Integrand size = 23, antiderivative size = 93 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx=\frac {D x}{b^3}-\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{2 b^4 (a+b x)^2}-\frac {b^2 B-2 a b C+3 a^2 D}{b^4 (a+b x)}+\frac {(b C-3 a D) \log (a+b x)}{b^4} \] Output:
D*x/b^3-1/2*(A*b^3-a*(B*b^2-C*a*b+D*a^2))/b^4/(b*x+a)^2-(B*b^2-2*C*a*b+3*D *a^2)/b^4/(b*x+a)+(C*b-3*D*a)*ln(b*x+a)/b^4
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx=\frac {2 b D x+\frac {-A b^3+a \left (b^2 B-a b C+a^2 D\right )}{(a+b x)^2}-\frac {2 \left (b^2 B-2 a b C+3 a^2 D\right )}{a+b x}+2 (b C-3 a D) \log (a+b x)}{2 b^4} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^3,x]
Output:
(2*b*D*x + (-(A*b^3) + a*(b^2*B - a*b*C + a^2*D))/(a + b*x)^2 - (2*(b^2*B - 2*a*b*C + 3*a^2*D))/(a + b*x) + 2*(b*C - 3*a*D)*Log[a + b*x])/(2*b^4)
Time = 0.28 (sec) , antiderivative size = 93, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b^3 (a+b x)^3}+\frac {3 a^2 D-2 a b C+b^2 B}{b^3 (a+b x)^2}+\frac {b C-3 a D}{b^3 (a+b x)}+\frac {D}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^4 (a+b x)^2}-\frac {3 a^2 D-2 a b C+b^2 B}{b^4 (a+b x)}+\frac {(b C-3 a D) \log (a+b x)}{b^4}+\frac {D x}{b^3}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^3,x]
Output:
(D*x)/b^3 - (A*b^3 - a*(b^2*B - a*b*C + a^2*D))/(2*b^4*(a + b*x)^2) - (b^2 *B - 2*a*b*C + 3*a^2*D)/(b^4*(a + b*x)) + ((b*C - 3*a*D)*Log[a + b*x])/b^4
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 = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97
method | result | size |
norman | \(\frac {\frac {D x^{3}}{b}-\frac {b^{3} A +a \,b^{2} B -3 a^{2} b C +9 a^{3} D}{2 b^{4}}-\frac {\left (B \,b^{2}-2 C a b +6 D a^{2}\right ) x}{b^{3}}}{\left (b x +a \right )^{2}}+\frac {\left (C b -3 D a \right ) \ln \left (b x +a \right )}{b^{4}}\) | \(90\) |
default | \(\frac {D x}{b^{3}}-\frac {b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D}{2 b^{4} \left (b x +a \right )^{2}}-\frac {B \,b^{2}-2 C a b +3 D a^{2}}{b^{4} \left (b x +a \right )}+\frac {\left (C b -3 D a \right ) \ln \left (b x +a \right )}{b^{4}}\) | \(92\) |
parallelrisch | \(-\frac {-2 C \ln \left (b x +a \right ) x^{2} b^{3}+6 D \ln \left (b x +a \right ) x^{2} a \,b^{2}-2 D x^{3} b^{3}-4 C \ln \left (b x +a \right ) x a \,b^{2}+12 D \ln \left (b x +a \right ) x \,a^{2} b +2 b^{3} B x -2 C \ln \left (b x +a \right ) a^{2} b -4 C x a \,b^{2}+6 D \ln \left (b x +a \right ) a^{3}+12 D x \,a^{2} b +b^{3} A +a \,b^{2} B -3 a^{2} b C +9 a^{3} D}{2 b^{4} \left (b x +a \right )^{2}}\) | \(154\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
(D*x^3/b-1/2*(A*b^3+B*a*b^2-3*C*a^2*b+9*D*a^3)/b^4-(B*b^2-2*C*a*b+6*D*a^2) /b^3*x)/(b*x+a)^2+(C*b-3*D*a)*ln(b*x+a)/b^4
Time = 0.07 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.65 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx=\frac {2 \, D b^{3} x^{3} + 4 \, D a b^{2} x^{2} - 5 \, D a^{3} + 3 \, C a^{2} b - B a b^{2} - A b^{3} - 2 \, {\left (2 \, D a^{2} b - 2 \, C a b^{2} + B b^{3}\right )} x - 2 \, {\left (3 \, D a^{3} - C a^{2} b + {\left (3 \, D a b^{2} - C b^{3}\right )} x^{2} + 2 \, {\left (3 \, D a^{2} b - C a b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3,x, algorithm="fricas")
Output:
1/2*(2*D*b^3*x^3 + 4*D*a*b^2*x^2 - 5*D*a^3 + 3*C*a^2*b - B*a*b^2 - A*b^3 - 2*(2*D*a^2*b - 2*C*a*b^2 + B*b^3)*x - 2*(3*D*a^3 - C*a^2*b + (3*D*a*b^2 - C*b^3)*x^2 + 2*(3*D*a^2*b - C*a*b^2)*x)*log(b*x + a))/(b^6*x^2 + 2*a*b^5* x + a^2*b^4)
Time = 0.65 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx=\frac {D x}{b^{3}} + \frac {- A b^{3} - B a b^{2} + 3 C a^{2} b - 5 D a^{3} + x \left (- 2 B b^{3} + 4 C a b^{2} - 6 D a^{2} b\right )}{2 a^{2} b^{4} + 4 a b^{5} x + 2 b^{6} x^{2}} - \frac {\left (- C b + 3 D a\right ) \log {\left (a + b x \right )}}{b^{4}} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**3,x)
Output:
D*x/b**3 + (-A*b**3 - B*a*b**2 + 3*C*a**2*b - 5*D*a**3 + x*(-2*B*b**3 + 4* C*a*b**2 - 6*D*a**2*b))/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) - (-C*b + 3*D*a)*log(a + b*x)/b**4
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx=-\frac {5 \, D a^{3} - 3 \, C a^{2} b + B a b^{2} + A b^{3} + 2 \, {\left (3 \, D a^{2} b - 2 \, C a b^{2} + B b^{3}\right )} x}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {D x}{b^{3}} - \frac {{\left (3 \, D a - C b\right )} \log \left (b x + a\right )}{b^{4}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3,x, algorithm="maxima")
Output:
-1/2*(5*D*a^3 - 3*C*a^2*b + B*a*b^2 + A*b^3 + 2*(3*D*a^2*b - 2*C*a*b^2 + B *b^3)*x)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4) + D*x/b^3 - (3*D*a - C*b)*log(b*x + a)/b^4
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx=\frac {D x}{b^{3}} - \frac {{\left (3 \, D a - C b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac {5 \, D a^{3} - 3 \, C a^{2} b + B a b^{2} + A b^{3} + 2 \, {\left (3 \, D a^{2} b - 2 \, C a b^{2} + B b^{3}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{4}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3,x, algorithm="giac")
Output:
D*x/b^3 - (3*D*a - C*b)*log(abs(b*x + a))/b^4 - 1/2*(5*D*a^3 - 3*C*a^2*b + B*a*b^2 + A*b^3 + 2*(3*D*a^2*b - 2*C*a*b^2 + B*b^3)*x)/((b*x + a)^2*b^4)
Time = 3.18 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.62 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx=\frac {\frac {3\,C\,a^2}{2\,b^3}+\frac {2\,C\,a\,x}{b^2}}{a^2+2\,a\,b\,x+b^2\,x^2}-\frac {\frac {B\,a}{2\,b^2}+\frac {B\,x}{b}}{a^2+2\,a\,b\,x+b^2\,x^2}-\frac {A}{2\,b\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}-\frac {D\,\left (3\,a\,\ln \left (a+b\,x\right )-b\,x+\frac {3\,a^2}{a+b\,x}-\frac {a^3}{2\,{\left (a+b\,x\right )}^2}\right )}{b^4}+\frac {C\,\ln \left (a+b\,x\right )}{b^3} \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^3,x)
Output:
((3*C*a^2)/(2*b^3) + (2*C*a*x)/b^2)/(a^2 + b^2*x^2 + 2*a*b*x) - ((B*a)/(2* b^2) + (B*x)/b)/(a^2 + b^2*x^2 + 2*a*b*x) - A/(2*b*(a^2 + b^2*x^2 + 2*a*b* x)) - (D*(3*a*log(a + b*x) - b*x + (3*a^2)/(a + b*x) - a^3/(2*(a + b*x)^2) ))/b^4 + (C*log(a + b*x))/b^3
Time = 0.14 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.88 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3} \, dx=\frac {-6 \,\mathrm {log}\left (b x +a \right ) a^{4} d +2 \,\mathrm {log}\left (b x +a \right ) a^{3} b c -12 \,\mathrm {log}\left (b x +a \right ) a^{3} b d x +4 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} c x -6 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} d \,x^{2}+2 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} c \,x^{2}-3 a^{4} d +a^{3} b c -a^{2} b^{3}+6 a^{2} b^{2} d \,x^{2}-2 a \,b^{3} c \,x^{2}+2 a \,b^{3} d \,x^{3}+b^{5} x^{2}}{2 a \,b^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^3,x)
Output:
( - 6*log(a + b*x)*a**4*d + 2*log(a + b*x)*a**3*b*c - 12*log(a + b*x)*a**3 *b*d*x + 4*log(a + b*x)*a**2*b**2*c*x - 6*log(a + b*x)*a**2*b**2*d*x**2 + 2*log(a + b*x)*a*b**3*c*x**2 - 3*a**4*d + a**3*b*c - a**2*b**3 + 6*a**2*b* *2*d*x**2 - 2*a*b**3*c*x**2 + 2*a*b**3*d*x**3 + b**5*x**2)/(2*a*b**4*(a**2 + 2*a*b*x + b**2*x**2))