\(\int x^6 (A+B x) (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\) [312]

Optimal result

Integrand size = 29, antiderivative size = 303 \[ \int x^6 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {a^5 A x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {a^4 (5 A b+a B) x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (a+b x)}+\frac {5 a^3 b (2 A b+a B) x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {a^2 b^2 (A b+a B) x^{10} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a b^3 (A b+2 a B) x^{11} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {b^4 (A b+5 a B) x^{12} \sqrt {a^2+2 a b x+b^2 x^2}}{12 (a+b x)}+\frac {b^5 B x^{13} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)} \] Output:

a^5*A*x^7*((b*x+a)^2)^(1/2)/(7*b*x+7*a)+a^4*(5*A*b+B*a)*x^8*((b*x+a)^2)^(1 
/2)/(8*b*x+8*a)+5*a^3*b*(2*A*b+B*a)*x^9*((b*x+a)^2)^(1/2)/(9*b*x+9*a)+a^2* 
b^2*(A*b+B*a)*x^10*((b*x+a)^2)^(1/2)/(b*x+a)+5*a*b^3*(A*b+2*B*a)*x^11*((b* 
x+a)^2)^(1/2)/(11*b*x+11*a)+b^4*(A*b+5*B*a)*x^12*((b*x+a)^2)^(1/2)/(12*b*x 
+12*a)+b^5*B*x^13*((b*x+a)^2)^(1/2)/(13*b*x+13*a)
 

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.41 \[ \int x^6 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {x^7 \sqrt {(a+b x)^2} \left (1287 a^5 (8 A+7 B x)+5005 a^4 b x (9 A+8 B x)+8008 a^3 b^2 x^2 (10 A+9 B x)+6552 a^2 b^3 x^3 (11 A+10 B x)+2730 a b^4 x^4 (12 A+11 B x)+462 b^5 x^5 (13 A+12 B x)\right )}{72072 (a+b x)} \] Input:

Integrate[x^6*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(x^7*Sqrt[(a + b*x)^2]*(1287*a^5*(8*A + 7*B*x) + 5005*a^4*b*x*(9*A + 8*B*x 
) + 8008*a^3*b^2*x^2*(10*A + 9*B*x) + 6552*a^2*b^3*x^3*(11*A + 10*B*x) + 2 
730*a*b^4*x^4*(12*A + 11*B*x) + 462*b^5*x^5*(13*A + 12*B*x)))/(72072*(a + 
b*x))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.47, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1187, 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.

\(\displaystyle \int x^6 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x) \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^5 x^6 (a+b x)^5 (A+B x)dx}{b^5 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^6 (a+b x)^5 (A+B x)dx}{a+b x}\)

\(\Big \downarrow \) 85

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{7} a^5 A x^7+\frac {1}{8} a^4 x^8 (a B+5 A b)+\frac {5}{9} a^3 b x^9 (a B+2 A b)+a^2 b^2 x^{10} (a B+A b)+\frac {1}{12} b^4 x^{12} (5 a B+A b)+\frac {5}{11} a b^3 x^{11} (2 a B+A b)+\frac {1}{13} b^5 B x^{13}\right )}{a+b x}\)

Input:

Int[x^6*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((a^5*A*x^7)/7 + (a^4*(5*A*b + a*B)*x^8)/8 
+ (5*a^3*b*(2*A*b + a*B)*x^9)/9 + a^2*b^2*(A*b + a*B)*x^10 + (5*a*b^3*(A*b 
 + 2*a*B)*x^11)/11 + (b^4*(A*b + 5*a*B)*x^12)/12 + (b^5*B*x^13)/13))/(a + 
b*x)
 

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 1187

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ 
IntPart[p]*(b/2 + c*x)^(2*FracPart[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, p}, x] && EqQ[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 = 1.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.46

method	result	size
gosper	\(\frac {x^{7} \left (5544 B \,b^{5} x^{6}+6006 A \,b^{5} x^{5}+30030 B a \,b^{4} x^{5}+32760 A a \,b^{4} x^{4}+65520 B \,a^{2} b^{3} x^{4}+72072 A \,a^{2} b^{3} x^{3}+72072 B \,a^{3} b^{2} x^{3}+80080 A \,a^{3} b^{2} x^{2}+40040 B \,a^{4} b \,x^{2}+45045 A \,a^{4} b x +9009 B \,a^{5} x +10296 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{72072 \left (b x +a \right )^{5}}\)	\(140\)
default	\(\frac {x^{7} \left (5544 B \,b^{5} x^{6}+6006 A \,b^{5} x^{5}+30030 B a \,b^{4} x^{5}+32760 A a \,b^{4} x^{4}+65520 B \,a^{2} b^{3} x^{4}+72072 A \,a^{2} b^{3} x^{3}+72072 B \,a^{3} b^{2} x^{3}+80080 A \,a^{3} b^{2} x^{2}+40040 B \,a^{4} b \,x^{2}+45045 A \,a^{4} b x +9009 B \,a^{5} x +10296 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{72072 \left (b x +a \right )^{5}}\)	\(140\)
orering	\(\frac {x^{7} \left (5544 B \,b^{5} x^{6}+6006 A \,b^{5} x^{5}+30030 B a \,b^{4} x^{5}+32760 A a \,b^{4} x^{4}+65520 B \,a^{2} b^{3} x^{4}+72072 A \,a^{2} b^{3} x^{3}+72072 B \,a^{3} b^{2} x^{3}+80080 A \,a^{3} b^{2} x^{2}+40040 B \,a^{4} b \,x^{2}+45045 A \,a^{4} b x +9009 B \,a^{5} x +10296 A \,a^{5}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}{72072 \left (b x +a \right )^{5}}\)	\(149\)
risch	\(\frac {b^{5} B \,x^{13} \sqrt {\left (b x +a \right )^{2}}}{13 b x +13 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{5}+5 B a \,b^{4}\right ) x^{12}}{12 b x +12 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A a \,b^{4}+10 B \,a^{2} b^{3}\right ) x^{11}}{11 b x +11 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (10 A \,a^{2} b^{3}+10 B \,a^{3} b^{2}\right ) x^{10}}{10 b x +10 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (10 a^{3} A \,b^{2}+5 B \,a^{4} b \right ) x^{9}}{9 b x +9 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A \,a^{4} b +B \,a^{5}\right ) x^{8}}{8 b x +8 a}+\frac {a^{5} A \,x^{7} \sqrt {\left (b x +a \right )^{2}}}{7 b x +7 a}\)	\(236\)

Input:

int(x^6*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/72072*x^7*(5544*B*b^5*x^6+6006*A*b^5*x^5+30030*B*a*b^4*x^5+32760*A*a*b^4 
*x^4+65520*B*a^2*b^3*x^4+72072*A*a^2*b^3*x^3+72072*B*a^3*b^2*x^3+80080*A*a 
^3*b^2*x^2+40040*B*a^4*b*x^2+45045*A*a^4*b*x+9009*B*a^5*x+10296*A*a^5)*((b 
*x+a)^2)^(5/2)/(b*x+a)^5
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.39 \[ \int x^6 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{13} \, B b^{5} x^{13} + \frac {1}{7} \, A a^{5} x^{7} + \frac {1}{12} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{12} + \frac {5}{11} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{11} + {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{10} + \frac {5}{9} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{9} + \frac {1}{8} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{8} \] Input:

integrate(x^6*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
 

Output:

1/13*B*b^5*x^13 + 1/7*A*a^5*x^7 + 1/12*(5*B*a*b^4 + A*b^5)*x^12 + 5/11*(2* 
B*a^2*b^3 + A*a*b^4)*x^11 + (B*a^3*b^2 + A*a^2*b^3)*x^10 + 5/9*(B*a^4*b + 
2*A*a^3*b^2)*x^9 + 1/8*(B*a^5 + 5*A*a^4*b)*x^8
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40368 vs. \(2 (228) = 456\).

Time = 1.25 (sec) , antiderivative size = 40368, normalized size of antiderivative = 133.23 \[ \int x^6 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\text {Too large to display} \] Input:

integrate(x**6*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
 

Output:

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(B*b**4*x**12/13 + x**11*(A*b* 
*6 + 53*B*a*b**5/13)/(12*b**2) + x**10*(6*A*a*b**5 + 183*B*a**2*b**4/13 - 
23*a*(A*b**6 + 53*B*a*b**5/13)/(12*b))/(11*b**2) + x**9*(15*A*a**2*b**4 + 
20*B*a**3*b**3 - 11*a**2*(A*b**6 + 53*B*a*b**5/13)/(12*b**2) - 21*a*(6*A*a 
*b**5 + 183*B*a**2*b**4/13 - 23*a*(A*b**6 + 53*B*a*b**5/13)/(12*b))/(11*b) 
)/(10*b**2) + x**8*(20*A*a**3*b**3 + 15*B*a**4*b**2 - 10*a**2*(6*A*a*b**5 
+ 183*B*a**2*b**4/13 - 23*a*(A*b**6 + 53*B*a*b**5/13)/(12*b))/(11*b**2) - 
19*a*(15*A*a**2*b**4 + 20*B*a**3*b**3 - 11*a**2*(A*b**6 + 53*B*a*b**5/13)/ 
(12*b**2) - 21*a*(6*A*a*b**5 + 183*B*a**2*b**4/13 - 23*a*(A*b**6 + 53*B*a* 
b**5/13)/(12*b))/(11*b))/(10*b))/(9*b**2) + x**7*(15*A*a**4*b**2 + 6*B*a** 
5*b - 9*a**2*(15*A*a**2*b**4 + 20*B*a**3*b**3 - 11*a**2*(A*b**6 + 53*B*a*b 
**5/13)/(12*b**2) - 21*a*(6*A*a*b**5 + 183*B*a**2*b**4/13 - 23*a*(A*b**6 + 
 53*B*a*b**5/13)/(12*b))/(11*b))/(10*b**2) - 17*a*(20*A*a**3*b**3 + 15*B*a 
**4*b**2 - 10*a**2*(6*A*a*b**5 + 183*B*a**2*b**4/13 - 23*a*(A*b**6 + 53*B* 
a*b**5/13)/(12*b))/(11*b**2) - 19*a*(15*A*a**2*b**4 + 20*B*a**3*b**3 - 11* 
a**2*(A*b**6 + 53*B*a*b**5/13)/(12*b**2) - 21*a*(6*A*a*b**5 + 183*B*a**2*b 
**4/13 - 23*a*(A*b**6 + 53*B*a*b**5/13)/(12*b))/(11*b))/(10*b))/(9*b))/(8* 
b**2) + x**6*(6*A*a**5*b + B*a**6 - 8*a**2*(20*A*a**3*b**3 + 15*B*a**4*b** 
2 - 10*a**2*(6*A*a*b**5 + 183*B*a**2*b**4/13 - 23*a*(A*b**6 + 53*B*a*b**5/ 
13)/(12*b))/(11*b**2) - 19*a*(15*A*a**2*b**4 + 20*B*a**3*b**3 - 11*a**2...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (214) = 428\).

Time = 0.05 (sec) , antiderivative size = 481, normalized size of antiderivative = 1.59 \[ \int x^6 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B x^{6}}{13 \, b^{2}} - \frac {19 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a x^{5}}{156 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A x^{5}}{12 \, b^{2}} + \frac {251 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{2} x^{4}}{1716 \, b^{4}} - \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a x^{4}}{132 \, b^{3}} - \frac {68 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{3} x^{3}}{429 \, b^{5}} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{2} x^{3}}{33 \, b^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{7} x}{6 \, b^{7}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{6} x}{6 \, b^{6}} + \frac {211 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{4} x^{2}}{1287 \, b^{6}} - \frac {16 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{3} x^{2}}{99 \, b^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{8}}{6 \, b^{8}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{7}}{6 \, b^{7}} - \frac {1709 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{5} x}{10296 \, b^{7}} + \frac {131 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{4} x}{792 \, b^{6}} + \frac {1715 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{6}}{10296 \, b^{8}} - \frac {923 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{5}}{5544 \, b^{7}} \] Input:

integrate(x^6*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
 

Output:

1/13*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*x^6/b^2 - 19/156*(b^2*x^2 + 2*a*b*x 
 + a^2)^(7/2)*B*a*x^5/b^3 + 1/12*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*x^5/b^2 
 + 251/1716*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^2*x^4/b^4 - 17/132*(b^2*x^ 
2 + 2*a*b*x + a^2)^(7/2)*A*a*x^4/b^3 - 68/429*(b^2*x^2 + 2*a*b*x + a^2)^(7 
/2)*B*a^3*x^3/b^5 + 5/33*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a^2*x^3/b^4 - 1 
/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^7*x/b^7 + 1/6*(b^2*x^2 + 2*a*b*x + 
a^2)^(5/2)*A*a^6*x/b^6 + 211/1287*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^4*x^ 
2/b^6 - 16/99*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a^3*x^2/b^5 - 1/6*(b^2*x^2 
 + 2*a*b*x + a^2)^(5/2)*B*a^8/b^8 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A* 
a^7/b^7 - 1709/10296*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^5*x/b^7 + 131/792 
*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a^4*x/b^6 + 1715/10296*(b^2*x^2 + 2*a*b 
*x + a^2)^(7/2)*B*a^6/b^8 - 923/5544*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a^5 
/b^7
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.73 \[ \int x^6 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{13} \, B b^{5} x^{13} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{12} \, B a b^{4} x^{12} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{12} \, A b^{5} x^{12} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{11} \, B a^{2} b^{3} x^{11} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{11} \, A a b^{4} x^{11} \mathrm {sgn}\left (b x + a\right ) + B a^{3} b^{2} x^{10} \mathrm {sgn}\left (b x + a\right ) + A a^{2} b^{3} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{9} \, B a^{4} b x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{9} \, A a^{3} b^{2} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{8} \, B a^{5} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{8} \, A a^{4} b x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{7} \, A a^{5} x^{7} \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (7 \, B a^{13} - 13 \, A a^{12} b\right )} \mathrm {sgn}\left (b x + a\right )}{72072 \, b^{8}} \] Input:

integrate(x^6*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
 

Output:

1/13*B*b^5*x^13*sgn(b*x + a) + 5/12*B*a*b^4*x^12*sgn(b*x + a) + 1/12*A*b^5 
*x^12*sgn(b*x + a) + 10/11*B*a^2*b^3*x^11*sgn(b*x + a) + 5/11*A*a*b^4*x^11 
*sgn(b*x + a) + B*a^3*b^2*x^10*sgn(b*x + a) + A*a^2*b^3*x^10*sgn(b*x + a) 
+ 5/9*B*a^4*b*x^9*sgn(b*x + a) + 10/9*A*a^3*b^2*x^9*sgn(b*x + a) + 1/8*B*a 
^5*x^8*sgn(b*x + a) + 5/8*A*a^4*b*x^8*sgn(b*x + a) + 1/7*A*a^5*x^7*sgn(b*x 
 + a) - 1/72072*(7*B*a^13 - 13*A*a^12*b)*sgn(b*x + a)/b^8
 

Mupad [F(-1)]

Timed out. \[ \int x^6 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\int x^6\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \] Input:

int(x^6*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
 

Output:

int(x^6*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.22 \[ \int x^6 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {x^{7} \left (924 b^{6} x^{6}+6006 a \,b^{5} x^{5}+16380 a^{2} b^{4} x^{4}+24024 a^{3} b^{3} x^{3}+20020 a^{4} b^{2} x^{2}+9009 a^{5} b x +1716 a^{6}\right )}{12012} \] Input:

int(x^6*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Output:

(x**7*(1716*a**6 + 9009*a**5*b*x + 20020*a**4*b**2*x**2 + 24024*a**3*b**3* 
x**3 + 16380*a**2*b**4*x**4 + 6006*a*b**5*x**5 + 924*b**6*x**6))/12012