Integrand size = 29, antiderivative size = 120 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^6} \, dx=\frac {(b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^5}-\frac {(b d-a e) (3 b B d-2 A b e-a B e)}{4 e^4 (d+e x)^4}+\frac {b (3 b B d-A b e-2 a B e)}{3 e^4 (d+e x)^3}-\frac {b^2 B}{2 e^4 (d+e x)^2} \] Output:
1/5*(-a*e+b*d)^2*(-A*e+B*d)/e^4/(e*x+d)^5-1/4*(-a*e+b*d)*(-2*A*b*e-B*a*e+3 *B*b*d)/e^4/(e*x+d)^4+1/3*b*(-A*b*e-2*B*a*e+3*B*b*d)/e^4/(e*x+d)^3-1/2*b^2 *B/e^4/(e*x+d)^2
Time = 0.07 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^6} \, dx=-\frac {3 a^2 e^2 (4 A e+B (d+5 e x))+2 a b e \left (3 A e (d+5 e x)+2 B \left (d^2+5 d e x+10 e^2 x^2\right )\right )+b^2 \left (2 A e \left (d^2+5 d e x+10 e^2 x^2\right )+3 B \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )\right )}{60 e^4 (d+e x)^5} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x)^6,x]
Output:
-1/60*(3*a^2*e^2*(4*A*e + B*(d + 5*e*x)) + 2*a*b*e*(3*A*e*(d + 5*e*x) + 2* B*(d^2 + 5*d*e*x + 10*e^2*x^2)) + b^2*(2*A*e*(d^2 + 5*d*e*x + 10*e^2*x^2) + 3*B*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3)))/(e^4*(d + e*x)^5)
Time = 0.52 (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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right ) (A+B x)}{(d+e x)^6} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^2 (A+B x)}{(d+e x)^6}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^6}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {b (2 a B e+A b e-3 b B d)}{e^3 (d+e x)^4}+\frac {(a e-b d) (a B e+2 A b e-3 b B d)}{e^3 (d+e x)^5}+\frac {(a e-b d)^2 (A e-B d)}{e^3 (d+e x)^6}+\frac {b^2 B}{e^3 (d+e x)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b (-2 a B e-A b e+3 b B d)}{3 e^4 (d+e x)^3}-\frac {(b d-a e) (-a B e-2 A b e+3 b B d)}{4 e^4 (d+e x)^4}+\frac {(b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^5}-\frac {b^2 B}{2 e^4 (d+e x)^2}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x)^6,x]
Output:
((b*d - a*e)^2*(B*d - A*e))/(5*e^4*(d + e*x)^5) - ((b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e))/(4*e^4*(d + e*x)^4) + (b*(3*b*B*d - A*b*e - 2*a*B*e))/(3 *e^4*(d + e*x)^3) - (b^2*B)/(2*e^4*(d + e*x)^2)
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 = 0.73 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\frac {-\frac {B \,b^{2} x^{3}}{2 e}-\frac {b \left (2 A b e +4 B a e +3 B b d \right ) x^{2}}{6 e^{2}}-\frac {\left (6 A a b \,e^{2}+2 A \,b^{2} d e +3 B \,e^{2} a^{2}+4 B a b d e +3 B \,b^{2} d^{2}\right ) x}{12 e^{3}}-\frac {12 a^{2} A \,e^{3}+6 A a b d \,e^{2}+2 A \,b^{2} d^{2} e +3 B \,a^{2} d \,e^{2}+4 B a b \,d^{2} e +3 B \,b^{2} d^{3}}{60 e^{4}}}{\left (e x +d \right )^{5}}\) | \(157\) |
default | \(-\frac {b \left (A b e +2 B a e -3 B b d \right )}{3 e^{4} \left (e x +d \right )^{3}}-\frac {2 A a b \,e^{2}-2 A \,b^{2} d e +B \,e^{2} a^{2}-4 B a b d e +3 B \,b^{2} d^{2}}{4 e^{4} \left (e x +d \right )^{4}}-\frac {a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -B \,b^{2} d^{3}}{5 e^{4} \left (e x +d \right )^{5}}-\frac {b^{2} B}{2 e^{4} \left (e x +d \right )^{2}}\) | \(166\) |
gosper | \(-\frac {30 B \,b^{2} x^{3} e^{3}+20 A \,b^{2} e^{3} x^{2}+40 B \,x^{2} a b \,e^{3}+30 B \,x^{2} b^{2} d \,e^{2}+30 A x a b \,e^{3}+10 A x \,b^{2} d \,e^{2}+15 B x \,a^{2} e^{3}+20 B x a b d \,e^{2}+15 B x \,b^{2} d^{2} e +12 a^{2} A \,e^{3}+6 A a b d \,e^{2}+2 A \,b^{2} d^{2} e +3 B \,a^{2} d \,e^{2}+4 B a b \,d^{2} e +3 B \,b^{2} d^{3}}{60 e^{4} \left (e x +d \right )^{5}}\) | \(169\) |
norman | \(\frac {-\frac {B \,b^{2} x^{3}}{2 e}-\frac {\left (2 A \,b^{2} e^{2}+4 B a b \,e^{2}+3 B \,b^{2} d e \right ) x^{2}}{6 e^{3}}-\frac {\left (6 A a b \,e^{3}+2 A \,b^{2} d \,e^{2}+3 B \,e^{3} a^{2}+4 B a b d \,e^{2}+3 B \,b^{2} d^{2} e \right ) x}{12 e^{4}}-\frac {12 A \,a^{2} e^{4}+6 A a b d \,e^{3}+2 A \,b^{2} d^{2} e^{2}+3 B \,a^{2} d \,e^{3}+4 B a b \,d^{2} e^{2}+3 B \,b^{2} d^{3} e}{60 e^{5}}}{\left (e x +d \right )^{5}}\) | \(176\) |
parallelrisch | \(-\frac {30 B \,b^{2} e^{4} x^{3}+20 A \,b^{2} e^{4} x^{2}+40 B a b \,e^{4} x^{2}+30 B \,b^{2} d \,e^{3} x^{2}+30 A a b \,e^{4} x +10 A \,b^{2} d \,e^{3} x +15 B \,a^{2} e^{4} x +20 B a b d \,e^{3} x +15 B \,b^{2} d^{2} e^{2} x +12 A \,a^{2} e^{4}+6 A a b d \,e^{3}+2 A \,b^{2} d^{2} e^{2}+3 B \,a^{2} d \,e^{3}+4 B a b \,d^{2} e^{2}+3 B \,b^{2} d^{3} e}{60 e^{5} \left (e x +d \right )^{5}}\) | \(176\) |
orering | \(-\frac {\left (30 B \,b^{2} x^{3} e^{3}+20 A \,b^{2} e^{3} x^{2}+40 B \,x^{2} a b \,e^{3}+30 B \,x^{2} b^{2} d \,e^{2}+30 A x a b \,e^{3}+10 A x \,b^{2} d \,e^{2}+15 B x \,a^{2} e^{3}+20 B x a b d \,e^{2}+15 B x \,b^{2} d^{2} e +12 a^{2} A \,e^{3}+6 A a b d \,e^{2}+2 A \,b^{2} d^{2} e +3 B \,a^{2} d \,e^{2}+4 B a b \,d^{2} e +3 B \,b^{2} d^{3}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}{60 e^{4} \left (b x +a \right )^{2} \left (e x +d \right )^{5}}\) | \(192\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^6,x,method=_RETURNVERBOSE)
Output:
(-1/2*B/e*b^2*x^3-1/6*b/e^2*(2*A*b*e+4*B*a*e+3*B*b*d)*x^2-1/12/e^3*(6*A*a* b*e^2+2*A*b^2*d*e+3*B*a^2*e^2+4*B*a*b*d*e+3*B*b^2*d^2)*x-1/60/e^4*(12*A*a^ 2*e^3+6*A*a*b*d*e^2+2*A*b^2*d^2*e+3*B*a^2*d*e^2+4*B*a*b*d^2*e+3*B*b^2*d^3) )/(e*x+d)^5
Time = 0.09 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.69 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B b^{2} e^{3} x^{3} + 3 \, B b^{2} d^{3} + 12 \, A a^{2} e^{3} + 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 3 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 10 \, {\left (3 \, B b^{2} d e^{2} + 2 \, {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 5 \, {\left (3 \, B b^{2} d^{2} e + 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{60 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^6,x, algorithm="fricas")
Output:
-1/60*(30*B*b^2*e^3*x^3 + 3*B*b^2*d^3 + 12*A*a^2*e^3 + 2*(2*B*a*b + A*b^2) *d^2*e + 3*(B*a^2 + 2*A*a*b)*d*e^2 + 10*(3*B*b^2*d*e^2 + 2*(2*B*a*b + A*b^ 2)*e^3)*x^2 + 5*(3*B*b^2*d^2*e + 2*(2*B*a*b + A*b^2)*d*e^2 + 3*(B*a^2 + 2* A*a*b)*e^3)*x)/(e^9*x^5 + 5*d*e^8*x^4 + 10*d^2*e^7*x^3 + 10*d^3*e^6*x^2 + 5*d^4*e^5*x + d^5*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (114) = 228\).
Time = 9.43 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.98 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^6} \, dx=\frac {- 12 A a^{2} e^{3} - 6 A a b d e^{2} - 2 A b^{2} d^{2} e - 3 B a^{2} d e^{2} - 4 B a b d^{2} e - 3 B b^{2} d^{3} - 30 B b^{2} e^{3} x^{3} + x^{2} \left (- 20 A b^{2} e^{3} - 40 B a b e^{3} - 30 B b^{2} d e^{2}\right ) + x \left (- 30 A a b e^{3} - 10 A b^{2} d e^{2} - 15 B a^{2} e^{3} - 20 B a b d e^{2} - 15 B b^{2} d^{2} e\right )}{60 d^{5} e^{4} + 300 d^{4} e^{5} x + 600 d^{3} e^{6} x^{2} + 600 d^{2} e^{7} x^{3} + 300 d e^{8} x^{4} + 60 e^{9} x^{5}} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)/(e*x+d)**6,x)
Output:
(-12*A*a**2*e**3 - 6*A*a*b*d*e**2 - 2*A*b**2*d**2*e - 3*B*a**2*d*e**2 - 4* B*a*b*d**2*e - 3*B*b**2*d**3 - 30*B*b**2*e**3*x**3 + x**2*(-20*A*b**2*e**3 - 40*B*a*b*e**3 - 30*B*b**2*d*e**2) + x*(-30*A*a*b*e**3 - 10*A*b**2*d*e** 2 - 15*B*a**2*e**3 - 20*B*a*b*d*e**2 - 15*B*b**2*d**2*e))/(60*d**5*e**4 + 300*d**4*e**5*x + 600*d**3*e**6*x**2 + 600*d**2*e**7*x**3 + 300*d*e**8*x** 4 + 60*e**9*x**5)
Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.69 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B b^{2} e^{3} x^{3} + 3 \, B b^{2} d^{3} + 12 \, A a^{2} e^{3} + 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 3 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 10 \, {\left (3 \, B b^{2} d e^{2} + 2 \, {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 5 \, {\left (3 \, B b^{2} d^{2} e + 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{60 \, {\left (e^{9} x^{5} + 5 \, d e^{8} x^{4} + 10 \, d^{2} e^{7} x^{3} + 10 \, d^{3} e^{6} x^{2} + 5 \, d^{4} e^{5} x + d^{5} e^{4}\right )}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^6,x, algorithm="maxima")
Output:
-1/60*(30*B*b^2*e^3*x^3 + 3*B*b^2*d^3 + 12*A*a^2*e^3 + 2*(2*B*a*b + A*b^2) *d^2*e + 3*(B*a^2 + 2*A*a*b)*d*e^2 + 10*(3*B*b^2*d*e^2 + 2*(2*B*a*b + A*b^ 2)*e^3)*x^2 + 5*(3*B*b^2*d^2*e + 2*(2*B*a*b + A*b^2)*d*e^2 + 3*(B*a^2 + 2* A*a*b)*e^3)*x)/(e^9*x^5 + 5*d*e^8*x^4 + 10*d^2*e^7*x^3 + 10*d^3*e^6*x^2 + 5*d^4*e^5*x + d^5*e^4)
Time = 0.21 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.40 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^6} \, dx=-\frac {30 \, B b^{2} e^{3} x^{3} + 30 \, B b^{2} d e^{2} x^{2} + 40 \, B a b e^{3} x^{2} + 20 \, A b^{2} e^{3} x^{2} + 15 \, B b^{2} d^{2} e x + 20 \, B a b d e^{2} x + 10 \, A b^{2} d e^{2} x + 15 \, B a^{2} e^{3} x + 30 \, A a b e^{3} x + 3 \, B b^{2} d^{3} + 4 \, B a b d^{2} e + 2 \, A b^{2} d^{2} e + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} + 12 \, A a^{2} e^{3}}{60 \, {\left (e x + d\right )}^{5} e^{4}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^6,x, algorithm="giac")
Output:
-1/60*(30*B*b^2*e^3*x^3 + 30*B*b^2*d*e^2*x^2 + 40*B*a*b*e^3*x^2 + 20*A*b^2 *e^3*x^2 + 15*B*b^2*d^2*e*x + 20*B*a*b*d*e^2*x + 10*A*b^2*d*e^2*x + 15*B*a ^2*e^3*x + 30*A*a*b*e^3*x + 3*B*b^2*d^3 + 4*B*a*b*d^2*e + 2*A*b^2*d^2*e + 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 + 12*A*a^2*e^3)/((e*x + d)^5*e^4)
Time = 0.10 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.68 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^6} \, dx=-\frac {\frac {3\,B\,a^2\,d\,e^2+12\,A\,a^2\,e^3+4\,B\,a\,b\,d^2\,e+6\,A\,a\,b\,d\,e^2+3\,B\,b^2\,d^3+2\,A\,b^2\,d^2\,e}{60\,e^4}+\frac {x\,\left (3\,B\,a^2\,e^2+4\,B\,a\,b\,d\,e+6\,A\,a\,b\,e^2+3\,B\,b^2\,d^2+2\,A\,b^2\,d\,e\right )}{12\,e^3}+\frac {b\,x^2\,\left (2\,A\,b\,e+4\,B\,a\,e+3\,B\,b\,d\right )}{6\,e^2}+\frac {B\,b^2\,x^3}{2\,e}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x))/(d + e*x)^6,x)
Output:
-((12*A*a^2*e^3 + 3*B*b^2*d^3 + 2*A*b^2*d^2*e + 3*B*a^2*d*e^2 + 6*A*a*b*d* e^2 + 4*B*a*b*d^2*e)/(60*e^4) + (x*(3*B*a^2*e^2 + 3*B*b^2*d^2 + 6*A*a*b*e^ 2 + 2*A*b^2*d*e + 4*B*a*b*d*e))/(12*e^3) + (b*x^2*(2*A*b*e + 4*B*a*e + 3*B *b*d))/(6*e^2) + (B*b^2*x^3)/(2*e))/(d^5 + e^5*x^5 + 5*d*e^4*x^4 + 10*d^3* e^2*x^2 + 10*d^2*e^3*x^3 + 5*d^4*e*x)
Time = 0.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^6} \, dx=\frac {-10 b^{3} e^{3} x^{3}-20 a \,b^{2} e^{3} x^{2}-10 b^{3} d \,e^{2} x^{2}-15 a^{2} b \,e^{3} x -10 a \,b^{2} d \,e^{2} x -5 b^{3} d^{2} e x -4 a^{3} e^{3}-3 a^{2} b d \,e^{2}-2 a \,b^{2} d^{2} e -b^{3} d^{3}}{20 e^{4} \left (e^{5} x^{5}+5 d \,e^{4} x^{4}+10 d^{2} e^{3} x^{3}+10 d^{3} e^{2} x^{2}+5 d^{4} e x +d^{5}\right )} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^6,x)
Output:
( - 4*a**3*e**3 - 3*a**2*b*d*e**2 - 15*a**2*b*e**3*x - 2*a*b**2*d**2*e - 1 0*a*b**2*d*e**2*x - 20*a*b**2*e**3*x**2 - b**3*d**3 - 5*b**3*d**2*e*x - 10 *b**3*d*e**2*x**2 - 10*b**3*e**3*x**3)/(20*e**4*(d**5 + 5*d**4*e*x + 10*d* *3*e**2*x**2 + 10*d**2*e**3*x**3 + 5*d*e**4*x**4 + e**5*x**5))