Integrand size = 27, antiderivative size = 99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^8} \, dx=-\frac {a^4 A}{7 x^7}-\frac {a^3 (4 A b+a B)}{6 x^6}-\frac {2 a^2 b (3 A b+2 a B)}{5 x^5}-\frac {a b^2 (2 A b+3 a B)}{2 x^4}-\frac {b^3 (A b+4 a B)}{3 x^3}-\frac {b^4 B}{2 x^2} \]
-1/7*a^4*A/x^7-1/6*a^3*(4*A*b+B*a)/x^6-2/5*a^2*b*(3*A*b+2*B*a)/x^5-1/2*a*b ^2*(2*A*b+3*B*a)/x^4-1/3*b^3*(A*b+4*B*a)/x^3-1/2*b^4*B/x^2
Time = 0.02 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^8} \, dx=-\frac {35 b^4 x^4 (2 A+3 B x)+70 a b^3 x^3 (3 A+4 B x)+63 a^2 b^2 x^2 (4 A+5 B x)+28 a^3 b x (5 A+6 B x)+5 a^4 (6 A+7 B x)}{210 x^7} \]
-1/210*(35*b^4*x^4*(2*A + 3*B*x) + 70*a*b^3*x^3*(3*A + 4*B*x) + 63*a^2*b^2 *x^2*(4*A + 5*B*x) + 28*a^3*b*x*(5*A + 6*B*x) + 5*a^4*(6*A + 7*B*x))/x^7
Time = 0.25 (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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x^8} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x^8}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x^8}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {a^4 A}{x^8}+\frac {a^3 (a B+4 A b)}{x^7}+\frac {2 a^2 b (2 a B+3 A b)}{x^6}+\frac {b^3 (4 a B+A b)}{x^4}+\frac {2 a b^2 (3 a B+2 A b)}{x^5}+\frac {b^4 B}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 A}{7 x^7}-\frac {a^3 (a B+4 A b)}{6 x^6}-\frac {2 a^2 b (2 a B+3 A b)}{5 x^5}-\frac {b^3 (4 a B+A b)}{3 x^3}-\frac {a b^2 (3 a B+2 A b)}{2 x^4}-\frac {b^4 B}{2 x^2}\) |
-1/7*(a^4*A)/x^7 - (a^3*(4*A*b + a*B))/(6*x^6) - (2*a^2*b*(3*A*b + 2*a*B)) /(5*x^5) - (a*b^2*(2*A*b + 3*a*B))/(2*x^4) - (b^3*(A*b + 4*a*B))/(3*x^3) - (b^4*B)/(2*x^2)
3.6.35.3.1 Defintions of rubi rules used
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 = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {a^{4} A}{7 x^{7}}-\frac {a^{3} \left (4 A b +B a \right )}{6 x^{6}}-\frac {2 a^{2} b \left (3 A b +2 B a \right )}{5 x^{5}}-\frac {a \,b^{2} \left (2 A b +3 B a \right )}{2 x^{4}}-\frac {b^{3} \left (A b +4 B a \right )}{3 x^{3}}-\frac {b^{4} B}{2 x^{2}}\) | \(88\) |
norman | \(\frac {-\frac {b^{4} B \,x^{5}}{2}+\left (-\frac {1}{3} A \,b^{4}-\frac {4}{3} B \,b^{3} a \right ) x^{4}+\left (-A \,b^{3} a -\frac {3}{2} B \,a^{2} b^{2}\right ) x^{3}+\left (-\frac {6}{5} A \,a^{2} b^{2}-\frac {4}{5} B \,a^{3} b \right ) x^{2}+\left (-\frac {2}{3} A \,a^{3} b -\frac {1}{6} B \,a^{4}\right ) x -\frac {A \,a^{4}}{7}}{x^{7}}\) | \(97\) |
risch | \(\frac {-\frac {b^{4} B \,x^{5}}{2}+\left (-\frac {1}{3} A \,b^{4}-\frac {4}{3} B \,b^{3} a \right ) x^{4}+\left (-A \,b^{3} a -\frac {3}{2} B \,a^{2} b^{2}\right ) x^{3}+\left (-\frac {6}{5} A \,a^{2} b^{2}-\frac {4}{5} B \,a^{3} b \right ) x^{2}+\left (-\frac {2}{3} A \,a^{3} b -\frac {1}{6} B \,a^{4}\right ) x -\frac {A \,a^{4}}{7}}{x^{7}}\) | \(97\) |
gosper | \(-\frac {105 b^{4} B \,x^{5}+70 A \,b^{4} x^{4}+280 x^{4} B \,b^{3} a +210 a A \,b^{3} x^{3}+315 x^{3} B \,a^{2} b^{2}+252 a^{2} A \,b^{2} x^{2}+168 x^{2} B \,a^{3} b +140 a^{3} A b x +35 a^{4} B x +30 A \,a^{4}}{210 x^{7}}\) | \(100\) |
parallelrisch | \(-\frac {105 b^{4} B \,x^{5}+70 A \,b^{4} x^{4}+280 x^{4} B \,b^{3} a +210 a A \,b^{3} x^{3}+315 x^{3} B \,a^{2} b^{2}+252 a^{2} A \,b^{2} x^{2}+168 x^{2} B \,a^{3} b +140 a^{3} A b x +35 a^{4} B x +30 A \,a^{4}}{210 x^{7}}\) | \(100\) |
-1/7*a^4*A/x^7-1/6*a^3*(4*A*b+B*a)/x^6-2/5*a^2*b*(3*A*b+2*B*a)/x^5-1/2*a*b ^2*(2*A*b+3*B*a)/x^4-1/3*b^3*(A*b+4*B*a)/x^3-1/2*b^4*B/x^2
Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^8} \, dx=-\frac {105 \, B b^{4} x^{5} + 30 \, A a^{4} + 70 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 105 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 84 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 35 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{210 \, x^{7}} \]
-1/210*(105*B*b^4*x^5 + 30*A*a^4 + 70*(4*B*a*b^3 + A*b^4)*x^4 + 105*(3*B*a ^2*b^2 + 2*A*a*b^3)*x^3 + 84*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 35*(B*a^4 + 4 *A*a^3*b)*x)/x^7
Time = 1.55 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^8} \, dx=\frac {- 30 A a^{4} - 105 B b^{4} x^{5} + x^{4} \left (- 70 A b^{4} - 280 B a b^{3}\right ) + x^{3} \left (- 210 A a b^{3} - 315 B a^{2} b^{2}\right ) + x^{2} \left (- 252 A a^{2} b^{2} - 168 B a^{3} b\right ) + x \left (- 140 A a^{3} b - 35 B a^{4}\right )}{210 x^{7}} \]
(-30*A*a**4 - 105*B*b**4*x**5 + x**4*(-70*A*b**4 - 280*B*a*b**3) + x**3*(- 210*A*a*b**3 - 315*B*a**2*b**2) + x**2*(-252*A*a**2*b**2 - 168*B*a**3*b) + x*(-140*A*a**3*b - 35*B*a**4))/(210*x**7)
Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^8} \, dx=-\frac {105 \, B b^{4} x^{5} + 30 \, A a^{4} + 70 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 105 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 84 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 35 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{210 \, x^{7}} \]
-1/210*(105*B*b^4*x^5 + 30*A*a^4 + 70*(4*B*a*b^3 + A*b^4)*x^4 + 105*(3*B*a ^2*b^2 + 2*A*a*b^3)*x^3 + 84*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 35*(B*a^4 + 4 *A*a^3*b)*x)/x^7
Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^8} \, dx=-\frac {105 \, B b^{4} x^{5} + 280 \, B a b^{3} x^{4} + 70 \, A b^{4} x^{4} + 315 \, B a^{2} b^{2} x^{3} + 210 \, A a b^{3} x^{3} + 168 \, B a^{3} b x^{2} + 252 \, A a^{2} b^{2} x^{2} + 35 \, B a^{4} x + 140 \, A a^{3} b x + 30 \, A a^{4}}{210 \, x^{7}} \]
-1/210*(105*B*b^4*x^5 + 280*B*a*b^3*x^4 + 70*A*b^4*x^4 + 315*B*a^2*b^2*x^3 + 210*A*a*b^3*x^3 + 168*B*a^3*b*x^2 + 252*A*a^2*b^2*x^2 + 35*B*a^4*x + 14 0*A*a^3*b*x + 30*A*a^4)/x^7
Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^8} \, dx=-\frac {x\,\left (\frac {B\,a^4}{6}+\frac {2\,A\,b\,a^3}{3}\right )+\frac {A\,a^4}{7}+x^3\,\left (\frac {3\,B\,a^2\,b^2}{2}+A\,a\,b^3\right )+x^2\,\left (\frac {4\,B\,a^3\,b}{5}+\frac {6\,A\,a^2\,b^2}{5}\right )+x^4\,\left (\frac {A\,b^4}{3}+\frac {4\,B\,a\,b^3}{3}\right )+\frac {B\,b^4\,x^5}{2}}{x^7} \]