\(\int \sqrt {x} (A+B x) (a^2+2 a b x+b^2 x^2)^3 \, dx\) [379]

Optimal result

Integrand size = 29, antiderivative size = 159 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{3} a^6 A x^{3/2}+\frac {2}{5} a^5 (6 A b+a B) x^{5/2}+\frac {6}{7} a^4 b (5 A b+2 a B) x^{7/2}+\frac {10}{9} a^3 b^2 (4 A b+3 a B) x^{9/2}+\frac {10}{11} a^2 b^3 (3 A b+4 a B) x^{11/2}+\frac {6}{13} a b^4 (2 A b+5 a B) x^{13/2}+\frac {2}{15} b^5 (A b+6 a B) x^{15/2}+\frac {2}{17} b^6 B x^{17/2} \] Output:

2/3*a^6*A*x^(3/2)+2/5*a^5*(6*A*b+B*a)*x^(5/2)+6/7*a^4*b*(5*A*b+2*B*a)*x^(7 
/2)+10/9*a^3*b^2*(4*A*b+3*B*a)*x^(9/2)+10/11*a^2*b^3*(3*A*b+4*B*a)*x^(11/2 
)+6/13*a*b^4*(2*A*b+5*B*a)*x^(13/2)+2/15*b^5*(A*b+6*B*a)*x^(15/2)+2/17*b^6 
*B*x^(17/2)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.81 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2 x^{3/2} \left (51051 a^6 (5 A+3 B x)+131274 a^5 b x (7 A+5 B x)+182325 a^4 b^2 x^2 (9 A+7 B x)+154700 a^3 b^3 x^3 (11 A+9 B x)+80325 a^2 b^4 x^4 (13 A+11 B x)+23562 a b^5 x^5 (15 A+13 B x)+3003 b^6 x^6 (17 A+15 B x)\right )}{765765} \] Input:

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

Output:

(2*x^(3/2)*(51051*a^6*(5*A + 3*B*x) + 131274*a^5*b*x*(7*A + 5*B*x) + 18232 
5*a^4*b^2*x^2*(9*A + 7*B*x) + 154700*a^3*b^3*x^3*(11*A + 9*B*x) + 80325*a^ 
2*b^4*x^4*(13*A + 11*B*x) + 23562*a*b^5*x^5*(15*A + 13*B*x) + 3003*b^6*x^6 
*(17*A + 15*B*x)))/765765
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, 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.

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

\(\Big \downarrow \) 1184

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 85

\(\displaystyle \int \left (a^6 A \sqrt {x}+a^5 x^{3/2} (a B+6 A b)+3 a^4 b x^{5/2} (2 a B+5 A b)+5 a^3 b^2 x^{7/2} (3 a B+4 A b)+5 a^2 b^3 x^{9/2} (4 a B+3 A b)+b^5 x^{13/2} (6 a B+A b)+3 a b^4 x^{11/2} (5 a B+2 A b)+b^6 B x^{15/2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2}{3} a^6 A x^{3/2}+\frac {2}{5} a^5 x^{5/2} (a B+6 A b)+\frac {6}{7} a^4 b x^{7/2} (2 a B+5 A b)+\frac {10}{9} a^3 b^2 x^{9/2} (3 a B+4 A b)+\frac {10}{11} a^2 b^3 x^{11/2} (4 a B+3 A b)+\frac {2}{15} b^5 x^{15/2} (6 a B+A b)+\frac {6}{13} a b^4 x^{13/2} (5 a B+2 A b)+\frac {2}{17} b^6 B x^{17/2}\)

Input:

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

Output:

(2*a^6*A*x^(3/2))/3 + (2*a^5*(6*A*b + a*B)*x^(5/2))/5 + (6*a^4*b*(5*A*b + 
2*a*B)*x^(7/2))/7 + (10*a^3*b^2*(4*A*b + 3*a*B)*x^(9/2))/9 + (10*a^2*b^3*( 
3*A*b + 4*a*B)*x^(11/2))/11 + (6*a*b^4*(2*A*b + 5*a*B)*x^(13/2))/13 + (2*b 
^5*(A*b + 6*a*B)*x^(15/2))/15 + (2*b^6*B*x^(17/2))/17
 

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 = 0.88 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.93

method	result	size
gosper	\(\frac {2 x^{\frac {3}{2}} \left (45045 b^{6} B \,x^{7}+51051 A \,b^{6} x^{6}+306306 B a \,b^{5} x^{6}+353430 A a \,b^{5} x^{5}+883575 B \,a^{2} b^{4} x^{5}+1044225 A \,a^{2} b^{4} x^{4}+1392300 B \,a^{3} b^{3} x^{4}+1701700 A \,a^{3} b^{3} x^{3}+1276275 B \,a^{4} b^{2} x^{3}+1640925 A \,a^{4} b^{2} x^{2}+656370 B \,a^{5} b \,x^{2}+918918 A \,a^{5} b x +153153 B \,a^{6} x +255255 A \,a^{6}\right )}{765765}\)	\(148\)
derivativedivides	\(\frac {2 b^{6} B \,x^{\frac {17}{2}}}{17}+\frac {2 \left (A \,b^{6}+6 B a \,b^{5}\right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (6 A a \,b^{5}+15 B \,a^{2} b^{4}\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (15 A \,a^{2} b^{4}+20 B \,a^{3} b^{3}\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (20 A \,a^{3} b^{3}+15 B \,a^{4} b^{2}\right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (15 A \,a^{4} b^{2}+6 B \,a^{5} b \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (6 A \,a^{5} b +B \,a^{6}\right ) x^{\frac {5}{2}}}{5}+\frac {2 a^{6} A \,x^{\frac {3}{2}}}{3}\)	\(148\)
default	\(\frac {2 b^{6} B \,x^{\frac {17}{2}}}{17}+\frac {2 \left (A \,b^{6}+6 B a \,b^{5}\right ) x^{\frac {15}{2}}}{15}+\frac {2 \left (6 A a \,b^{5}+15 B \,a^{2} b^{4}\right ) x^{\frac {13}{2}}}{13}+\frac {2 \left (15 A \,a^{2} b^{4}+20 B \,a^{3} b^{3}\right ) x^{\frac {11}{2}}}{11}+\frac {2 \left (20 A \,a^{3} b^{3}+15 B \,a^{4} b^{2}\right ) x^{\frac {9}{2}}}{9}+\frac {2 \left (15 A \,a^{4} b^{2}+6 B \,a^{5} b \right ) x^{\frac {7}{2}}}{7}+\frac {2 \left (6 A \,a^{5} b +B \,a^{6}\right ) x^{\frac {5}{2}}}{5}+\frac {2 a^{6} A \,x^{\frac {3}{2}}}{3}\)	\(148\)
trager	\(\frac {2 x^{\frac {3}{2}} \left (45045 b^{6} B \,x^{7}+51051 A \,b^{6} x^{6}+306306 B a \,b^{5} x^{6}+353430 A a \,b^{5} x^{5}+883575 B \,a^{2} b^{4} x^{5}+1044225 A \,a^{2} b^{4} x^{4}+1392300 B \,a^{3} b^{3} x^{4}+1701700 A \,a^{3} b^{3} x^{3}+1276275 B \,a^{4} b^{2} x^{3}+1640925 A \,a^{4} b^{2} x^{2}+656370 B \,a^{5} b \,x^{2}+918918 A \,a^{5} b x +153153 B \,a^{6} x +255255 A \,a^{6}\right )}{765765}\)	\(148\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (45045 b^{6} B \,x^{7}+51051 A \,b^{6} x^{6}+306306 B a \,b^{5} x^{6}+353430 A a \,b^{5} x^{5}+883575 B \,a^{2} b^{4} x^{5}+1044225 A \,a^{2} b^{4} x^{4}+1392300 B \,a^{3} b^{3} x^{4}+1701700 A \,a^{3} b^{3} x^{3}+1276275 B \,a^{4} b^{2} x^{3}+1640925 A \,a^{4} b^{2} x^{2}+656370 B \,a^{5} b \,x^{2}+918918 A \,a^{5} b x +153153 B \,a^{6} x +255255 A \,a^{6}\right )}{765765}\)	\(148\)
orering	\(\frac {2 x^{\frac {3}{2}} \left (45045 b^{6} B \,x^{7}+51051 A \,b^{6} x^{6}+306306 B a \,b^{5} x^{6}+353430 A a \,b^{5} x^{5}+883575 B \,a^{2} b^{4} x^{5}+1044225 A \,a^{2} b^{4} x^{4}+1392300 B \,a^{3} b^{3} x^{4}+1701700 A \,a^{3} b^{3} x^{3}+1276275 B \,a^{4} b^{2} x^{3}+1640925 A \,a^{4} b^{2} x^{2}+656370 B \,a^{5} b \,x^{2}+918918 A \,a^{5} b x +153153 B \,a^{6} x +255255 A \,a^{6}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{3}}{765765 \left (b x +a \right )^{6}}\)	\(173\)

Input:

int(x^(1/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2/765765*x^(3/2)*(45045*B*b^6*x^7+51051*A*b^6*x^6+306306*B*a*b^5*x^6+35343 
0*A*a*b^5*x^5+883575*B*a^2*b^4*x^5+1044225*A*a^2*b^4*x^4+1392300*B*a^3*b^3 
*x^4+1701700*A*a^3*b^3*x^3+1276275*B*a^4*b^2*x^3+1640925*A*a^4*b^2*x^2+656 
370*B*a^5*b*x^2+918918*A*a^5*b*x+153153*B*a^6*x+255255*A*a^6)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{765765} \, {\left (45045 \, B b^{6} x^{8} + 255255 \, A a^{6} x + 51051 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{7} + 176715 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{6} + 348075 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{5} + 425425 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{4} + 328185 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{3} + 153153 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x^{2}\right )} \sqrt {x} \] Input:

integrate(x^(1/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")
 

Output:

2/765765*(45045*B*b^6*x^8 + 255255*A*a^6*x + 51051*(6*B*a*b^5 + A*b^6)*x^7 
 + 176715*(5*B*a^2*b^4 + 2*A*a*b^5)*x^6 + 348075*(4*B*a^3*b^3 + 3*A*a^2*b^ 
4)*x^5 + 425425*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^4 + 328185*(2*B*a^5*b + 5*A* 
a^4*b^2)*x^3 + 153153*(B*a^6 + 6*A*a^5*b)*x^2)*sqrt(x)
 

Sympy [A] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.14 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2 A a^{6} x^{\frac {3}{2}}}{3} + \frac {2 B b^{6} x^{\frac {17}{2}}}{17} + \frac {2 x^{\frac {15}{2}} \left (A b^{6} + 6 B a b^{5}\right )}{15} + \frac {2 x^{\frac {13}{2}} \cdot \left (6 A a b^{5} + 15 B a^{2} b^{4}\right )}{13} + \frac {2 x^{\frac {11}{2}} \cdot \left (15 A a^{2} b^{4} + 20 B a^{3} b^{3}\right )}{11} + \frac {2 x^{\frac {9}{2}} \cdot \left (20 A a^{3} b^{3} + 15 B a^{4} b^{2}\right )}{9} + \frac {2 x^{\frac {7}{2}} \cdot \left (15 A a^{4} b^{2} + 6 B a^{5} b\right )}{7} + \frac {2 x^{\frac {5}{2}} \cdot \left (6 A a^{5} b + B a^{6}\right )}{5} \] Input:

integrate(x**(1/2)*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**3,x)
 

Output:

2*A*a**6*x**(3/2)/3 + 2*B*b**6*x**(17/2)/17 + 2*x**(15/2)*(A*b**6 + 6*B*a* 
b**5)/15 + 2*x**(13/2)*(6*A*a*b**5 + 15*B*a**2*b**4)/13 + 2*x**(11/2)*(15* 
A*a**2*b**4 + 20*B*a**3*b**3)/11 + 2*x**(9/2)*(20*A*a**3*b**3 + 15*B*a**4* 
b**2)/9 + 2*x**(7/2)*(15*A*a**4*b**2 + 6*B*a**5*b)/7 + 2*x**(5/2)*(6*A*a** 
5*b + B*a**6)/5
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{17} \, B b^{6} x^{\frac {17}{2}} + \frac {2}{3} \, A a^{6} x^{\frac {3}{2}} + \frac {2}{15} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac {15}{2}} + \frac {6}{13} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac {13}{2}} + \frac {10}{11} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac {11}{2}} + \frac {10}{9} \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{\frac {9}{2}} + \frac {6}{7} \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x^{\frac {5}{2}} \] Input:

integrate(x^(1/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")
 

Output:

2/17*B*b^6*x^(17/2) + 2/3*A*a^6*x^(3/2) + 2/15*(6*B*a*b^5 + A*b^6)*x^(15/2 
) + 6/13*(5*B*a^2*b^4 + 2*A*a*b^5)*x^(13/2) + 10/11*(4*B*a^3*b^3 + 3*A*a^2 
*b^4)*x^(11/2) + 10/9*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^(9/2) + 6/7*(2*B*a^5*b 
 + 5*A*a^4*b^2)*x^(7/2) + 2/5*(B*a^6 + 6*A*a^5*b)*x^(5/2)
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.94 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2}{17} \, B b^{6} x^{\frac {17}{2}} + \frac {4}{5} \, B a b^{5} x^{\frac {15}{2}} + \frac {2}{15} \, A b^{6} x^{\frac {15}{2}} + \frac {30}{13} \, B a^{2} b^{4} x^{\frac {13}{2}} + \frac {12}{13} \, A a b^{5} x^{\frac {13}{2}} + \frac {40}{11} \, B a^{3} b^{3} x^{\frac {11}{2}} + \frac {30}{11} \, A a^{2} b^{4} x^{\frac {11}{2}} + \frac {10}{3} \, B a^{4} b^{2} x^{\frac {9}{2}} + \frac {40}{9} \, A a^{3} b^{3} x^{\frac {9}{2}} + \frac {12}{7} \, B a^{5} b x^{\frac {7}{2}} + \frac {30}{7} \, A a^{4} b^{2} x^{\frac {7}{2}} + \frac {2}{5} \, B a^{6} x^{\frac {5}{2}} + \frac {12}{5} \, A a^{5} b x^{\frac {5}{2}} + \frac {2}{3} \, A a^{6} x^{\frac {3}{2}} \] Input:

integrate(x^(1/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")
 

Output:

2/17*B*b^6*x^(17/2) + 4/5*B*a*b^5*x^(15/2) + 2/15*A*b^6*x^(15/2) + 30/13*B 
*a^2*b^4*x^(13/2) + 12/13*A*a*b^5*x^(13/2) + 40/11*B*a^3*b^3*x^(11/2) + 30 
/11*A*a^2*b^4*x^(11/2) + 10/3*B*a^4*b^2*x^(9/2) + 40/9*A*a^3*b^3*x^(9/2) + 
 12/7*B*a^5*b*x^(7/2) + 30/7*A*a^4*b^2*x^(7/2) + 2/5*B*a^6*x^(5/2) + 12/5* 
A*a^5*b*x^(5/2) + 2/3*A*a^6*x^(3/2)
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.82 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=x^{5/2}\,\left (\frac {2\,B\,a^6}{5}+\frac {12\,A\,b\,a^5}{5}\right )+x^{15/2}\,\left (\frac {2\,A\,b^6}{15}+\frac {4\,B\,a\,b^5}{5}\right )+\frac {2\,A\,a^6\,x^{3/2}}{3}+\frac {2\,B\,b^6\,x^{17/2}}{17}+\frac {10\,a^3\,b^2\,x^{9/2}\,\left (4\,A\,b+3\,B\,a\right )}{9}+\frac {10\,a^2\,b^3\,x^{11/2}\,\left (3\,A\,b+4\,B\,a\right )}{11}+\frac {6\,a^4\,b\,x^{7/2}\,\left (5\,A\,b+2\,B\,a\right )}{7}+\frac {6\,a\,b^4\,x^{13/2}\,\left (2\,A\,b+5\,B\,a\right )}{13} \] Input:

int(x^(1/2)*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^3,x)
 

Output:

x^(5/2)*((2*B*a^6)/5 + (12*A*a^5*b)/5) + x^(15/2)*((2*A*b^6)/15 + (4*B*a*b 
^5)/5) + (2*A*a^6*x^(3/2))/3 + (2*B*b^6*x^(17/2))/17 + (10*a^3*b^2*x^(9/2) 
*(4*A*b + 3*B*a))/9 + (10*a^2*b^3*x^(11/2)*(3*A*b + 4*B*a))/11 + (6*a^4*b* 
x^(7/2)*(5*A*b + 2*B*a))/7 + (6*a*b^4*x^(13/2)*(2*A*b + 5*B*a))/13
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.50 \[ \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3 \, dx=\frac {2 \sqrt {x}\, x \left (6435 b^{7} x^{7}+51051 a \,b^{6} x^{6}+176715 a^{2} b^{5} x^{5}+348075 a^{3} b^{4} x^{4}+425425 a^{4} b^{3} x^{3}+328185 a^{5} b^{2} x^{2}+153153 a^{6} b x +36465 a^{7}\right )}{109395} \] Input:

int(x^(1/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^3,x)
 

Output:

(2*sqrt(x)*x*(36465*a**7 + 153153*a**6*b*x + 328185*a**5*b**2*x**2 + 42542 
5*a**4*b**3*x**3 + 348075*a**3*b**4*x**4 + 176715*a**2*b**5*x**5 + 51051*a 
*b**6*x**6 + 6435*b**7*x**7))/109395