Integrand size = 27, antiderivative size = 99 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{5} a^4 A x^5+\frac {1}{6} a^3 (4 A b+a B) x^6+\frac {2}{7} a^2 b (3 A b+2 a B) x^7+\frac {1}{4} a b^2 (2 A b+3 a B) x^8+\frac {1}{9} b^3 (A b+4 a B) x^9+\frac {1}{10} b^4 B x^{10} \] Output:
1/5*a^4*A*x^5+1/6*a^3*(4*A*b+B*a)*x^6+2/7*a^2*b*(3*A*b+2*B*a)*x^7+1/4*a*b^ 2*(2*A*b+3*B*a)*x^8+1/9*b^3*(A*b+4*B*a)*x^9+1/10*b^4*B*x^10
Time = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{5} a^4 A x^5+\frac {1}{6} a^3 (4 A b+a B) x^6+\frac {2}{7} a^2 b (3 A b+2 a B) x^7+\frac {1}{4} a b^2 (2 A b+3 a B) x^8+\frac {1}{9} b^3 (A b+4 a B) x^9+\frac {1}{10} b^4 B x^{10} \] Input:
Integrate[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(a^4*A*x^5)/5 + (a^3*(4*A*b + a*B)*x^6)/6 + (2*a^2*b*(3*A*b + 2*a*B)*x^7)/ 7 + (a*b^2*(2*A*b + 3*a*B)*x^8)/4 + (b^3*(A*b + 4*a*B)*x^9)/9 + (b^4*B*x^1 0)/10
Time = 0.51 (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.
\(\displaystyle \int x^4 \left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x) \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int b^4 x^4 (a+b x)^4 (A+B x)dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int x^4 (a+b x)^4 (A+B x)dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (a^4 A x^4+a^3 x^5 (a B+4 A b)+2 a^2 b x^6 (2 a B+3 A b)+b^3 x^8 (4 a B+A b)+2 a b^2 x^7 (3 a B+2 A b)+b^4 B x^9\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} a^4 A x^5+\frac {1}{6} a^3 x^6 (a B+4 A b)+\frac {2}{7} a^2 b x^7 (2 a B+3 A b)+\frac {1}{9} b^3 x^9 (4 a B+A b)+\frac {1}{4} a b^2 x^8 (3 a B+2 A b)+\frac {1}{10} b^4 B x^{10}\) |
Input:
Int[x^4*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(a^4*A*x^5)/5 + (a^3*(4*A*b + a*B)*x^6)/6 + (2*a^2*b*(3*A*b + 2*a*B)*x^7)/ 7 + (a*b^2*(2*A*b + 3*a*B)*x^8)/4 + (b^3*(A*b + 4*a*B)*x^9)/9 + (b^4*B*x^1 0)/10
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 = 0.97 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.99
method | result | size |
norman | \(\frac {b^{4} B \,x^{10}}{10}+\left (\frac {1}{9} A \,b^{4}+\frac {4}{9} B a \,b^{3}\right ) x^{9}+\left (\frac {1}{2} A a \,b^{3}+\frac {3}{4} B \,a^{2} b^{2}\right ) x^{8}+\left (\frac {6}{7} a^{2} A \,b^{2}+\frac {4}{7} B \,a^{3} b \right ) x^{7}+\left (\frac {2}{3} A \,a^{3} b +\frac {1}{6} a^{4} B \right ) x^{6}+\frac {a^{4} A \,x^{5}}{5}\) | \(98\) |
gosper | \(\frac {x^{5} \left (126 b^{4} B \,x^{5}+140 A \,b^{4} x^{4}+560 B a \,b^{3} x^{4}+630 A a \,b^{3} x^{3}+945 B \,a^{2} b^{2} x^{3}+1080 A \,a^{2} b^{2} x^{2}+720 B \,a^{3} b \,x^{2}+840 A \,a^{3} b x +210 a^{4} B x +252 a^{4} A \right )}{1260}\) | \(100\) |
default | \(\frac {b^{4} B \,x^{10}}{10}+\frac {\left (A \,b^{4}+4 B a \,b^{3}\right ) x^{9}}{9}+\frac {\left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) x^{8}}{8}+\frac {\left (6 a^{2} A \,b^{2}+4 B \,a^{3} b \right ) x^{7}}{7}+\frac {\left (4 A \,a^{3} b +a^{4} B \right ) x^{6}}{6}+\frac {a^{4} A \,x^{5}}{5}\) | \(100\) |
risch | \(\frac {1}{10} b^{4} B \,x^{10}+\frac {1}{9} x^{9} A \,b^{4}+\frac {4}{9} x^{9} B a \,b^{3}+\frac {1}{2} x^{8} A a \,b^{3}+\frac {3}{4} x^{8} B \,a^{2} b^{2}+\frac {6}{7} x^{7} a^{2} A \,b^{2}+\frac {4}{7} x^{7} B \,a^{3} b +\frac {2}{3} x^{6} A \,a^{3} b +\frac {1}{6} x^{6} a^{4} B +\frac {1}{5} a^{4} A \,x^{5}\) | \(102\) |
parallelrisch | \(\frac {1}{10} b^{4} B \,x^{10}+\frac {1}{9} x^{9} A \,b^{4}+\frac {4}{9} x^{9} B a \,b^{3}+\frac {1}{2} x^{8} A a \,b^{3}+\frac {3}{4} x^{8} B \,a^{2} b^{2}+\frac {6}{7} x^{7} a^{2} A \,b^{2}+\frac {4}{7} x^{7} B \,a^{3} b +\frac {2}{3} x^{6} A \,a^{3} b +\frac {1}{6} x^{6} a^{4} B +\frac {1}{5} a^{4} A \,x^{5}\) | \(102\) |
orering | \(\frac {x^{5} \left (126 b^{4} B \,x^{5}+140 A \,b^{4} x^{4}+560 B a \,b^{3} x^{4}+630 A a \,b^{3} x^{3}+945 B \,a^{2} b^{2} x^{3}+1080 A \,a^{2} b^{2} x^{2}+720 B \,a^{3} b \,x^{2}+840 A \,a^{3} b x +210 a^{4} B x +252 a^{4} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}{1260 \left (b x +a \right )^{4}}\) | \(125\) |
Input:
int(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
1/10*b^4*B*x^10+(1/9*A*b^4+4/9*B*a*b^3)*x^9+(1/2*A*a*b^3+3/4*B*a^2*b^2)*x^ 8+(6/7*a^2*A*b^2+4/7*B*a^3*b)*x^7+(2/3*A*a^3*b+1/6*a^4*B)*x^6+1/5*a^4*A*x^ 5
Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{10} \, B b^{4} x^{10} + \frac {1}{5} \, A a^{4} x^{5} + \frac {1}{9} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{9} + \frac {1}{4} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{8} + \frac {2}{7} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{6} \] Input:
integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
1/10*B*b^4*x^10 + 1/5*A*a^4*x^5 + 1/9*(4*B*a*b^3 + A*b^4)*x^9 + 1/4*(3*B*a ^2*b^2 + 2*A*a*b^3)*x^8 + 2/7*(2*B*a^3*b + 3*A*a^2*b^2)*x^7 + 1/6*(B*a^4 + 4*A*a^3*b)*x^6
Time = 0.02 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {A a^{4} x^{5}}{5} + \frac {B b^{4} x^{10}}{10} + x^{9} \left (\frac {A b^{4}}{9} + \frac {4 B a b^{3}}{9}\right ) + x^{8} \left (\frac {A a b^{3}}{2} + \frac {3 B a^{2} b^{2}}{4}\right ) + x^{7} \cdot \left (\frac {6 A a^{2} b^{2}}{7} + \frac {4 B a^{3} b}{7}\right ) + x^{6} \cdot \left (\frac {2 A a^{3} b}{3} + \frac {B a^{4}}{6}\right ) \] Input:
integrate(x**4*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
A*a**4*x**5/5 + B*b**4*x**10/10 + x**9*(A*b**4/9 + 4*B*a*b**3/9) + x**8*(A *a*b**3/2 + 3*B*a**2*b**2/4) + x**7*(6*A*a**2*b**2/7 + 4*B*a**3*b/7) + x** 6*(2*A*a**3*b/3 + B*a**4/6)
Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{10} \, B b^{4} x^{10} + \frac {1}{5} \, A a^{4} x^{5} + \frac {1}{9} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{9} + \frac {1}{4} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{8} + \frac {2}{7} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{6} \] Input:
integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
1/10*B*b^4*x^10 + 1/5*A*a^4*x^5 + 1/9*(4*B*a*b^3 + A*b^4)*x^9 + 1/4*(3*B*a ^2*b^2 + 2*A*a*b^3)*x^8 + 2/7*(2*B*a^3*b + 3*A*a^2*b^2)*x^7 + 1/6*(B*a^4 + 4*A*a^3*b)*x^6
Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.02 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {1}{10} \, B b^{4} x^{10} + \frac {4}{9} \, B a b^{3} x^{9} + \frac {1}{9} \, A b^{4} x^{9} + \frac {3}{4} \, B a^{2} b^{2} x^{8} + \frac {1}{2} \, A a b^{3} x^{8} + \frac {4}{7} \, B a^{3} b x^{7} + \frac {6}{7} \, A a^{2} b^{2} x^{7} + \frac {1}{6} \, B a^{4} x^{6} + \frac {2}{3} \, A a^{3} b x^{6} + \frac {1}{5} \, A a^{4} x^{5} \] Input:
integrate(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
1/10*B*b^4*x^10 + 4/9*B*a*b^3*x^9 + 1/9*A*b^4*x^9 + 3/4*B*a^2*b^2*x^8 + 1/ 2*A*a*b^3*x^8 + 4/7*B*a^3*b*x^7 + 6/7*A*a^2*b^2*x^7 + 1/6*B*a^4*x^6 + 2/3* A*a^3*b*x^6 + 1/5*A*a^4*x^5
Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.92 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=x^6\,\left (\frac {B\,a^4}{6}+\frac {2\,A\,b\,a^3}{3}\right )+x^9\,\left (\frac {A\,b^4}{9}+\frac {4\,B\,a\,b^3}{9}\right )+\frac {A\,a^4\,x^5}{5}+\frac {B\,b^4\,x^{10}}{10}+\frac {2\,a^2\,b\,x^7\,\left (3\,A\,b+2\,B\,a\right )}{7}+\frac {a\,b^2\,x^8\,\left (2\,A\,b+3\,B\,a\right )}{4} \] Input:
int(x^4*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
x^6*((B*a^4)/6 + (2*A*a^3*b)/3) + x^9*((A*b^4)/9 + (4*B*a*b^3)/9) + (A*a^4 *x^5)/5 + (B*b^4*x^10)/10 + (2*a^2*b*x^7*(3*A*b + 2*B*a))/7 + (a*b^2*x^8*( 2*A*b + 3*B*a))/4
Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx=\frac {x^{5} \left (126 b^{5} x^{5}+700 a \,b^{4} x^{4}+1575 a^{2} b^{3} x^{3}+1800 a^{3} b^{2} x^{2}+1050 a^{4} b x +252 a^{5}\right )}{1260} \] Input:
int(x^4*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
(x**5*(252*a**5 + 1050*a**4*b*x + 1800*a**3*b**2*x**2 + 1575*a**2*b**3*x** 3 + 700*a*b**4*x**4 + 126*b**5*x**5))/1260