3.8.52 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^3}{x^{3/2}} \, dx\) [752]

3.8.52.1 Optimal result

Integrand size = 29, antiderivative size = 151 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{3/2}} \, dx=-\frac {2 a^6 A}{\sqrt {x}}+2 a^5 (6 A b+a B) \sqrt {x}+2 a^4 b (5 A b+2 a B) x^{3/2}+2 a^3 b^2 (4 A b+3 a B) x^{5/2}+\frac {10}{7} a^2 b^3 (3 A b+4 a B) x^{7/2}+\frac {2}{3} a b^4 (2 A b+5 a B) x^{9/2}+\frac {2}{11} b^5 (A b+6 a B) x^{11/2}+\frac {2}{13} b^6 B x^{13/2} \]

output

2*a^4*b*(5*A*b+2*B*a)*x^(3/2)+2*a^3*b^2*(4*A*b+3*B*a)*x^(5/2)+10/7*a^2*b^3 
*(3*A*b+4*B*a)*x^(7/2)+2/3*a*b^4*(2*A*b+5*B*a)*x^(9/2)+2/11*b^5*(A*b+6*B*a 
)*x^(11/2)+2/13*b^6*B*x^(13/2)-2*a^6*A/x^(1/2)+2*a^5*(6*A*b+B*a)*x^(1/2)
 

3.8.52.2 Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{3/2}} \, dx=\frac {2 \left (-3003 a^6 (A-B x)+6006 a^5 b x (3 A+B x)+3003 a^4 b^2 x^2 (5 A+3 B x)+1716 a^3 b^3 x^3 (7 A+5 B x)+715 a^2 b^4 x^4 (9 A+7 B x)+182 a b^5 x^5 (11 A+9 B x)+21 b^6 x^6 (13 A+11 B x)\right )}{3003 \sqrt {x}} \]

input

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

output

(2*(-3003*a^6*(A - B*x) + 6006*a^5*b*x*(3*A + B*x) + 3003*a^4*b^2*x^2*(5*A 
 + 3*B*x) + 1716*a^3*b^3*x^3*(7*A + 5*B*x) + 715*a^2*b^4*x^4*(9*A + 7*B*x) 
 + 182*a*b^5*x^5*(11*A + 9*B*x) + 21*b^6*x^6*(13*A + 11*B*x)))/(3003*Sqrt[ 
x])
 

3.8.52.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 151, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle \frac {\int \frac {b^6 (a+b x)^6 (A+B x)}{x^{3/2}}dx}{b^6}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {(a+b x)^6 (A+B x)}{x^{3/2}}dx\)

\(\Big \downarrow \) 85

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

\(\Big \downarrow \) 2009

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

input

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

output

(-2*a^6*A)/Sqrt[x] + 2*a^5*(6*A*b + a*B)*Sqrt[x] + 2*a^4*b*(5*A*b + 2*a*B) 
*x^(3/2) + 2*a^3*b^2*(4*A*b + 3*a*B)*x^(5/2) + (10*a^2*b^3*(3*A*b + 4*a*B) 
*x^(7/2))/7 + (2*a*b^4*(2*A*b + 5*a*B)*x^(9/2))/3 + (2*b^5*(A*b + 6*a*B)*x 
^(11/2))/11 + (2*b^6*B*x^(13/2))/13
 

3.8.52.3.1 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]
 

3.8.52.4 Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.98

method	result	size
gosper	\(-\frac {2 \left (-231 b^{6} B \,x^{7}-273 A \,b^{6} x^{6}-1638 x^{6} B a \,b^{5}-2002 a A \,b^{5} x^{5}-5005 x^{5} B \,b^{4} a^{2}-6435 a^{2} A \,b^{4} x^{4}-8580 x^{4} B \,a^{3} b^{3}-12012 a^{3} A \,b^{3} x^{3}-9009 x^{3} B \,a^{4} b^{2}-15015 a^{4} A \,b^{2} x^{2}-6006 x^{2} B \,a^{5} b -18018 a^{5} A b x -3003 x B \,a^{6}+3003 A \,a^{6}\right )}{3003 \sqrt {x}}\)	\(148\)
trager	\(-\frac {2 \left (-231 b^{6} B \,x^{7}-273 A \,b^{6} x^{6}-1638 x^{6} B a \,b^{5}-2002 a A \,b^{5} x^{5}-5005 x^{5} B \,b^{4} a^{2}-6435 a^{2} A \,b^{4} x^{4}-8580 x^{4} B \,a^{3} b^{3}-12012 a^{3} A \,b^{3} x^{3}-9009 x^{3} B \,a^{4} b^{2}-15015 a^{4} A \,b^{2} x^{2}-6006 x^{2} B \,a^{5} b -18018 a^{5} A b x -3003 x B \,a^{6}+3003 A \,a^{6}\right )}{3003 \sqrt {x}}\)	\(148\)
risch	\(-\frac {2 \left (-231 b^{6} B \,x^{7}-273 A \,b^{6} x^{6}-1638 x^{6} B a \,b^{5}-2002 a A \,b^{5} x^{5}-5005 x^{5} B \,b^{4} a^{2}-6435 a^{2} A \,b^{4} x^{4}-8580 x^{4} B \,a^{3} b^{3}-12012 a^{3} A \,b^{3} x^{3}-9009 x^{3} B \,a^{4} b^{2}-15015 a^{4} A \,b^{2} x^{2}-6006 x^{2} B \,a^{5} b -18018 a^{5} A b x -3003 x B \,a^{6}+3003 A \,a^{6}\right )}{3003 \sqrt {x}}\)	\(148\)
derivativedivides	\(\frac {2 b^{6} B \,x^{\frac {13}{2}}}{13}+\frac {2 A \,b^{6} x^{\frac {11}{2}}}{11}+\frac {12 B a \,b^{5} x^{\frac {11}{2}}}{11}+\frac {4 A a \,b^{5} x^{\frac {9}{2}}}{3}+\frac {10 B \,a^{2} b^{4} x^{\frac {9}{2}}}{3}+\frac {30 A \,a^{2} b^{4} x^{\frac {7}{2}}}{7}+\frac {40 B \,a^{3} b^{3} x^{\frac {7}{2}}}{7}+8 A \,a^{3} b^{3} x^{\frac {5}{2}}+6 B \,a^{4} b^{2} x^{\frac {5}{2}}+10 A \,a^{4} b^{2} x^{\frac {3}{2}}+4 B \,a^{5} b \,x^{\frac {3}{2}}+12 A \,a^{5} b \sqrt {x}+2 B \,a^{6} \sqrt {x}-\frac {2 a^{6} A}{\sqrt {x}}\)	\(150\)
default	\(\frac {2 b^{6} B \,x^{\frac {13}{2}}}{13}+\frac {2 A \,b^{6} x^{\frac {11}{2}}}{11}+\frac {12 B a \,b^{5} x^{\frac {11}{2}}}{11}+\frac {4 A a \,b^{5} x^{\frac {9}{2}}}{3}+\frac {10 B \,a^{2} b^{4} x^{\frac {9}{2}}}{3}+\frac {30 A \,a^{2} b^{4} x^{\frac {7}{2}}}{7}+\frac {40 B \,a^{3} b^{3} x^{\frac {7}{2}}}{7}+8 A \,a^{3} b^{3} x^{\frac {5}{2}}+6 B \,a^{4} b^{2} x^{\frac {5}{2}}+10 A \,a^{4} b^{2} x^{\frac {3}{2}}+4 B \,a^{5} b \,x^{\frac {3}{2}}+12 A \,a^{5} b \sqrt {x}+2 B \,a^{6} \sqrt {x}-\frac {2 a^{6} A}{\sqrt {x}}\)	\(150\)

input

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

output

-2/3003*(-231*B*b^6*x^7-273*A*b^6*x^6-1638*B*a*b^5*x^6-2002*A*a*b^5*x^5-50 
05*B*a^2*b^4*x^5-6435*A*a^2*b^4*x^4-8580*B*a^3*b^3*x^4-12012*A*a^3*b^3*x^3 
-9009*B*a^4*b^2*x^3-15015*A*a^4*b^2*x^2-6006*B*a^5*b*x^2-18018*A*a^5*b*x-3 
003*B*a^6*x+3003*A*a^6)/x^(1/2)
 

3.8.52.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{3/2}} \, dx=\frac {2 \, {\left (231 \, B b^{6} x^{7} - 3003 \, A a^{6} + 273 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 1001 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 2145 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 3003 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 3003 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 3003 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{3003 \, \sqrt {x}} \]

input

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

output

2/3003*(231*B*b^6*x^7 - 3003*A*a^6 + 273*(6*B*a*b^5 + A*b^6)*x^6 + 1001*(5 
*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 2145*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 3003* 
(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 + 3003*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 + 300 
3*(B*a^6 + 6*A*a^5*b)*x)/sqrt(x)
 

3.8.52.6 Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.35 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{3/2}} \, dx=- \frac {2 A a^{6}}{\sqrt {x}} + 12 A a^{5} b \sqrt {x} + 10 A a^{4} b^{2} x^{\frac {3}{2}} + 8 A a^{3} b^{3} x^{\frac {5}{2}} + \frac {30 A a^{2} b^{4} x^{\frac {7}{2}}}{7} + \frac {4 A a b^{5} x^{\frac {9}{2}}}{3} + \frac {2 A b^{6} x^{\frac {11}{2}}}{11} + 2 B a^{6} \sqrt {x} + 4 B a^{5} b x^{\frac {3}{2}} + 6 B a^{4} b^{2} x^{\frac {5}{2}} + \frac {40 B a^{3} b^{3} x^{\frac {7}{2}}}{7} + \frac {10 B a^{2} b^{4} x^{\frac {9}{2}}}{3} + \frac {12 B a b^{5} x^{\frac {11}{2}}}{11} + \frac {2 B b^{6} x^{\frac {13}{2}}}{13} \]

input

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

output

-2*A*a**6/sqrt(x) + 12*A*a**5*b*sqrt(x) + 10*A*a**4*b**2*x**(3/2) + 8*A*a* 
*3*b**3*x**(5/2) + 30*A*a**2*b**4*x**(7/2)/7 + 4*A*a*b**5*x**(9/2)/3 + 2*A 
*b**6*x**(11/2)/11 + 2*B*a**6*sqrt(x) + 4*B*a**5*b*x**(3/2) + 6*B*a**4*b** 
2*x**(5/2) + 40*B*a**3*b**3*x**(7/2)/7 + 10*B*a**2*b**4*x**(9/2)/3 + 12*B* 
a*b**5*x**(11/2)/11 + 2*B*b**6*x**(13/2)/13
 

3.8.52.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{3/2}} \, dx=\frac {2}{13} \, B b^{6} x^{\frac {13}{2}} - \frac {2 \, A a^{6}}{\sqrt {x}} + \frac {2}{11} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac {11}{2}} + \frac {2}{3} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac {9}{2}} + \frac {10}{7} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac {7}{2}} + 2 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{\frac {5}{2}} + 2 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{\frac {3}{2}} + 2 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} \sqrt {x} \]

input

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

output

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

3.8.52.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{3/2}} \, dx=\frac {2}{13} \, B b^{6} x^{\frac {13}{2}} + \frac {12}{11} \, B a b^{5} x^{\frac {11}{2}} + \frac {2}{11} \, A b^{6} x^{\frac {11}{2}} + \frac {10}{3} \, B a^{2} b^{4} x^{\frac {9}{2}} + \frac {4}{3} \, A a b^{5} x^{\frac {9}{2}} + \frac {40}{7} \, B a^{3} b^{3} x^{\frac {7}{2}} + \frac {30}{7} \, A a^{2} b^{4} x^{\frac {7}{2}} + 6 \, B a^{4} b^{2} x^{\frac {5}{2}} + 8 \, A a^{3} b^{3} x^{\frac {5}{2}} + 4 \, B a^{5} b x^{\frac {3}{2}} + 10 \, A a^{4} b^{2} x^{\frac {3}{2}} + 2 \, B a^{6} \sqrt {x} + 12 \, A a^{5} b \sqrt {x} - \frac {2 \, A a^{6}}{\sqrt {x}} \]

input

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

output

2/13*B*b^6*x^(13/2) + 12/11*B*a*b^5*x^(11/2) + 2/11*A*b^6*x^(11/2) + 10/3* 
B*a^2*b^4*x^(9/2) + 4/3*A*a*b^5*x^(9/2) + 40/7*B*a^3*b^3*x^(7/2) + 30/7*A* 
a^2*b^4*x^(7/2) + 6*B*a^4*b^2*x^(5/2) + 8*A*a^3*b^3*x^(5/2) + 4*B*a^5*b*x^ 
(3/2) + 10*A*a^4*b^2*x^(3/2) + 2*B*a^6*sqrt(x) + 12*A*a^5*b*sqrt(x) - 2*A* 
a^6/sqrt(x)
 

3.8.52.9 Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{x^{3/2}} \, dx=\sqrt {x}\,\left (2\,B\,a^6+12\,A\,b\,a^5\right )+x^{11/2}\,\left (\frac {2\,A\,b^6}{11}+\frac {12\,B\,a\,b^5}{11}\right )-\frac {2\,A\,a^6}{\sqrt {x}}+\frac {2\,B\,b^6\,x^{13/2}}{13}+2\,a^3\,b^2\,x^{5/2}\,\left (4\,A\,b+3\,B\,a\right )+\frac {10\,a^2\,b^3\,x^{7/2}\,\left (3\,A\,b+4\,B\,a\right )}{7}+2\,a^4\,b\,x^{3/2}\,\left (5\,A\,b+2\,B\,a\right )+\frac {2\,a\,b^4\,x^{9/2}\,\left (2\,A\,b+5\,B\,a\right )}{3} \]

input

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

output

x^(1/2)*(2*B*a^6 + 12*A*a^5*b) + x^(11/2)*((2*A*b^6)/11 + (12*B*a*b^5)/11) 
 - (2*A*a^6)/x^(1/2) + (2*B*b^6*x^(13/2))/13 + 2*a^3*b^2*x^(5/2)*(4*A*b + 
3*B*a) + (10*a^2*b^3*x^(7/2)*(3*A*b + 4*B*a))/7 + 2*a^4*b*x^(3/2)*(5*A*b + 
 2*B*a) + (2*a*b^4*x^(9/2)*(2*A*b + 5*B*a))/3