Integrand size = 23, antiderivative size = 106 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx=-\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{4 b^4 (a+b x)^4}-\frac {b^2 B-2 a b C+3 a^2 D}{3 b^4 (a+b x)^3}-\frac {b C-3 a D}{2 b^4 (a+b x)^2}-\frac {D}{b^4 (a+b x)} \] Output:
-1/4*(A*b^3-a*(B*b^2-C*a*b+D*a^2))/b^4/(b*x+a)^4-1/3*(B*b^2-2*C*a*b+3*D*a^ 2)/b^4/(b*x+a)^3-1/2*(C*b-3*D*a)/b^4/(b*x+a)^2-D/b^4/(b*x+a)
Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx=-\frac {3 A b^3+3 a^3 D+a^2 b (C+12 D x)+2 b^3 x (2 B+3 x (C+2 D x))+a b^2 (B+2 x (2 C+9 D x))}{12 b^4 (a+b x)^4} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^5,x]
Output:
-1/12*(3*A*b^3 + 3*a^3*D + a^2*b*(C + 12*D*x) + 2*b^3*x*(2*B + 3*x*(C + 2* D*x)) + a*b^2*(B + 2*x*(2*C + 9*D*x)))/(b^4*(a + b*x)^4)
Time = 0.31 (sec) , antiderivative size = 106, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b^3 (a+b x)^5}+\frac {3 a^2 D-2 a b C+b^2 B}{b^3 (a+b x)^4}+\frac {b C-3 a D}{b^3 (a+b x)^3}+\frac {D}{b^3 (a+b x)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{4 b^4 (a+b x)^4}-\frac {3 a^2 D-2 a b C+b^2 B}{3 b^4 (a+b x)^3}-\frac {b C-3 a D}{2 b^4 (a+b x)^2}-\frac {D}{b^4 (a+b x)}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^5,x]
Output:
-1/4*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))/(b^4*(a + b*x)^4) - (b^2*B - 2*a* b*C + 3*a^2*D)/(3*b^4*(a + b*x)^3) - (b*C - 3*a*D)/(2*b^4*(a + b*x)^2) - D /(b^4*(a + b*x))
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 = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82
method | result | size |
norman | \(\frac {-\frac {D x^{3}}{b}-\frac {\left (C b +3 D a \right ) x^{2}}{2 b^{2}}-\frac {\left (B \,b^{2}+C a b +3 D a^{2}\right ) x}{3 b^{3}}-\frac {3 b^{3} A +a \,b^{2} B +a^{2} b C +3 a^{3} D}{12 b^{4}}}{\left (b x +a \right )^{4}}\) | \(87\) |
gosper | \(-\frac {12 D x^{3} b^{3}+6 C \,x^{2} b^{3}+18 D x^{2} a \,b^{2}+4 b^{3} B x +4 C x a \,b^{2}+12 D x \,a^{2} b +3 b^{3} A +a \,b^{2} B +a^{2} b C +3 a^{3} D}{12 \left (b x +a \right )^{4} b^{4}}\) | \(89\) |
parallelrisch | \(-\frac {12 D x^{3} b^{3}+6 C \,x^{2} b^{3}+18 D x^{2} a \,b^{2}+4 b^{3} B x +4 C x a \,b^{2}+12 D x \,a^{2} b +3 b^{3} A +a \,b^{2} B +a^{2} b C +3 a^{3} D}{12 \left (b x +a \right )^{4} b^{4}}\) | \(89\) |
orering | \(-\frac {12 D x^{3} b^{3}+6 C \,x^{2} b^{3}+18 D x^{2} a \,b^{2}+4 b^{3} B x +4 C x a \,b^{2}+12 D x \,a^{2} b +3 b^{3} A +a \,b^{2} B +a^{2} b C +3 a^{3} D}{12 \left (b x +a \right )^{4} b^{4}}\) | \(89\) |
default | \(-\frac {b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D}{4 b^{4} \left (b x +a \right )^{4}}-\frac {C b -3 D a}{2 b^{4} \left (b x +a \right )^{2}}-\frac {D}{b^{4} \left (b x +a \right )}-\frac {B \,b^{2}-2 C a b +3 D a^{2}}{3 b^{4} \left (b x +a \right )^{3}}\) | \(101\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^5,x,method=_RETURNVERBOSE)
Output:
1/(b*x+a)^4*(-D*x^3/b-1/2*(C*b+3*D*a)/b^2*x^2-1/3*(B*b^2+C*a*b+3*D*a^2)/b^ 3*x-1/12*(3*A*b^3+B*a*b^2+C*a^2*b+3*D*a^3)/b^4)
Time = 0.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx=-\frac {12 \, D b^{3} x^{3} + 3 \, D a^{3} + C a^{2} b + B a b^{2} + 3 \, A b^{3} + 6 \, {\left (3 \, D a b^{2} + C b^{3}\right )} x^{2} + 4 \, {\left (3 \, D a^{2} b + C a b^{2} + B b^{3}\right )} x}{12 \, {\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^5,x, algorithm="fricas")
Output:
-1/12*(12*D*b^3*x^3 + 3*D*a^3 + C*a^2*b + B*a*b^2 + 3*A*b^3 + 6*(3*D*a*b^2 + C*b^3)*x^2 + 4*(3*D*a^2*b + C*a*b^2 + B*b^3)*x)/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^4)
Time = 3.62 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx=\frac {- 3 A b^{3} - B a b^{2} - C a^{2} b - 3 D a^{3} - 12 D b^{3} x^{3} + x^{2} \left (- 6 C b^{3} - 18 D a b^{2}\right ) + x \left (- 4 B b^{3} - 4 C a b^{2} - 12 D a^{2} b\right )}{12 a^{4} b^{4} + 48 a^{3} b^{5} x + 72 a^{2} b^{6} x^{2} + 48 a b^{7} x^{3} + 12 b^{8} x^{4}} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**5,x)
Output:
(-3*A*b**3 - B*a*b**2 - C*a**2*b - 3*D*a**3 - 12*D*b**3*x**3 + x**2*(-6*C* b**3 - 18*D*a*b**2) + x*(-4*B*b**3 - 4*C*a*b**2 - 12*D*a**2*b))/(12*a**4*b **4 + 48*a**3*b**5*x + 72*a**2*b**6*x**2 + 48*a*b**7*x**3 + 12*b**8*x**4)
Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx=-\frac {12 \, D b^{3} x^{3} + 3 \, D a^{3} + C a^{2} b + B a b^{2} + 3 \, A b^{3} + 6 \, {\left (3 \, D a b^{2} + C b^{3}\right )} x^{2} + 4 \, {\left (3 \, D a^{2} b + C a b^{2} + B b^{3}\right )} x}{12 \, {\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^5,x, algorithm="maxima")
Output:
-1/12*(12*D*b^3*x^3 + 3*D*a^3 + C*a^2*b + B*a*b^2 + 3*A*b^3 + 6*(3*D*a*b^2 + C*b^3)*x^2 + 4*(3*D*a^2*b + C*a*b^2 + B*b^3)*x)/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^4)
Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx=-\frac {\frac {6 \, C}{{\left (b x + a\right )}^{2}} - \frac {8 \, C a}{{\left (b x + a\right )}^{3}} + \frac {3 \, C a^{2}}{{\left (b x + a\right )}^{4}} + \frac {12 \, D}{{\left (b x + a\right )} b} - \frac {18 \, D a}{{\left (b x + a\right )}^{2} b} + \frac {12 \, D a^{2}}{{\left (b x + a\right )}^{3} b} - \frac {3 \, D a^{3}}{{\left (b x + a\right )}^{4} b} + \frac {4 \, B b}{{\left (b x + a\right )}^{3}} - \frac {3 \, B a b}{{\left (b x + a\right )}^{4}} + \frac {3 \, A b^{2}}{{\left (b x + a\right )}^{4}}}{12 \, b^{3}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^5,x, algorithm="giac")
Output:
-1/12*(6*C/(b*x + a)^2 - 8*C*a/(b*x + a)^3 + 3*C*a^2/(b*x + a)^4 + 12*D/(( b*x + a)*b) - 18*D*a/((b*x + a)^2*b) + 12*D*a^2/((b*x + a)^3*b) - 3*D*a^3/ ((b*x + a)^4*b) + 4*B*b/(b*x + a)^3 - 3*B*a*b/(b*x + a)^4 + 3*A*b^2/(b*x + a)^4)/b^3
Time = 3.18 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.66 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx=\frac {D\,\left (\frac {3\,a}{2\,{\left (a+b\,x\right )}^2}-\frac {1}{a+b\,x}-\frac {a^2}{{\left (a+b\,x\right )}^3}+\frac {a^3}{4\,{\left (a+b\,x\right )}^4}\right )}{b^4}-\frac {A}{4\,b\,\left (a^4+4\,a^3\,b\,x+6\,a^2\,b^2\,x^2+4\,a\,b^3\,x^3+b^4\,x^4\right )}-\frac {\frac {B\,a}{12\,b^2}+\frac {B\,x}{3\,b}}{a^4+4\,a^3\,b\,x+6\,a^2\,b^2\,x^2+4\,a\,b^3\,x^3+b^4\,x^4}+\frac {C\,x^3\,\left (4\,a+b\,x\right )}{12\,a^2\,{\left (a+b\,x\right )}^4} \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^5,x)
Output:
(D*((3*a)/(2*(a + b*x)^2) - 1/(a + b*x) - a^2/(a + b*x)^3 + a^3/(4*(a + b* x)^4)))/b^4 - A/(4*b*(a^4 + b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3* b*x)) - ((B*a)/(12*b^2) + (B*x)/(3*b))/(a^4 + b^4*x^4 + 4*a*b^3*x^3 + 6*a^ 2*b^2*x^2 + 4*a^3*b*x) + (C*x^3*(4*a + b*x))/(12*a^2*(a + b*x)^4)
Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^5} \, dx=\frac {3 b^{3} d \,x^{4}-6 a \,b^{2} c \,x^{2}-4 a^{2} b c x -4 a \,b^{3} x -a^{3} c -4 a^{2} b^{2}}{12 a \,b^{3} \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^5,x)
Output:
( - a**3*c - 4*a**2*b**2 - 4*a**2*b*c*x - 4*a*b**3*x - 6*a*b**2*c*x**2 + 3 *b**3*d*x**4)/(12*a*b**3*(a**4 + 4*a**3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3* x**3 + b**4*x**4))