Integrand size = 27, antiderivative size = 99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{11}} \, dx=-\frac {a^4 A}{10 x^{10}}-\frac {a^3 (4 A b+a B)}{9 x^9}-\frac {a^2 b (3 A b+2 a B)}{4 x^8}-\frac {2 a b^2 (2 A b+3 a B)}{7 x^7}-\frac {b^3 (A b+4 a B)}{6 x^6}-\frac {b^4 B}{5 x^5} \] Output:
-1/10*a^4*A/x^10-1/9*a^3*(4*A*b+B*a)/x^9-1/4*a^2*b*(3*A*b+2*B*a)/x^8-2/7*a *b^2*(2*A*b+3*B*a)/x^7-1/6*b^3*(A*b+4*B*a)/x^6-1/5*b^4*B/x^5
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{11}} \, dx=-\frac {42 b^4 x^4 (5 A+6 B x)+120 a b^3 x^3 (6 A+7 B x)+135 a^2 b^2 x^2 (7 A+8 B x)+70 a^3 b x (8 A+9 B x)+14 a^4 (9 A+10 B x)}{1260 x^{10}} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^11,x]
Output:
-1/1260*(42*b^4*x^4*(5*A + 6*B*x) + 120*a*b^3*x^3*(6*A + 7*B*x) + 135*a^2* b^2*x^2*(7*A + 8*B*x) + 70*a^3*b*x*(8*A + 9*B*x) + 14*a^4*(9*A + 10*B*x))/ x^10
Time = 0.41 (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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^{11}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^{11}}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^{11}}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^4 A}{x^{11}}+\frac {a^3 (a B+4 A b)}{x^{10}}+\frac {2 a^2 b (2 a B+3 A b)}{x^9}+\frac {b^3 (4 a B+A b)}{x^7}+\frac {2 a b^2 (3 a B+2 A b)}{x^8}+\frac {b^4 B}{x^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 A}{10 x^{10}}-\frac {a^3 (a B+4 A b)}{9 x^9}-\frac {a^2 b (2 a B+3 A b)}{4 x^8}-\frac {b^3 (4 a B+A b)}{6 x^6}-\frac {2 a b^2 (3 a B+2 A b)}{7 x^7}-\frac {b^4 B}{5 x^5}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x^11,x]
Output:
-1/10*(a^4*A)/x^10 - (a^3*(4*A*b + a*B))/(9*x^9) - (a^2*b*(3*A*b + 2*a*B)) /(4*x^8) - (2*a*b^2*(2*A*b + 3*a*B))/(7*x^7) - (b^3*(A*b + 4*a*B))/(6*x^6) - (b^4*B)/(5*x^5)
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 = 0.91 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {a^{4} A}{10 x^{10}}-\frac {a^{3} \left (4 A b +B a \right )}{9 x^{9}}-\frac {a^{2} b \left (3 A b +2 B a \right )}{4 x^{8}}-\frac {2 a \,b^{2} \left (2 A b +3 B a \right )}{7 x^{7}}-\frac {b^{3} \left (A b +4 B a \right )}{6 x^{6}}-\frac {b^{4} B}{5 x^{5}}\) | \(88\) |
norman | \(\frac {-\frac {b^{4} B \,x^{5}}{5}+\left (-\frac {1}{6} A \,b^{4}-\frac {2}{3} B a \,b^{3}\right ) x^{4}+\left (-\frac {4}{7} A a \,b^{3}-\frac {6}{7} B \,a^{2} b^{2}\right ) x^{3}+\left (-\frac {3}{4} a^{2} A \,b^{2}-\frac {1}{2} B \,a^{3} b \right ) x^{2}+\left (-\frac {4}{9} A \,a^{3} b -\frac {1}{9} a^{4} B \right ) x -\frac {a^{4} A}{10}}{x^{10}}\) | \(97\) |
risch | \(\frac {-\frac {b^{4} B \,x^{5}}{5}+\left (-\frac {1}{6} A \,b^{4}-\frac {2}{3} B a \,b^{3}\right ) x^{4}+\left (-\frac {4}{7} A a \,b^{3}-\frac {6}{7} B \,a^{2} b^{2}\right ) x^{3}+\left (-\frac {3}{4} a^{2} A \,b^{2}-\frac {1}{2} B \,a^{3} b \right ) x^{2}+\left (-\frac {4}{9} A \,a^{3} b -\frac {1}{9} a^{4} B \right ) x -\frac {a^{4} A}{10}}{x^{10}}\) | \(97\) |
gosper | \(-\frac {252 b^{4} B \,x^{5}+210 A \,b^{4} x^{4}+840 B a \,b^{3} x^{4}+720 A a \,b^{3} x^{3}+1080 B \,a^{2} b^{2} x^{3}+945 A \,a^{2} b^{2} x^{2}+630 B \,a^{3} b \,x^{2}+560 A \,a^{3} b x +140 a^{4} B x +126 a^{4} A}{1260 x^{10}}\) | \(100\) |
parallelrisch | \(-\frac {252 b^{4} B \,x^{5}+210 A \,b^{4} x^{4}+840 B a \,b^{3} x^{4}+720 A a \,b^{3} x^{3}+1080 B \,a^{2} b^{2} x^{3}+945 A \,a^{2} b^{2} x^{2}+630 B \,a^{3} b \,x^{2}+560 A \,a^{3} b x +140 a^{4} B x +126 a^{4} A}{1260 x^{10}}\) | \(100\) |
orering | \(-\frac {\left (252 b^{4} B \,x^{5}+210 A \,b^{4} x^{4}+840 B a \,b^{3} x^{4}+720 A a \,b^{3} x^{3}+1080 B \,a^{2} b^{2} x^{3}+945 A \,a^{2} b^{2} x^{2}+630 B \,a^{3} b \,x^{2}+560 A \,a^{3} b x +140 a^{4} B x +126 a^{4} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}{1260 x^{10} \left (b x +a \right )^{4}}\) | \(125\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^11,x,method=_RETURNVERBOSE)
Output:
-1/10*a^4*A/x^10-1/9*a^3*(4*A*b+B*a)/x^9-1/4*a^2*b*(3*A*b+2*B*a)/x^8-2/7*a *b^2*(2*A*b+3*B*a)/x^7-1/6*b^3*(A*b+4*B*a)/x^6-1/5*b^4*B/x^5
Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{11}} \, dx=-\frac {252 \, B b^{4} x^{5} + 126 \, A a^{4} + 210 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 360 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 315 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 140 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{1260 \, x^{10}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^11,x, algorithm="fricas")
Output:
-1/1260*(252*B*b^4*x^5 + 126*A*a^4 + 210*(4*B*a*b^3 + A*b^4)*x^4 + 360*(3* B*a^2*b^2 + 2*A*a*b^3)*x^3 + 315*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 140*(B*a^ 4 + 4*A*a^3*b)*x)/x^10
Time = 2.79 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{11}} \, dx=\frac {- 126 A a^{4} - 252 B b^{4} x^{5} + x^{4} \left (- 210 A b^{4} - 840 B a b^{3}\right ) + x^{3} \left (- 720 A a b^{3} - 1080 B a^{2} b^{2}\right ) + x^{2} \left (- 945 A a^{2} b^{2} - 630 B a^{3} b\right ) + x \left (- 560 A a^{3} b - 140 B a^{4}\right )}{1260 x^{10}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2/x**11,x)
Output:
(-126*A*a**4 - 252*B*b**4*x**5 + x**4*(-210*A*b**4 - 840*B*a*b**3) + x**3* (-720*A*a*b**3 - 1080*B*a**2*b**2) + x**2*(-945*A*a**2*b**2 - 630*B*a**3*b ) + x*(-560*A*a**3*b - 140*B*a**4))/(1260*x**10)
Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{11}} \, dx=-\frac {252 \, B b^{4} x^{5} + 126 \, A a^{4} + 210 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 360 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 315 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 140 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{1260 \, x^{10}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^11,x, algorithm="maxima")
Output:
-1/1260*(252*B*b^4*x^5 + 126*A*a^4 + 210*(4*B*a*b^3 + A*b^4)*x^4 + 360*(3* B*a^2*b^2 + 2*A*a*b^3)*x^3 + 315*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 140*(B*a^ 4 + 4*A*a^3*b)*x)/x^10
Time = 0.23 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{11}} \, dx=-\frac {252 \, B b^{4} x^{5} + 840 \, B a b^{3} x^{4} + 210 \, A b^{4} x^{4} + 1080 \, B a^{2} b^{2} x^{3} + 720 \, A a b^{3} x^{3} + 630 \, B a^{3} b x^{2} + 945 \, A a^{2} b^{2} x^{2} + 140 \, B a^{4} x + 560 \, A a^{3} b x + 126 \, A a^{4}}{1260 \, x^{10}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^11,x, algorithm="giac")
Output:
-1/1260*(252*B*b^4*x^5 + 840*B*a*b^3*x^4 + 210*A*b^4*x^4 + 1080*B*a^2*b^2* x^3 + 720*A*a*b^3*x^3 + 630*B*a^3*b*x^2 + 945*A*a^2*b^2*x^2 + 140*B*a^4*x + 560*A*a^3*b*x + 126*A*a^4)/x^10
Time = 10.57 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{11}} \, dx=-\frac {x\,\left (\frac {B\,a^4}{9}+\frac {4\,A\,b\,a^3}{9}\right )+\frac {A\,a^4}{10}+x^2\,\left (\frac {B\,a^3\,b}{2}+\frac {3\,A\,a^2\,b^2}{4}\right )+x^3\,\left (\frac {6\,B\,a^2\,b^2}{7}+\frac {4\,A\,a\,b^3}{7}\right )+x^4\,\left (\frac {A\,b^4}{6}+\frac {2\,B\,a\,b^3}{3}\right )+\frac {B\,b^4\,x^5}{5}}{x^{10}} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2)/x^11,x)
Output:
-(x*((B*a^4)/9 + (4*A*a^3*b)/9) + (A*a^4)/10 + x^2*((3*A*a^2*b^2)/4 + (B*a ^3*b)/2) + x^3*((6*B*a^2*b^2)/7 + (4*A*a*b^3)/7) + x^4*((A*b^4)/6 + (2*B*a *b^3)/3) + (B*b^4*x^5)/5)/x^10
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^{11}} \, dx=\frac {-252 b^{5} x^{5}-1050 a \,b^{4} x^{4}-1800 a^{2} b^{3} x^{3}-1575 a^{3} b^{2} x^{2}-700 a^{4} b x -126 a^{5}}{1260 x^{10}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x^11,x)
Output:
( - 126*a**5 - 700*a**4*b*x - 1575*a**3*b**2*x**2 - 1800*a**2*b**3*x**3 - 1050*a*b**4*x**4 - 252*b**5*x**5)/(1260*x**10)