Integrand size = 27, antiderivative size = 99 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{4} a^4 A x^4+\frac {1}{5} a^3 (4 A b+a B) x^5+\frac {1}{3} a^2 b (3 A b+2 a B) x^6+\frac {2}{7} a b^2 (2 A b+3 a B) x^7+\frac {1}{8} b^3 (A b+4 a B) x^8+\frac {1}{9} b^4 B x^9 \] Output:
1/4*a^4*A*x^4+1/5*a^3*(4*A*b+B*a)*x^5+1/3*a^2*b*(3*A*b+2*B*a)*x^6+2/7*a*b^ 2*(2*A*b+3*B*a)*x^7+1/8*b^3*(A*b+4*B*a)*x^8+1/9*b^4*B*x^9
Time = 0.01 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{4} a^4 A x^4+\frac {1}{5} a^3 (4 A b+a B) x^5+\frac {1}{3} a^2 b (3 A b+2 a B) x^6+\frac {2}{7} a b^2 (2 A b+3 a B) x^7+\frac {1}{8} b^3 (A b+4 a B) x^8+\frac {1}{9} b^4 B x^9 \] Input:
Integrate[x^3*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(a^4*A*x^4)/4 + (a^3*(4*A*b + a*B)*x^5)/5 + (a^2*b*(3*A*b + 2*a*B)*x^6)/3 + (2*a*b^2*(2*A*b + 3*a*B)*x^7)/7 + (b^3*(A*b + 4*a*B)*x^8)/8 + (b^4*B*x^9 )/9
Time = 0.46 (sec) , antiderivative size = 99, 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.
\(\displaystyle \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x) \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int b^4 x^3 (a+b x)^4 (A+B x)dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int x^3 (a+b x)^4 (A+B x)dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (a^4 A x^3+a^3 x^4 (a B+4 A b)+2 a^2 b x^5 (2 a B+3 A b)+b^3 x^7 (4 a B+A b)+2 a b^2 x^6 (3 a B+2 A b)+b^4 B x^8\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} a^4 A x^4+\frac {1}{5} a^3 x^5 (a B+4 A b)+\frac {1}{3} a^2 b x^6 (2 a B+3 A b)+\frac {1}{8} b^3 x^8 (4 a B+A b)+\frac {2}{7} a b^2 x^7 (3 a B+2 A b)+\frac {1}{9} b^4 B x^9\) |
Input:
Int[x^3*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(a^4*A*x^4)/4 + (a^3*(4*A*b + a*B)*x^5)/5 + (a^2*b*(3*A*b + 2*a*B)*x^6)/3 + (2*a*b^2*(2*A*b + 3*a*B)*x^7)/7 + (b^3*(A*b + 4*a*B)*x^8)/8 + (b^4*B*x^9 )/9
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
Time = 0.96 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98
method | result | size |
norman | \(\frac {B \,b^{4} x^{9}}{9}+\left (\frac {1}{8} A \,b^{4}+\frac {1}{2} B a \,b^{3}\right ) x^{8}+\left (\frac {4}{7} A a \,b^{3}+\frac {6}{7} B \,a^{2} b^{2}\right ) x^{7}+\left (a^{2} A \,b^{2}+\frac {2}{3} B \,a^{3} b \right ) x^{6}+\left (\frac {4}{5} A \,a^{3} b +\frac {1}{5} a^{4} B \right ) x^{5}+\frac {a^{4} A \,x^{4}}{4}\) | \(97\) |
gosper | \(\frac {x^{4} \left (280 b^{4} B \,x^{5}+315 A \,b^{4} x^{4}+1260 B a \,b^{3} x^{4}+1440 A a \,b^{3} x^{3}+2160 B \,a^{2} b^{2} x^{3}+2520 A \,a^{2} b^{2} x^{2}+1680 B \,a^{3} b \,x^{2}+2016 A \,a^{3} b x +504 a^{4} B x +630 a^{4} A \right )}{2520}\) | \(100\) |
default | \(\frac {B \,b^{4} x^{9}}{9}+\frac {\left (A \,b^{4}+4 B a \,b^{3}\right ) x^{8}}{8}+\frac {\left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) x^{7}}{7}+\frac {\left (6 a^{2} A \,b^{2}+4 B \,a^{3} b \right ) x^{6}}{6}+\frac {\left (4 A \,a^{3} b +a^{4} B \right ) x^{5}}{5}+\frac {a^{4} A \,x^{4}}{4}\) | \(100\) |
risch | \(\frac {1}{9} B \,b^{4} x^{9}+\frac {1}{8} b^{4} A \,x^{8}+\frac {1}{2} x^{8} B a \,b^{3}+\frac {4}{7} x^{7} A a \,b^{3}+\frac {6}{7} x^{7} B \,a^{2} b^{2}+x^{6} a^{2} A \,b^{2}+\frac {2}{3} x^{6} B \,a^{3} b +\frac {4}{5} x^{5} A \,a^{3} b +\frac {1}{5} x^{5} a^{4} B +\frac {1}{4} a^{4} A \,x^{4}\) | \(101\) |
parallelrisch | \(\frac {1}{9} B \,b^{4} x^{9}+\frac {1}{8} b^{4} A \,x^{8}+\frac {1}{2} x^{8} B a \,b^{3}+\frac {4}{7} x^{7} A a \,b^{3}+\frac {6}{7} x^{7} B \,a^{2} b^{2}+x^{6} a^{2} A \,b^{2}+\frac {2}{3} x^{6} B \,a^{3} b +\frac {4}{5} x^{5} A \,a^{3} b +\frac {1}{5} x^{5} a^{4} B +\frac {1}{4} a^{4} A \,x^{4}\) | \(101\) |
orering | \(\frac {x^{4} \left (280 b^{4} B \,x^{5}+315 A \,b^{4} x^{4}+1260 B a \,b^{3} x^{4}+1440 A a \,b^{3} x^{3}+2160 B \,a^{2} b^{2} x^{3}+2520 A \,a^{2} b^{2} x^{2}+1680 B \,a^{3} b \,x^{2}+2016 A \,a^{3} b x +504 a^{4} B x +630 a^{4} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}{2520 \left (b x +a \right )^{4}}\) | \(125\) |
Input:
int(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
1/9*B*b^4*x^9+(1/8*A*b^4+1/2*B*a*b^3)*x^8+(4/7*A*a*b^3+6/7*B*a^2*b^2)*x^7+ (a^2*A*b^2+2/3*B*a^3*b)*x^6+(4/5*A*a^3*b+1/5*a^4*B)*x^5+1/4*a^4*A*x^4
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{9} \, B b^{4} x^{9} + \frac {1}{4} \, A a^{4} x^{4} + \frac {1}{8} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{8} + \frac {2}{7} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{7} + \frac {1}{3} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{5} \] Input:
integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
1/9*B*b^4*x^9 + 1/4*A*a^4*x^4 + 1/8*(4*B*a*b^3 + A*b^4)*x^8 + 2/7*(3*B*a^2 *b^2 + 2*A*a*b^3)*x^7 + 1/3*(2*B*a^3*b + 3*A*a^2*b^2)*x^6 + 1/5*(B*a^4 + 4 *A*a^3*b)*x^5
Time = 0.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {A a^{4} x^{4}}{4} + \frac {B b^{4} x^{9}}{9} + x^{8} \left (\frac {A b^{4}}{8} + \frac {B a b^{3}}{2}\right ) + x^{7} \cdot \left (\frac {4 A a b^{3}}{7} + \frac {6 B a^{2} b^{2}}{7}\right ) + x^{6} \left (A a^{2} b^{2} + \frac {2 B a^{3} b}{3}\right ) + x^{5} \cdot \left (\frac {4 A a^{3} b}{5} + \frac {B a^{4}}{5}\right ) \] Input:
integrate(x**3*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
A*a**4*x**4/4 + B*b**4*x**9/9 + x**8*(A*b**4/8 + B*a*b**3/2) + x**7*(4*A*a *b**3/7 + 6*B*a**2*b**2/7) + x**6*(A*a**2*b**2 + 2*B*a**3*b/3) + x**5*(4*A *a**3*b/5 + B*a**4/5)
Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{9} \, B b^{4} x^{9} + \frac {1}{4} \, A a^{4} x^{4} + \frac {1}{8} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{8} + \frac {2}{7} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{7} + \frac {1}{3} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{5} \] Input:
integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
1/9*B*b^4*x^9 + 1/4*A*a^4*x^4 + 1/8*(4*B*a*b^3 + A*b^4)*x^8 + 2/7*(3*B*a^2 *b^2 + 2*A*a*b^3)*x^7 + 1/3*(2*B*a^3*b + 3*A*a^2*b^2)*x^6 + 1/5*(B*a^4 + 4 *A*a^3*b)*x^5
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{9} \, B b^{4} x^{9} + \frac {1}{2} \, B a b^{3} x^{8} + \frac {1}{8} \, A b^{4} x^{8} + \frac {6}{7} \, B a^{2} b^{2} x^{7} + \frac {4}{7} \, A a b^{3} x^{7} + \frac {2}{3} \, B a^{3} b x^{6} + A a^{2} b^{2} x^{6} + \frac {1}{5} \, B a^{4} x^{5} + \frac {4}{5} \, A a^{3} b x^{5} + \frac {1}{4} \, A a^{4} x^{4} \] Input:
integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
1/9*B*b^4*x^9 + 1/2*B*a*b^3*x^8 + 1/8*A*b^4*x^8 + 6/7*B*a^2*b^2*x^7 + 4/7* A*a*b^3*x^7 + 2/3*B*a^3*b*x^6 + A*a^2*b^2*x^6 + 1/5*B*a^4*x^5 + 4/5*A*a^3* b*x^5 + 1/4*A*a^4*x^4
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.92 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=x^5\,\left (\frac {B\,a^4}{5}+\frac {4\,A\,b\,a^3}{5}\right )+x^8\,\left (\frac {A\,b^4}{8}+\frac {B\,a\,b^3}{2}\right )+\frac {A\,a^4\,x^4}{4}+\frac {B\,b^4\,x^9}{9}+\frac {a^2\,b\,x^6\,\left (3\,A\,b+2\,B\,a\right )}{3}+\frac {2\,a\,b^2\,x^7\,\left (2\,A\,b+3\,B\,a\right )}{7} \] Input:
int(x^3*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
x^5*((B*a^4)/5 + (4*A*a^3*b)/5) + x^8*((A*b^4)/8 + (B*a*b^3)/2) + (A*a^4*x ^4)/4 + (B*b^4*x^9)/9 + (a^2*b*x^6*(3*A*b + 2*B*a))/3 + (2*a*b^2*x^7*(2*A* b + 3*B*a))/7
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {x^{4} \left (56 b^{5} x^{5}+315 a \,b^{4} x^{4}+720 a^{2} b^{3} x^{3}+840 a^{3} b^{2} x^{2}+504 a^{4} b x +126 a^{5}\right )}{504} \] Input:
int(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
(x**4*(126*a**5 + 504*a**4*b*x + 840*a**3*b**2*x**2 + 720*a**2*b**3*x**3 + 315*a*b**4*x**4 + 56*b**5*x**5))/504