Integrand size = 27, antiderivative size = 99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{10}} \, dx=-\frac {a^4 A}{9 x^9}-\frac {a^3 (4 A b+a B)}{8 x^8}-\frac {2 a^2 b (3 A b+2 a B)}{7 x^7}-\frac {a b^2 (2 A b+3 a B)}{3 x^6}-\frac {b^3 (A b+4 a B)}{5 x^5}-\frac {b^4 B}{4 x^4} \] Output:
-1/9*a^4*A/x^9-1/8*a^3*(4*A*b+B*a)/x^8-2/7*a^2*b*(3*A*b+2*B*a)/x^7-1/3*a*b ^2*(2*A*b+3*B*a)/x^6-1/5*b^3*(A*b+4*B*a)/x^5-1/4*b^4*B/x^4
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{10}} \, dx=-\frac {126 b^4 x^4 (4 A+5 B x)+336 a b^3 x^3 (5 A+6 B x)+360 a^2 b^2 x^2 (6 A+7 B x)+180 a^3 b x (7 A+8 B x)+35 a^4 (8 A+9 B x)}{2520 x^9} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^10,x]
Output:
-1/2520*(126*b^4*x^4*(4*A + 5*B*x) + 336*a*b^3*x^3*(5*A + 6*B*x) + 360*a^2 *b^2*x^2*(6*A + 7*B*x) + 180*a^3*b*x*(7*A + 8*B*x) + 35*a^4*(8*A + 9*B*x)) /x^9
Time = 0.42 (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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^{10}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^{10}}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^{10}}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^4 A}{x^{10}}+\frac {a^3 (a B+4 A b)}{x^9}+\frac {2 a^2 b (2 a B+3 A b)}{x^8}+\frac {b^3 (4 a B+A b)}{x^6}+\frac {2 a b^2 (3 a B+2 A b)}{x^7}+\frac {b^4 B}{x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 A}{9 x^9}-\frac {a^3 (a B+4 A b)}{8 x^8}-\frac {2 a^2 b (2 a B+3 A b)}{7 x^7}-\frac {b^3 (4 a B+A b)}{5 x^5}-\frac {a b^2 (3 a B+2 A b)}{3 x^6}-\frac {b^4 B}{4 x^4}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^10,x]
Output:
-1/9*(a^4*A)/x^9 - (a^3*(4*A*b + a*B))/(8*x^8) - (2*a^2*b*(3*A*b + 2*a*B)) /(7*x^7) - (a*b^2*(2*A*b + 3*a*B))/(3*x^6) - (b^3*(A*b + 4*a*B))/(5*x^5) - (b^4*B)/(4*x^4)
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 = 0.86 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {a^{4} A}{9 x^{9}}-\frac {a^{3} \left (4 A b +B a \right )}{8 x^{8}}-\frac {2 a^{2} b \left (3 A b +2 B a \right )}{7 x^{7}}-\frac {a \,b^{2} \left (2 A b +3 B a \right )}{3 x^{6}}-\frac {b^{3} \left (A b +4 B a \right )}{5 x^{5}}-\frac {b^{4} B}{4 x^{4}}\) | \(88\) |
norman | \(\frac {-\frac {b^{4} B \,x^{5}}{4}+\left (-\frac {1}{5} A \,b^{4}-\frac {4}{5} B a \,b^{3}\right ) x^{4}+\left (-\frac {2}{3} A a \,b^{3}-B \,a^{2} b^{2}\right ) x^{3}+\left (-\frac {6}{7} a^{2} A \,b^{2}-\frac {4}{7} B \,a^{3} b \right ) x^{2}+\left (-\frac {1}{2} A \,a^{3} b -\frac {1}{8} a^{4} B \right ) x -\frac {a^{4} A}{9}}{x^{9}}\) | \(97\) |
risch | \(\frac {-\frac {b^{4} B \,x^{5}}{4}+\left (-\frac {1}{5} A \,b^{4}-\frac {4}{5} B a \,b^{3}\right ) x^{4}+\left (-\frac {2}{3} A a \,b^{3}-B \,a^{2} b^{2}\right ) x^{3}+\left (-\frac {6}{7} a^{2} A \,b^{2}-\frac {4}{7} B \,a^{3} b \right ) x^{2}+\left (-\frac {1}{2} A \,a^{3} b -\frac {1}{8} a^{4} B \right ) x -\frac {a^{4} A}{9}}{x^{9}}\) | \(97\) |
gosper | \(-\frac {630 b^{4} B \,x^{5}+504 A \,b^{4} x^{4}+2016 B a \,b^{3} x^{4}+1680 A a \,b^{3} x^{3}+2520 B \,a^{2} b^{2} x^{3}+2160 A \,a^{2} b^{2} x^{2}+1440 B \,a^{3} b \,x^{2}+1260 A \,a^{3} b x +315 a^{4} B x +280 a^{4} A}{2520 x^{9}}\) | \(100\) |
parallelrisch | \(-\frac {630 b^{4} B \,x^{5}+504 A \,b^{4} x^{4}+2016 B a \,b^{3} x^{4}+1680 A a \,b^{3} x^{3}+2520 B \,a^{2} b^{2} x^{3}+2160 A \,a^{2} b^{2} x^{2}+1440 B \,a^{3} b \,x^{2}+1260 A \,a^{3} b x +315 a^{4} B x +280 a^{4} A}{2520 x^{9}}\) | \(100\) |
orering | \(-\frac {\left (630 b^{4} B \,x^{5}+504 A \,b^{4} x^{4}+2016 B a \,b^{3} x^{4}+1680 A a \,b^{3} x^{3}+2520 B \,a^{2} b^{2} x^{3}+2160 A \,a^{2} b^{2} x^{2}+1440 B \,a^{3} b \,x^{2}+1260 A \,a^{3} b x +315 a^{4} B x +280 a^{4} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}{2520 x^{9} \left (b x +a \right )^{4}}\) | \(125\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^10,x,method=_RETURNVERBOSE)
Output:
-1/9*a^4*A/x^9-1/8*a^3*(4*A*b+B*a)/x^8-2/7*a^2*b*(3*A*b+2*B*a)/x^7-1/3*a*b ^2*(2*A*b+3*B*a)/x^6-1/5*b^3*(A*b+4*B*a)/x^5-1/4*b^4*B/x^4
Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{10}} \, dx=-\frac {630 \, B b^{4} x^{5} + 280 \, A a^{4} + 504 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 840 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 720 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 315 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{2520 \, x^{9}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^10,x, algorithm="fricas")
Output:
-1/2520*(630*B*b^4*x^5 + 280*A*a^4 + 504*(4*B*a*b^3 + A*b^4)*x^4 + 840*(3* B*a^2*b^2 + 2*A*a*b^3)*x^3 + 720*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 315*(B*a^ 4 + 4*A*a^3*b)*x)/x^9
Time = 2.39 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{10}} \, dx=\frac {- 280 A a^{4} - 630 B b^{4} x^{5} + x^{4} \left (- 504 A b^{4} - 2016 B a b^{3}\right ) + x^{3} \left (- 1680 A a b^{3} - 2520 B a^{2} b^{2}\right ) + x^{2} \left (- 2160 A a^{2} b^{2} - 1440 B a^{3} b\right ) + x \left (- 1260 A a^{3} b - 315 B a^{4}\right )}{2520 x^{9}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2/x**10,x)
Output:
(-280*A*a**4 - 630*B*b**4*x**5 + x**4*(-504*A*b**4 - 2016*B*a*b**3) + x**3 *(-1680*A*a*b**3 - 2520*B*a**2*b**2) + x**2*(-2160*A*a**2*b**2 - 1440*B*a* *3*b) + x*(-1260*A*a**3*b - 315*B*a**4))/(2520*x**9)
Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{10}} \, dx=-\frac {630 \, B b^{4} x^{5} + 280 \, A a^{4} + 504 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 840 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 720 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 315 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{2520 \, x^{9}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^10,x, algorithm="maxima")
Output:
-1/2520*(630*B*b^4*x^5 + 280*A*a^4 + 504*(4*B*a*b^3 + A*b^4)*x^4 + 840*(3* B*a^2*b^2 + 2*A*a*b^3)*x^3 + 720*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 315*(B*a^ 4 + 4*A*a^3*b)*x)/x^9
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{10}} \, dx=-\frac {630 \, B b^{4} x^{5} + 2016 \, B a b^{3} x^{4} + 504 \, A b^{4} x^{4} + 2520 \, B a^{2} b^{2} x^{3} + 1680 \, A a b^{3} x^{3} + 1440 \, B a^{3} b x^{2} + 2160 \, A a^{2} b^{2} x^{2} + 315 \, B a^{4} x + 1260 \, A a^{3} b x + 280 \, A a^{4}}{2520 \, x^{9}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^10,x, algorithm="giac")
Output:
-1/2520*(630*B*b^4*x^5 + 2016*B*a*b^3*x^4 + 504*A*b^4*x^4 + 2520*B*a^2*b^2 *x^3 + 1680*A*a*b^3*x^3 + 1440*B*a^3*b*x^2 + 2160*A*a^2*b^2*x^2 + 315*B*a^ 4*x + 1260*A*a^3*b*x + 280*A*a^4)/x^9
Time = 10.65 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{10}} \, dx=-\frac {x\,\left (\frac {B\,a^4}{8}+\frac {A\,b\,a^3}{2}\right )+\frac {A\,a^4}{9}+x^3\,\left (B\,a^2\,b^2+\frac {2\,A\,a\,b^3}{3}\right )+x^2\,\left (\frac {4\,B\,a^3\,b}{7}+\frac {6\,A\,a^2\,b^2}{7}\right )+x^4\,\left (\frac {A\,b^4}{5}+\frac {4\,B\,a\,b^3}{5}\right )+\frac {B\,b^4\,x^5}{4}}{x^9} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2)/x^10,x)
Output:
-(x*((B*a^4)/8 + (A*a^3*b)/2) + (A*a^4)/9 + x^3*(B*a^2*b^2 + (2*A*a*b^3)/3 ) + x^2*((6*A*a^2*b^2)/7 + (4*B*a^3*b)/7) + x^4*((A*b^4)/5 + (4*B*a*b^3)/5 ) + (B*b^4*x^5)/4)/x^9
Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{10}} \, dx=\frac {-126 b^{5} x^{5}-504 a \,b^{4} x^{4}-840 a^{2} b^{3} x^{3}-720 a^{3} b^{2} x^{2}-315 a^{4} b x -56 a^{5}}{504 x^{9}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^10,x)
Output:
( - 56*a**5 - 315*a**4*b*x - 720*a**3*b**2*x**2 - 840*a**2*b**3*x**3 - 504 *a*b**4*x**4 - 126*b**5*x**5)/(504*x**9)