Integrand size = 29, antiderivative size = 120 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^7} \, dx=\frac {(b d-a e)^2 (B d-A e)}{6 e^4 (d+e x)^6}-\frac {(b d-a e) (3 b B d-2 A b e-a B e)}{5 e^4 (d+e x)^5}+\frac {b (3 b B d-A b e-2 a B e)}{4 e^4 (d+e x)^4}-\frac {b^2 B}{3 e^4 (d+e x)^3} \]
1/6*(-a*e+b*d)^2*(-A*e+B*d)/e^4/(e*x+d)^6-1/5*(-a*e+b*d)*(-2*A*b*e-B*a*e+3 *B*b*d)/e^4/(e*x+d)^5+1/4*b*(-A*b*e-2*B*a*e+3*B*b*d)/e^4/(e*x+d)^4-1/3*b^2 *B/e^4/(e*x+d)^3
Time = 0.04 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^7} \, dx=-\frac {2 a^2 e^2 (5 A e+B (d+6 e x))+2 a b e \left (2 A e (d+6 e x)+B \left (d^2+6 d e x+15 e^2 x^2\right )\right )+b^2 \left (A e \left (d^2+6 d e x+15 e^2 x^2\right )+B \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )\right )}{60 e^4 (d+e x)^6} \]
-1/60*(2*a^2*e^2*(5*A*e + B*(d + 6*e*x)) + 2*a*b*e*(2*A*e*(d + 6*e*x) + B* (d^2 + 6*d*e*x + 15*e^2*x^2)) + b^2*(A*e*(d^2 + 6*d*e*x + 15*e^2*x^2) + B* (d^3 + 6*d^2*e*x + 15*d*e^2*x^2 + 20*e^3*x^3)))/(e^4*(d + e*x)^6)
Time = 0.30 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1184, 27, 86, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right ) (A+B x)}{(d+e x)^7} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^2 (A+B x)}{(d+e x)^7}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^7}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {b (2 a B e+A b e-3 b B d)}{e^3 (d+e x)^5}+\frac {(a e-b d) (a B e+2 A b e-3 b B d)}{e^3 (d+e x)^6}+\frac {(a e-b d)^2 (A e-B d)}{e^3 (d+e x)^7}+\frac {b^2 B}{e^3 (d+e x)^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b (-2 a B e-A b e+3 b B d)}{4 e^4 (d+e x)^4}-\frac {(b d-a e) (-a B e-2 A b e+3 b B d)}{5 e^4 (d+e x)^5}+\frac {(b d-a e)^2 (B d-A e)}{6 e^4 (d+e x)^6}-\frac {b^2 B}{3 e^4 (d+e x)^3}\) |
((b*d - a*e)^2*(B*d - A*e))/(6*e^4*(d + e*x)^6) - ((b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e))/(5*e^4*(d + e*x)^5) + (b*(3*b*B*d - A*b*e - 2*a*B*e))/(4 *e^4*(d + e*x)^4) - (b^2*B)/(3*e^4*(d + e*x)^3)
3.17.71.3.1 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[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 0.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.26
method | result | size |
risch | \(\frac {-\frac {B \,b^{2} x^{3}}{3 e}-\frac {b \left (A b e +2 B a e +B b d \right ) x^{2}}{4 e^{2}}-\frac {\left (4 A a b \,e^{2}+A \,b^{2} d e +2 a^{2} B \,e^{2}+2 B a b d e +B \,b^{2} d^{2}\right ) x}{10 e^{3}}-\frac {10 A \,a^{2} e^{3}+4 A a b d \,e^{2}+A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e +B \,b^{2} d^{3}}{60 e^{4}}}{\left (e x +d \right )^{6}}\) | \(151\) |
default | \(-\frac {A \,a^{2} e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -B \,b^{2} d^{3}}{6 e^{4} \left (e x +d \right )^{6}}-\frac {2 A a b \,e^{2}-2 A \,b^{2} d e +a^{2} B \,e^{2}-4 B a b d e +3 B \,b^{2} d^{2}}{5 e^{4} \left (e x +d \right )^{5}}-\frac {b^{2} B}{3 e^{4} \left (e x +d \right )^{3}}-\frac {b \left (A b e +2 B a e -3 B b d \right )}{4 e^{4} \left (e x +d \right )^{4}}\) | \(166\) |
gosper | \(-\frac {20 B \,b^{2} x^{3} e^{3}+15 A \,b^{2} e^{3} x^{2}+30 B \,x^{2} a b \,e^{3}+15 B \,x^{2} b^{2} d \,e^{2}+24 A x a b \,e^{3}+6 A x \,b^{2} d \,e^{2}+12 B x \,a^{2} e^{3}+12 B x a b d \,e^{2}+6 B x \,b^{2} d^{2} e +10 A \,a^{2} e^{3}+4 A a b d \,e^{2}+A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e +B \,b^{2} d^{3}}{60 e^{4} \left (e x +d \right )^{6}}\) | \(167\) |
norman | \(\frac {-\frac {B \,b^{2} x^{3}}{3 e}-\frac {\left (A \,b^{2} e^{3}+2 B a b \,e^{3}+B \,b^{2} d \,e^{2}\right ) x^{2}}{4 e^{4}}-\frac {\left (4 A a b \,e^{4}+A \,b^{2} d \,e^{3}+2 B \,e^{4} a^{2}+2 B a b d \,e^{3}+B \,b^{2} d^{2} e^{2}\right ) x}{10 e^{5}}-\frac {10 A \,a^{2} e^{5}+4 A a b d \,e^{4}+A \,b^{2} d^{2} e^{3}+2 B \,a^{2} d \,e^{4}+2 B a b \,d^{2} e^{3}+B \,b^{2} d^{3} e^{2}}{60 e^{6}}}{\left (e x +d \right )^{6}}\) | \(176\) |
parallelrisch | \(-\frac {20 B \,x^{3} b^{2} e^{5}+15 A \,x^{2} b^{2} e^{5}+30 B a b \,e^{5} x^{2}+15 B \,x^{2} b^{2} d \,e^{4}+24 A a b \,e^{5} x +6 A x \,b^{2} d \,e^{4}+12 B x \,a^{2} e^{5}+12 B a b d \,e^{4} x +6 B x \,b^{2} d^{2} e^{3}+10 A \,a^{2} e^{5}+4 A a b d \,e^{4}+A \,b^{2} d^{2} e^{3}+2 B \,a^{2} d \,e^{4}+2 B a b \,d^{2} e^{3}+B \,b^{2} d^{3} e^{2}}{60 e^{6} \left (e x +d \right )^{6}}\) | \(176\) |
(-1/3*B*b^2/e*x^3-1/4*b/e^2*(A*b*e+2*B*a*e+B*b*d)*x^2-1/10/e^3*(4*A*a*b*e^ 2+A*b^2*d*e+2*B*a^2*e^2+2*B*a*b*d*e+B*b^2*d^2)*x-1/60/e^4*(10*A*a^2*e^3+4* A*a*b*d*e^2+A*b^2*d^2*e+2*B*a^2*d*e^2+2*B*a*b*d^2*e+B*b^2*d^3))/(e*x+d)^6
Time = 0.39 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.73 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^7} \, dx=-\frac {20 \, B b^{2} e^{3} x^{3} + B b^{2} d^{3} + 10 \, A a^{2} e^{3} + {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 2 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 15 \, {\left (B b^{2} d e^{2} + {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} e + {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{60 \, {\left (e^{10} x^{6} + 6 \, d e^{9} x^{5} + 15 \, d^{2} e^{8} x^{4} + 20 \, d^{3} e^{7} x^{3} + 15 \, d^{4} e^{6} x^{2} + 6 \, d^{5} e^{5} x + d^{6} e^{4}\right )}} \]
-1/60*(20*B*b^2*e^3*x^3 + B*b^2*d^3 + 10*A*a^2*e^3 + (2*B*a*b + A*b^2)*d^2 *e + 2*(B*a^2 + 2*A*a*b)*d*e^2 + 15*(B*b^2*d*e^2 + (2*B*a*b + A*b^2)*e^3)* x^2 + 6*(B*b^2*d^2*e + (2*B*a*b + A*b^2)*d*e^2 + 2*(B*a^2 + 2*A*a*b)*e^3)* x)/(e^10*x^6 + 6*d*e^9*x^5 + 15*d^2*e^8*x^4 + 20*d^3*e^7*x^3 + 15*d^4*e^6* x^2 + 6*d^5*e^5*x + d^6*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (114) = 228\).
Time = 67.29 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.05 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^7} \, dx=\frac {- 10 A a^{2} e^{3} - 4 A a b d e^{2} - A b^{2} d^{2} e - 2 B a^{2} d e^{2} - 2 B a b d^{2} e - B b^{2} d^{3} - 20 B b^{2} e^{3} x^{3} + x^{2} \left (- 15 A b^{2} e^{3} - 30 B a b e^{3} - 15 B b^{2} d e^{2}\right ) + x \left (- 24 A a b e^{3} - 6 A b^{2} d e^{2} - 12 B a^{2} e^{3} - 12 B a b d e^{2} - 6 B b^{2} d^{2} e\right )}{60 d^{6} e^{4} + 360 d^{5} e^{5} x + 900 d^{4} e^{6} x^{2} + 1200 d^{3} e^{7} x^{3} + 900 d^{2} e^{8} x^{4} + 360 d e^{9} x^{5} + 60 e^{10} x^{6}} \]
(-10*A*a**2*e**3 - 4*A*a*b*d*e**2 - A*b**2*d**2*e - 2*B*a**2*d*e**2 - 2*B* a*b*d**2*e - B*b**2*d**3 - 20*B*b**2*e**3*x**3 + x**2*(-15*A*b**2*e**3 - 3 0*B*a*b*e**3 - 15*B*b**2*d*e**2) + x*(-24*A*a*b*e**3 - 6*A*b**2*d*e**2 - 1 2*B*a**2*e**3 - 12*B*a*b*d*e**2 - 6*B*b**2*d**2*e))/(60*d**6*e**4 + 360*d* *5*e**5*x + 900*d**4*e**6*x**2 + 1200*d**3*e**7*x**3 + 900*d**2*e**8*x**4 + 360*d*e**9*x**5 + 60*e**10*x**6)
Time = 0.20 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.73 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^7} \, dx=-\frac {20 \, B b^{2} e^{3} x^{3} + B b^{2} d^{3} + 10 \, A a^{2} e^{3} + {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 2 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 15 \, {\left (B b^{2} d e^{2} + {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} e + {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{60 \, {\left (e^{10} x^{6} + 6 \, d e^{9} x^{5} + 15 \, d^{2} e^{8} x^{4} + 20 \, d^{3} e^{7} x^{3} + 15 \, d^{4} e^{6} x^{2} + 6 \, d^{5} e^{5} x + d^{6} e^{4}\right )}} \]
-1/60*(20*B*b^2*e^3*x^3 + B*b^2*d^3 + 10*A*a^2*e^3 + (2*B*a*b + A*b^2)*d^2 *e + 2*(B*a^2 + 2*A*a*b)*d*e^2 + 15*(B*b^2*d*e^2 + (2*B*a*b + A*b^2)*e^3)* x^2 + 6*(B*b^2*d^2*e + (2*B*a*b + A*b^2)*d*e^2 + 2*(B*a^2 + 2*A*a*b)*e^3)* x)/(e^10*x^6 + 6*d*e^9*x^5 + 15*d^2*e^8*x^4 + 20*d^3*e^7*x^3 + 15*d^4*e^6* x^2 + 6*d^5*e^5*x + d^6*e^4)
Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^7} \, dx=-\frac {20 \, B b^{2} e^{3} x^{3} + 15 \, B b^{2} d e^{2} x^{2} + 30 \, B a b e^{3} x^{2} + 15 \, A b^{2} e^{3} x^{2} + 6 \, B b^{2} d^{2} e x + 12 \, B a b d e^{2} x + 6 \, A b^{2} d e^{2} x + 12 \, B a^{2} e^{3} x + 24 \, A a b e^{3} x + B b^{2} d^{3} + 2 \, B a b d^{2} e + A b^{2} d^{2} e + 2 \, B a^{2} d e^{2} + 4 \, A a b d e^{2} + 10 \, A a^{2} e^{3}}{60 \, {\left (e x + d\right )}^{6} e^{4}} \]
-1/60*(20*B*b^2*e^3*x^3 + 15*B*b^2*d*e^2*x^2 + 30*B*a*b*e^3*x^2 + 15*A*b^2 *e^3*x^2 + 6*B*b^2*d^2*e*x + 12*B*a*b*d*e^2*x + 6*A*b^2*d*e^2*x + 12*B*a^2 *e^3*x + 24*A*a*b*e^3*x + B*b^2*d^3 + 2*B*a*b*d^2*e + A*b^2*d^2*e + 2*B*a^ 2*d*e^2 + 4*A*a*b*d*e^2 + 10*A*a^2*e^3)/((e*x + d)^6*e^4)
Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.72 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^7} \, dx=-\frac {\frac {2\,B\,a^2\,d\,e^2+10\,A\,a^2\,e^3+2\,B\,a\,b\,d^2\,e+4\,A\,a\,b\,d\,e^2+B\,b^2\,d^3+A\,b^2\,d^2\,e}{60\,e^4}+\frac {x\,\left (2\,B\,a^2\,e^2+2\,B\,a\,b\,d\,e+4\,A\,a\,b\,e^2+B\,b^2\,d^2+A\,b^2\,d\,e\right )}{10\,e^3}+\frac {b\,x^2\,\left (A\,b\,e+2\,B\,a\,e+B\,b\,d\right )}{4\,e^2}+\frac {B\,b^2\,x^3}{3\,e}}{d^6+6\,d^5\,e\,x+15\,d^4\,e^2\,x^2+20\,d^3\,e^3\,x^3+15\,d^2\,e^4\,x^4+6\,d\,e^5\,x^5+e^6\,x^6} \]
-((10*A*a^2*e^3 + B*b^2*d^3 + A*b^2*d^2*e + 2*B*a^2*d*e^2 + 4*A*a*b*d*e^2 + 2*B*a*b*d^2*e)/(60*e^4) + (x*(2*B*a^2*e^2 + B*b^2*d^2 + 4*A*a*b*e^2 + A* b^2*d*e + 2*B*a*b*d*e))/(10*e^3) + (b*x^2*(A*b*e + 2*B*a*e + B*b*d))/(4*e^ 2) + (B*b^2*x^3)/(3*e))/(d^6 + e^6*x^6 + 6*d*e^5*x^5 + 15*d^4*e^2*x^2 + 20 *d^3*e^3*x^3 + 15*d^2*e^4*x^4 + 6*d^5*e*x)