Integrand size = 23, antiderivative size = 108 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx=-\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{5 b^4 (a+b x)^5}-\frac {b^2 B-2 a b C+3 a^2 D}{4 b^4 (a+b x)^4}-\frac {b C-3 a D}{3 b^4 (a+b x)^3}-\frac {D}{2 b^4 (a+b x)^2} \] Output:
-1/5*(A*b^3-a*(B*b^2-C*a*b+D*a^2))/b^4/(b*x+a)^5-1/4*(B*b^2-2*C*a*b+3*D*a^ 2)/b^4/(b*x+a)^4-1/3*(C*b-3*D*a)/b^4/(b*x+a)^3-1/2*D/b^4/(b*x+a)^2
Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx=-\frac {12 A b^3+3 a^3 D+a^2 b (2 C+15 D x)+5 b^3 x \left (3 B+4 C x+6 D x^2\right )+a b^2 (3 B+10 x (C+3 D x))}{60 b^4 (a+b x)^5} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^6,x]
Output:
-1/60*(12*A*b^3 + 3*a^3*D + a^2*b*(2*C + 15*D*x) + 5*b^3*x*(3*B + 4*C*x + 6*D*x^2) + a*b^2*(3*B + 10*x*(C + 3*D*x)))/(b^4*(a + b*x)^5)
Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2389, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b^3 (a+b x)^6}+\frac {3 a^2 D-2 a b C+b^2 B}{b^3 (a+b x)^5}+\frac {b C-3 a D}{b^3 (a+b x)^4}+\frac {D}{b^3 (a+b x)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{5 b^4 (a+b x)^5}-\frac {3 a^2 D-2 a b C+b^2 B}{4 b^4 (a+b x)^4}-\frac {b C-3 a D}{3 b^4 (a+b x)^3}-\frac {D}{2 b^4 (a+b x)^2}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^6,x]
Output:
-1/5*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))/(b^4*(a + b*x)^5) - (b^2*B - 2*a* b*C + 3*a^2*D)/(4*b^4*(a + b*x)^4) - (b*C - 3*a*D)/(3*b^4*(a + b*x)^3) - D /(2*b^4*(a + b*x)^2)
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.33 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.84
method | result | size |
gosper | \(-\frac {30 D x^{3} b^{3}+20 C \,x^{2} b^{3}+30 D x^{2} a \,b^{2}+15 b^{3} B x +10 C x a \,b^{2}+15 D x \,a^{2} b +12 b^{3} A +3 a \,b^{2} B +2 a^{2} b C +3 a^{3} D}{60 b^{4} \left (b x +a \right )^{5}}\) | \(91\) |
orering | \(-\frac {30 D x^{3} b^{3}+20 C \,x^{2} b^{3}+30 D x^{2} a \,b^{2}+15 b^{3} B x +10 C x a \,b^{2}+15 D x \,a^{2} b +12 b^{3} A +3 a \,b^{2} B +2 a^{2} b C +3 a^{3} D}{60 b^{4} \left (b x +a \right )^{5}}\) | \(91\) |
parallelrisch | \(-\frac {30 D x^{3} b^{4}+20 C \,b^{4} x^{2}+30 D a \,b^{3} x^{2}+15 B \,b^{4} x +10 C a \,b^{3} x +15 D a^{2} b^{2} x +12 A \,b^{4}+3 B a \,b^{3}+2 a^{2} b^{2} C +3 a^{3} b D}{60 b^{5} \left (b x +a \right )^{5}}\) | \(96\) |
default | \(-\frac {b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D}{5 b^{4} \left (b x +a \right )^{5}}-\frac {B \,b^{2}-2 C a b +3 D a^{2}}{4 b^{4} \left (b x +a \right )^{4}}-\frac {D}{2 b^{4} \left (b x +a \right )^{2}}-\frac {C b -3 D a}{3 b^{4} \left (b x +a \right )^{3}}\) | \(101\) |
norman | \(\frac {-\frac {D x^{3}}{2 b}-\frac {\left (2 b^{2} C +3 a b D\right ) x^{2}}{6 b^{3}}-\frac {\left (3 b^{3} B +2 a \,b^{2} C +3 a^{2} b D\right ) x}{12 b^{4}}-\frac {12 A \,b^{4}+3 B a \,b^{3}+2 a^{2} b^{2} C +3 a^{3} b D}{60 b^{5}}}{\left (b x +a \right )^{5}}\) | \(101\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^6,x,method=_RETURNVERBOSE)
Output:
-1/60/b^4*(30*D*b^3*x^3+20*C*b^3*x^2+30*D*a*b^2*x^2+15*B*b^3*x+10*C*a*b^2* x+15*D*a^2*b*x+12*A*b^3+3*B*a*b^2+2*C*a^2*b+3*D*a^3)/(b*x+a)^5
Time = 0.07 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.28 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx=-\frac {30 \, D b^{3} x^{3} + 3 \, D a^{3} + 2 \, C a^{2} b + 3 \, B a b^{2} + 12 \, A b^{3} + 10 \, {\left (3 \, D a b^{2} + 2 \, C b^{3}\right )} x^{2} + 5 \, {\left (3 \, D a^{2} b + 2 \, C a b^{2} + 3 \, B b^{3}\right )} x}{60 \, {\left (b^{9} x^{5} + 5 \, a b^{8} x^{4} + 10 \, a^{2} b^{7} x^{3} + 10 \, a^{3} b^{6} x^{2} + 5 \, a^{4} b^{5} x + a^{5} b^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^6,x, algorithm="fricas")
Output:
-1/60*(30*D*b^3*x^3 + 3*D*a^3 + 2*C*a^2*b + 3*B*a*b^2 + 12*A*b^3 + 10*(3*D *a*b^2 + 2*C*b^3)*x^2 + 5*(3*D*a^2*b + 2*C*a*b^2 + 3*B*b^3)*x)/(b^9*x^5 + 5*a*b^8*x^4 + 10*a^2*b^7*x^3 + 10*a^3*b^6*x^2 + 5*a^4*b^5*x + a^5*b^4)
Time = 7.92 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.37 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx=\frac {- 12 A b^{3} - 3 B a b^{2} - 2 C a^{2} b - 3 D a^{3} - 30 D b^{3} x^{3} + x^{2} \left (- 20 C b^{3} - 30 D a b^{2}\right ) + x \left (- 15 B b^{3} - 10 C a b^{2} - 15 D a^{2} b\right )}{60 a^{5} b^{4} + 300 a^{4} b^{5} x + 600 a^{3} b^{6} x^{2} + 600 a^{2} b^{7} x^{3} + 300 a b^{8} x^{4} + 60 b^{9} x^{5}} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**6,x)
Output:
(-12*A*b**3 - 3*B*a*b**2 - 2*C*a**2*b - 3*D*a**3 - 30*D*b**3*x**3 + x**2*( -20*C*b**3 - 30*D*a*b**2) + x*(-15*B*b**3 - 10*C*a*b**2 - 15*D*a**2*b))/(6 0*a**5*b**4 + 300*a**4*b**5*x + 600*a**3*b**6*x**2 + 600*a**2*b**7*x**3 + 300*a*b**8*x**4 + 60*b**9*x**5)
Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.28 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx=-\frac {30 \, D b^{3} x^{3} + 3 \, D a^{3} + 2 \, C a^{2} b + 3 \, B a b^{2} + 12 \, A b^{3} + 10 \, {\left (3 \, D a b^{2} + 2 \, C b^{3}\right )} x^{2} + 5 \, {\left (3 \, D a^{2} b + 2 \, C a b^{2} + 3 \, B b^{3}\right )} x}{60 \, {\left (b^{9} x^{5} + 5 \, a b^{8} x^{4} + 10 \, a^{2} b^{7} x^{3} + 10 \, a^{3} b^{6} x^{2} + 5 \, a^{4} b^{5} x + a^{5} b^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^6,x, algorithm="maxima")
Output:
-1/60*(30*D*b^3*x^3 + 3*D*a^3 + 2*C*a^2*b + 3*B*a*b^2 + 12*A*b^3 + 10*(3*D *a*b^2 + 2*C*b^3)*x^2 + 5*(3*D*a^2*b + 2*C*a*b^2 + 3*B*b^3)*x)/(b^9*x^5 + 5*a*b^8*x^4 + 10*a^2*b^7*x^3 + 10*a^3*b^6*x^2 + 5*a^4*b^5*x + a^5*b^4)
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx=-\frac {30 \, D b^{3} x^{3} + 30 \, D a b^{2} x^{2} + 20 \, C b^{3} x^{2} + 15 \, D a^{2} b x + 10 \, C a b^{2} x + 15 \, B b^{3} x + 3 \, D a^{3} + 2 \, C a^{2} b + 3 \, B a b^{2} + 12 \, A b^{3}}{60 \, {\left (b x + a\right )}^{5} b^{4}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^6,x, algorithm="giac")
Output:
-1/60*(30*D*b^3*x^3 + 30*D*a*b^2*x^2 + 20*C*b^3*x^2 + 15*D*a^2*b*x + 10*C* a*b^2*x + 15*B*b^3*x + 3*D*a^3 + 2*C*a^2*b + 3*B*a*b^2 + 12*A*b^3)/((b*x + a)^5*b^4)
Time = 3.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx=\frac {x^4\,\left (5\,a+b\,x\right )\,D}{20\,a^2\,{\left (a+b\,x\right )}^5}-\frac {B\,\left (a+5\,b\,x\right )}{20\,b^2\,{\left (a+b\,x\right )}^5}-\frac {C\,\left (8\,a^2+40\,a\,b\,x+80\,b^2\,x^2\right )}{240\,b^3\,{\left (a+b\,x\right )}^5}-\frac {A}{5\,b\,\left (a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5\right )} \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^6,x)
Output:
(x^4*(5*a + b*x)*D)/(20*a^2*(a + b*x)^5) - (B*(a + 5*b*x))/(20*b^2*(a + b* x)^5) - (C*(8*a^2 + 80*b^2*x^2 + 40*a*b*x))/(240*b^3*(a + b*x)^5) - A/(5*b *(a^5 + b^5*x^5 + 5*a*b^4*x^4 + 10*a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a^4*b* x))
Time = 0.14 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^6} \, dx=\frac {-30 b^{3} d \,x^{3}-30 a \,b^{2} d \,x^{2}-20 b^{3} c \,x^{2}-15 a^{2} b d x -10 a \,b^{2} c x -15 b^{4} x -3 a^{3} d -2 a^{2} b c -15 a \,b^{3}}{60 b^{4} \left (b^{5} x^{5}+5 a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}+10 a^{3} b^{2} x^{2}+5 a^{4} b x +a^{5}\right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^6,x)
Output:
( - 3*a**3*d - 2*a**2*b*c - 15*a**2*b*d*x - 15*a*b**3 - 10*a*b**2*c*x - 30 *a*b**2*d*x**2 - 15*b**4*x - 20*b**3*c*x**2 - 30*b**3*d*x**3)/(60*b**4*(a* *5 + 5*a**4*b*x + 10*a**3*b**2*x**2 + 10*a**2*b**3*x**3 + 5*a*b**4*x**4 + b**5*x**5))