\(\int x^3 (A+B x) (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\) [315]

Optimal result

Integrand size = 29, antiderivative size = 212 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=-\frac {a^3 (A b-a B) (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^5}+\frac {a^2 (3 A b-4 a B) (a+b x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{7 b^5}-\frac {3 a (A b-2 a B) (a+b x)^7 \sqrt {a^2+2 a b x+b^2 x^2}}{8 b^5}+\frac {(A b-4 a B) (a+b x)^8 \sqrt {a^2+2 a b x+b^2 x^2}}{9 b^5}+\frac {B (a+b x)^9 \sqrt {a^2+2 a b x+b^2 x^2}}{10 b^5} \] Output:

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

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.59 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {x^4 \sqrt {(a+b x)^2} \left (126 a^5 (5 A+4 B x)+420 a^4 b x (6 A+5 B x)+600 a^3 b^2 x^2 (7 A+6 B x)+450 a^2 b^3 x^3 (8 A+7 B x)+175 a b^4 x^4 (9 A+8 B x)+28 b^5 x^5 (10 A+9 B x)\right )}{2520 (a+b x)} \] Input:

Integrate[x^3*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(x^4*Sqrt[(a + b*x)^2]*(126*a^5*(5*A + 4*B*x) + 420*a^4*b*x*(6*A + 5*B*x) 
+ 600*a^3*b^2*x^2*(7*A + 6*B*x) + 450*a^2*b^3*x^3*(8*A + 7*B*x) + 175*a*b^ 
4*x^4*(9*A + 8*B*x) + 28*b^5*x^5*(10*A + 9*B*x)))/(2520*(a + b*x))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.66, 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.

\(\displaystyle \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (A+B x) \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^5 x^3 (a+b x)^5 (A+B x)dx}{b^5 (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^3 (a+b x)^5 (A+B x)dx}{a+b x}\)

\(\Big \downarrow \) 85

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^3*(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]*(-1/6*(a^3*(A*b - a*B)*(a + b*x)^6)/b^5 + ( 
a^2*(3*A*b - 4*a*B)*(a + b*x)^7)/(7*b^5) - (3*a*(A*b - 2*a*B)*(a + b*x)^8) 
/(8*b^5) + ((A*b - 4*a*B)*(a + b*x)^9)/(9*b^5) + (B*(a + b*x)^10)/(10*b^5) 
))/(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 = 1.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.66

method	result	size
gosper	\(\frac {x^{4} \left (252 B \,b^{5} x^{6}+280 A \,b^{5} x^{5}+1400 B a \,b^{4} x^{5}+1575 A a \,b^{4} x^{4}+3150 B \,a^{2} b^{3} x^{4}+3600 A \,a^{2} b^{3} x^{3}+3600 B \,a^{3} b^{2} x^{3}+4200 A \,a^{3} b^{2} x^{2}+2100 B \,a^{4} b \,x^{2}+2520 A \,a^{4} b x +504 B \,a^{5} x +630 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{2520 \left (b x +a \right )^{5}}\)	\(140\)
default	\(\frac {x^{4} \left (252 B \,b^{5} x^{6}+280 A \,b^{5} x^{5}+1400 B a \,b^{4} x^{5}+1575 A a \,b^{4} x^{4}+3150 B \,a^{2} b^{3} x^{4}+3600 A \,a^{2} b^{3} x^{3}+3600 B \,a^{3} b^{2} x^{3}+4200 A \,a^{3} b^{2} x^{2}+2100 B \,a^{4} b \,x^{2}+2520 A \,a^{4} b x +504 B \,a^{5} x +630 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{2520 \left (b x +a \right )^{5}}\)	\(140\)
orering	\(\frac {x^{4} \left (252 B \,b^{5} x^{6}+280 A \,b^{5} x^{5}+1400 B a \,b^{4} x^{5}+1575 A a \,b^{4} x^{4}+3150 B \,a^{2} b^{3} x^{4}+3600 A \,a^{2} b^{3} x^{3}+3600 B \,a^{3} b^{2} x^{3}+4200 A \,a^{3} b^{2} x^{2}+2100 B \,a^{4} b \,x^{2}+2520 A \,a^{4} b x +504 B \,a^{5} x +630 A \,a^{5}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}{2520 \left (b x +a \right )^{5}}\)	\(149\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{5} B \,x^{10}}{10 b x +10 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{5}+5 B a \,b^{4}\right ) x^{9}}{9 b x +9 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A a \,b^{4}+10 B \,a^{2} b^{3}\right ) x^{8}}{8 b x +8 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (10 A \,a^{2} b^{3}+10 B \,a^{3} b^{2}\right ) x^{7}}{7 b x +7 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (10 a^{3} A \,b^{2}+5 B \,a^{4} b \right ) x^{6}}{6 b x +6 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (5 A \,a^{4} b +B \,a^{5}\right ) x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, A \,a^{5} x^{4}}{4 b x +4 a}\)	\(236\)

Input:

int(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/2520*x^4*(252*B*b^5*x^6+280*A*b^5*x^5+1400*B*a*b^4*x^5+1575*A*a*b^4*x^4+ 
3150*B*a^2*b^3*x^4+3600*A*a^2*b^3*x^3+3600*B*a^3*b^2*x^3+4200*A*a^3*b^2*x^ 
2+2100*B*a^4*b*x^2+2520*A*a^4*b*x+504*B*a^5*x+630*A*a^5)*((b*x+a)^2)^(5/2) 
/(b*x+a)^5
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.56 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{10} \, B b^{5} x^{10} + \frac {1}{4} \, A a^{5} x^{4} + \frac {1}{9} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{9} + \frac {5}{8} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} + \frac {10}{7} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{7} + \frac {5}{6} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{5} \] Input:

integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 9493 vs. \(2 (156) = 312\).

Time = 0.98 (sec) , antiderivative size = 9493, normalized size of antiderivative = 44.78 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\text {Too large to display} \] Input:

integrate(x**3*(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**9/10 + x**8*(A*b**6 
 + 41*B*a*b**5/10)/(9*b**2) + x**7*(6*A*a*b**5 + 141*B*a**2*b**4/10 - 17*a 
*(A*b**6 + 41*B*a*b**5/10)/(9*b))/(8*b**2) + x**6*(15*A*a**2*b**4 + 20*B*a 
**3*b**3 - 8*a**2*(A*b**6 + 41*B*a*b**5/10)/(9*b**2) - 15*a*(6*A*a*b**5 + 
141*B*a**2*b**4/10 - 17*a*(A*b**6 + 41*B*a*b**5/10)/(9*b))/(8*b))/(7*b**2) 
 + x**5*(20*A*a**3*b**3 + 15*B*a**4*b**2 - 7*a**2*(6*A*a*b**5 + 141*B*a**2 
*b**4/10 - 17*a*(A*b**6 + 41*B*a*b**5/10)/(9*b))/(8*b**2) - 13*a*(15*A*a** 
2*b**4 + 20*B*a**3*b**3 - 8*a**2*(A*b**6 + 41*B*a*b**5/10)/(9*b**2) - 15*a 
*(6*A*a*b**5 + 141*B*a**2*b**4/10 - 17*a*(A*b**6 + 41*B*a*b**5/10)/(9*b))/ 
(8*b))/(7*b))/(6*b**2) + x**4*(15*A*a**4*b**2 + 6*B*a**5*b - 6*a**2*(15*A* 
a**2*b**4 + 20*B*a**3*b**3 - 8*a**2*(A*b**6 + 41*B*a*b**5/10)/(9*b**2) - 1 
5*a*(6*A*a*b**5 + 141*B*a**2*b**4/10 - 17*a*(A*b**6 + 41*B*a*b**5/10)/(9*b 
))/(8*b))/(7*b**2) - 11*a*(20*A*a**3*b**3 + 15*B*a**4*b**2 - 7*a**2*(6*A*a 
*b**5 + 141*B*a**2*b**4/10 - 17*a*(A*b**6 + 41*B*a*b**5/10)/(9*b))/(8*b**2 
) - 13*a*(15*A*a**2*b**4 + 20*B*a**3*b**3 - 8*a**2*(A*b**6 + 41*B*a*b**5/1 
0)/(9*b**2) - 15*a*(6*A*a*b**5 + 141*B*a**2*b**4/10 - 17*a*(A*b**6 + 41*B* 
a*b**5/10)/(9*b))/(8*b))/(7*b))/(6*b))/(5*b**2) + x**3*(6*A*a**5*b + B*a** 
6 - 5*a**2*(20*A*a**3*b**3 + 15*B*a**4*b**2 - 7*a**2*(6*A*a*b**5 + 141*B*a 
**2*b**4/10 - 17*a*(A*b**6 + 41*B*a*b**5/10)/(9*b))/(8*b**2) - 13*a*(15*A* 
a**2*b**4 + 20*B*a**3*b**3 - 8*a**2*(A*b**6 + 41*B*a*b**5/10)/(9*b**2) ...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (149) = 298\).

Time = 0.04 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.42 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B x^{3}}{10 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{4} x}{6 \, b^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{3} x}{6 \, b^{3}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a x^{2}}{90 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A x^{2}}{9 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{5}}{6 \, b^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{4}}{6 \, b^{4}} + \frac {29 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{2} x}{180 \, b^{4}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a x}{72 \, b^{3}} - \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B a^{3}}{1260 \, b^{5}} + \frac {83 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A a^{2}}{504 \, b^{4}} \] Input:

integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
 

Output:

1/10*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*x^3/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + 
a^2)^(5/2)*B*a^4*x/b^4 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^3*x/b^3 - 
 13/90*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a*x^2/b^3 + 1/9*(b^2*x^2 + 2*a*b* 
x + a^2)^(7/2)*A*x^2/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^5/b^5 - 
 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^4/b^4 + 29/180*(b^2*x^2 + 2*a*b*x 
 + a^2)^(7/2)*B*a^2*x/b^4 - 11/72*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*a*x/b^ 
3 - 209/1260*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*a^3/b^5 + 83/504*(b^2*x^2 + 
 2*a*b*x + a^2)^(7/2)*A*a^2/b^4
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.04 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{10} \, B b^{5} x^{10} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{9} \, B a b^{4} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{9} \, A b^{5} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, B a^{2} b^{3} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{8} \, A a b^{4} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, B a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{7} \, A a^{2} b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, B a^{4} b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, A a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B a^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + A a^{4} b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A a^{5} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (2 \, B a^{10} - 5 \, A a^{9} b\right )} \mathrm {sgn}\left (b x + a\right )}{2520 \, b^{5}} \] Input:

integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
 

Output:

1/10*B*b^5*x^10*sgn(b*x + a) + 5/9*B*a*b^4*x^9*sgn(b*x + a) + 1/9*A*b^5*x^ 
9*sgn(b*x + a) + 5/4*B*a^2*b^3*x^8*sgn(b*x + a) + 5/8*A*a*b^4*x^8*sgn(b*x 
+ a) + 10/7*B*a^3*b^2*x^7*sgn(b*x + a) + 10/7*A*a^2*b^3*x^7*sgn(b*x + a) + 
 5/6*B*a^4*b*x^6*sgn(b*x + a) + 5/3*A*a^3*b^2*x^6*sgn(b*x + a) + 1/5*B*a^5 
*x^5*sgn(b*x + a) + A*a^4*b*x^5*sgn(b*x + a) + 1/4*A*a^5*x^4*sgn(b*x + a) 
+ 1/2520*(2*B*a^10 - 5*A*a^9*b)*sgn(b*x + a)/b^5
 

Mupad [F(-1)]

Timed out. \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\int x^3\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \] Input:

int(x^3*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
 

Output:

int(x^3*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.32 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {x^{4} \left (84 b^{6} x^{6}+560 a \,b^{5} x^{5}+1575 a^{2} b^{4} x^{4}+2400 a^{3} b^{3} x^{3}+2100 a^{4} b^{2} x^{2}+1008 a^{5} b x +210 a^{6}\right )}{840} \] Input:

int(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Output:

(x**4*(210*a**6 + 1008*a**5*b*x + 2100*a**4*b**2*x**2 + 2400*a**3*b**3*x** 
3 + 1575*a**2*b**4*x**4 + 560*a*b**5*x**5 + 84*b**6*x**6))/840