right )}{3960} \] Input:

int(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x)
 

Output:

(x**5*(792*a**7 + 4620*a**6*b*x + 11880*a**5*b**2*x**2 + 17325*a**4*b**3*x 
**3 + 15400*a**3*b**4*x**4 + 8316*a**2*b**5*x**5 + 2520*a*b**6*x**6 + 330* 
b**7*x**7))/3960