Integrand size = 21, antiderivative size = 155 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx=-\frac {a^3 A}{3 x^3}-\frac {a^2 (3 A b+a B)}{2 x^2}-\frac {3 a \left (a b B+A \left (b^2+a c\right )\right )}{x}+\left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x+\frac {3}{2} c \left (b^2 B+A b c+a B c\right ) x^2+\frac {1}{3} c^2 (3 b B+A c) x^3+\frac {1}{4} B c^3 x^4+\left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) \log (x) \] Output:
-1/3*a^3*A/x^3-1/2*a^2*(3*A*b+B*a)/x^2-3*a*(a*b*B+A*(a*c+b^2))/x+(3*A*a*c^ 2+3*A*b^2*c+6*B*a*b*c+B*b^3)*x+3/2*c*(A*b*c+B*a*c+B*b^2)*x^2+1/3*c^2*(A*c+ 3*B*b)*x^3+1/4*B*c^3*x^4+(3*a*B*(a*c+b^2)+A*(6*a*b*c+b^3))*ln(x)
Time = 0.07 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx=\frac {-2 a^3 (2 A+3 B x)+x^4 \left (12 b^3 B+18 b^2 c (2 A+B x)+6 b c^2 x (3 A+2 B x)+c^3 x^2 (4 A+3 B x)\right )-18 a^2 x (2 b B x+A (b+2 c x))+18 a x^2 \left (B c x^2 (4 b+c x)-2 A \left (b^2-c^2 x^2\right )\right )+12 \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^3 \log (x)}{12 x^3} \] Input:
Integrate[((A + B*x)*(a + b*x + c*x^2)^3)/x^4,x]
Output:
(-2*a^3*(2*A + 3*B*x) + x^4*(12*b^3*B + 18*b^2*c*(2*A + B*x) + 6*b*c^2*x*( 3*A + 2*B*x) + c^3*x^2*(4*A + 3*B*x)) - 18*a^2*x*(2*b*B*x + A*(b + 2*c*x)) + 18*a*x^2*(B*c*x^2*(4*b + c*x) - 2*A*(b^2 - c^2*x^2)) + 12*(3*a*B*(b^2 + a*c) + A*(b^3 + 6*a*b*c))*x^3*Log[x])/(12*x^3)
Time = 0.36 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1195, 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.
| \(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \int \left (\frac {a^3 A}{x^4}+\frac {a^2 (a B+3 A b)}{x^3}+\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x^2}+3 c x \left (a B c+A b c+b^2 B\right )+\frac {A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )}{x}+b^3 B \left (\frac {3 c \left (A \left (a c+b^2\right )+2 a b B\right )}{b^3 B}+1\right )+c^2 x^2 (A c+3 b B)+B c^3 x^3\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 A}{3 x^3}-\frac {a^2 (a B+3 A b)}{2 x^2}+\frac {3}{2} c x^2 \left (a B c+A b c+b^2 B\right )-\frac {3 a \left (A \left (a c+b^2\right )+a b B\right )}{x}+x \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\log (x) \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {1}{3} c^2 x^3 (A c+3 b B)+\frac {1}{4} B c^3 x^4\) |
Input:
Int[((A + B*x)*(a + b*x + c*x^2)^3)/x^4,x]
Output:
-1/3*(a^3*A)/x^3 - (a^2*(3*A*b + a*B))/(2*x^2) - (3*a*(a*b*B + A*(b^2 + a* c)))/x + (b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x + (3*c*(b^2*B + A*b *c + a*B*c)*x^2)/2 + (c^2*(3*b*B + A*c)*x^3)/3 + (B*c^3*x^4)/4 + (3*a*B*(b ^2 + a*c) + A*(b^3 + 6*a*b*c))*Log[x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 1.02 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05
| method | result | size |
| default | \(\frac {B \,c^{3} x^{4}}{4}+\frac {A \,c^{3} x^{3}}{3}+x^{3} B b \,c^{2}+\frac {3 A b \,c^{2} x^{2}}{2}+\frac {3 B a \,c^{2} x^{2}}{2}+\frac {3 x^{2} B \,b^{2} c}{2}+3 A a \,c^{2} x +3 A \,b^{2} c x +6 B a b c x +x B \,b^{3}-\frac {a^{3} A}{3 x^{3}}-\frac {a^{2} \left (3 A b +B a \right )}{2 x^{2}}+\left (6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}\right ) \ln \left (x \right )-\frac {3 a \left (A a c +b^{2} A +a b B \right )}{x}\) | \(162\) |
| norman | \(\frac {\left (\frac {1}{3} A \,c^{3}+B b \,c^{2}\right ) x^{6}+\left (-\frac {3}{2} A \,a^{2} b -\frac {1}{2} B \,a^{3}\right ) x +\left (\frac {3}{2} A b \,c^{2}+\frac {3}{2} B a \,c^{2}+\frac {3}{2} B \,b^{2} c \right ) x^{5}+\left (-3 a^{2} A c -3 A a \,b^{2}-3 B \,a^{2} b \right ) x^{2}+\left (3 A a \,c^{2}+3 A \,b^{2} c +6 B a b c +B \,b^{3}\right ) x^{4}-\frac {a^{3} A}{3}+\frac {B \,c^{3} x^{7}}{4}}{x^{3}}+\left (6 A a b c +A \,b^{3}+3 B \,a^{2} c +3 B a \,b^{2}\right ) \ln \left (x \right )\) | \(166\) |
| risch | \(\frac {B \,c^{3} x^{4}}{4}+\frac {A \,c^{3} x^{3}}{3}+x^{3} B b \,c^{2}+\frac {3 A b \,c^{2} x^{2}}{2}+\frac {3 B a \,c^{2} x^{2}}{2}+\frac {3 x^{2} B \,b^{2} c}{2}+3 A a \,c^{2} x +3 A \,b^{2} c x +6 B a b c x +x B \,b^{3}+\frac {\left (-3 a^{2} A c -3 A a \,b^{2}-3 B \,a^{2} b \right ) x^{2}+\left (-\frac {3}{2} A \,a^{2} b -\frac {1}{2} B \,a^{3}\right ) x -\frac {a^{3} A}{3}}{x^{3}}+6 A \ln \left (x \right ) a b c +A \,b^{3} \ln \left (x \right )+3 B \ln \left (x \right ) a^{2} c +3 B \ln \left (x \right ) a \,b^{2}\) | \(174\) |
| parallelrisch | \(\frac {3 B \,c^{3} x^{7}+4 A \,c^{3} x^{6}+12 B b \,c^{2} x^{6}+18 A b \,c^{2} x^{5}+18 B a \,c^{2} x^{5}+18 B \,b^{2} c \,x^{5}+72 A \ln \left (x \right ) x^{3} a b c +12 A \ln \left (x \right ) x^{3} b^{3}+36 A a \,c^{2} x^{4}+36 A \,b^{2} c \,x^{4}+36 B \ln \left (x \right ) x^{3} a^{2} c +36 B \ln \left (x \right ) x^{3} a \,b^{2}+72 B a b c \,x^{4}+12 B \,b^{3} x^{4}-36 A \,a^{2} c \,x^{2}-36 A a \,b^{2} x^{2}-36 B \,a^{2} b \,x^{2}-18 A \,a^{2} b x -6 B \,a^{3} x -4 a^{3} A}{12 x^{3}}\) | \(200\) |
Input:
int((B*x+A)*(c*x^2+b*x+a)^3/x^4,x,method=_RETURNVERBOSE)
Output:
1/4*B*c^3*x^4+1/3*A*c^3*x^3+x^3*B*b*c^2+3/2*A*b*c^2*x^2+3/2*B*a*c^2*x^2+3/ 2*x^2*B*b^2*c+3*A*a*c^2*x+3*A*b^2*c*x+6*B*a*b*c*x+x*B*b^3-1/3*a^3*A/x^3-1/ 2*a^2*(3*A*b+B*a)/x^2+(6*A*a*b*c+A*b^3+3*B*a^2*c+3*B*a*b^2)*ln(x)-3*a*(A*a *c+A*b^2+B*a*b)/x
Time = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx=\frac {3 \, B c^{3} x^{7} + 4 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 18 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 12 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + 12 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} \log \left (x\right ) - 4 \, A a^{3} - 36 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{3}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^3/x^4,x, algorithm="fricas")
Output:
1/12*(3*B*c^3*x^7 + 4*(3*B*b*c^2 + A*c^3)*x^6 + 18*(B*b^2*c + (B*a + A*b)* c^2)*x^5 + 12*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 + 12*(3*B*a* b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3*log(x) - 4*A*a^3 - 36*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 - 6*(B*a^3 + 3*A*a^2*b)*x)/x^3
Time = 0.66 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx=\frac {B c^{3} x^{4}}{4} + x^{3} \left (\frac {A c^{3}}{3} + B b c^{2}\right ) + x^{2} \cdot \left (\frac {3 A b c^{2}}{2} + \frac {3 B a c^{2}}{2} + \frac {3 B b^{2} c}{2}\right ) + x \left (3 A a c^{2} + 3 A b^{2} c + 6 B a b c + B b^{3}\right ) + \left (6 A a b c + A b^{3} + 3 B a^{2} c + 3 B a b^{2}\right ) \log {\left (x \right )} + \frac {- 2 A a^{3} + x^{2} \left (- 18 A a^{2} c - 18 A a b^{2} - 18 B a^{2} b\right ) + x \left (- 9 A a^{2} b - 3 B a^{3}\right )}{6 x^{3}} \] Input:
integrate((B*x+A)*(c*x**2+b*x+a)**3/x**4,x)
Output:
B*c**3*x**4/4 + x**3*(A*c**3/3 + B*b*c**2) + x**2*(3*A*b*c**2/2 + 3*B*a*c* *2/2 + 3*B*b**2*c/2) + x*(3*A*a*c**2 + 3*A*b**2*c + 6*B*a*b*c + B*b**3) + (6*A*a*b*c + A*b**3 + 3*B*a**2*c + 3*B*a*b**2)*log(x) + (-2*A*a**3 + x**2* (-18*A*a**2*c - 18*A*a*b**2 - 18*B*a**2*b) + x*(-9*A*a**2*b - 3*B*a**3))/( 6*x**3)
Time = 0.04 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx=\frac {1}{4} \, B c^{3} x^{4} + \frac {1}{3} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + \frac {3}{2} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{2} + {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x + {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} \log \left (x\right ) - \frac {2 \, A a^{3} + 18 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^3/x^4,x, algorithm="maxima")
Output:
1/4*B*c^3*x^4 + 1/3*(3*B*b*c^2 + A*c^3)*x^3 + 3/2*(B*b^2*c + (B*a + A*b)*c ^2)*x^2 + (B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x + (3*B*a*b^2 + A*b ^3 + 3*(B*a^2 + 2*A*a*b)*c)*log(x) - 1/6*(2*A*a^3 + 18*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 3*(B*a^3 + 3*A*a^2*b)*x)/x^3
Time = 0.22 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx=\frac {1}{4} \, B c^{3} x^{4} + B b c^{2} x^{3} + \frac {1}{3} \, A c^{3} x^{3} + \frac {3}{2} \, B b^{2} c x^{2} + \frac {3}{2} \, B a c^{2} x^{2} + \frac {3}{2} \, A b c^{2} x^{2} + B b^{3} x + 6 \, B a b c x + 3 \, A b^{2} c x + 3 \, A a c^{2} x + {\left (3 \, B a b^{2} + A b^{3} + 3 \, B a^{2} c + 6 \, A a b c\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A a^{3} + 18 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{6 \, x^{3}} \] Input:
integrate((B*x+A)*(c*x^2+b*x+a)^3/x^4,x, algorithm="giac")
Output:
1/4*B*c^3*x^4 + B*b*c^2*x^3 + 1/3*A*c^3*x^3 + 3/2*B*b^2*c*x^2 + 3/2*B*a*c^ 2*x^2 + 3/2*A*b*c^2*x^2 + B*b^3*x + 6*B*a*b*c*x + 3*A*b^2*c*x + 3*A*a*c^2* x + (3*B*a*b^2 + A*b^3 + 3*B*a^2*c + 6*A*a*b*c)*log(abs(x)) - 1/6*(2*A*a^3 + 18*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 3*(B*a^3 + 3*A*a^2*b)*x)/x^3
Time = 10.59 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx=\ln \left (x\right )\,\left (3\,B\,c\,a^2+3\,B\,a\,b^2+6\,A\,c\,a\,b+A\,b^3\right )-\frac {x\,\left (\frac {B\,a^3}{2}+\frac {3\,A\,b\,a^2}{2}\right )+\frac {A\,a^3}{3}+x^2\,\left (3\,B\,a^2\,b+3\,A\,c\,a^2+3\,A\,a\,b^2\right )}{x^3}+x^3\,\left (\frac {A\,c^3}{3}+B\,b\,c^2\right )+x^2\,\left (\frac {3\,B\,b^2\,c}{2}+\frac {3\,A\,b\,c^2}{2}+\frac {3\,B\,a\,c^2}{2}\right )+x\,\left (B\,b^3+3\,A\,b^2\,c+6\,B\,a\,b\,c+3\,A\,a\,c^2\right )+\frac {B\,c^3\,x^4}{4} \] Input:
int(((A + B*x)*(a + b*x + c*x^2)^3)/x^4,x)
Output:
log(x)*(A*b^3 + 3*B*a*b^2 + 3*B*a^2*c + 6*A*a*b*c) - (x*((B*a^3)/2 + (3*A* a^2*b)/2) + (A*a^3)/3 + x^2*(3*A*a*b^2 + 3*A*a^2*c + 3*B*a^2*b))/x^3 + x^3 *((A*c^3)/3 + B*b*c^2) + x^2*((3*A*b*c^2)/2 + (3*B*a*c^2)/2 + (3*B*b^2*c)/ 2) + x*(B*b^3 + 3*A*a*c^2 + 3*A*b^2*c + 6*B*a*b*c) + (B*c^3*x^4)/4
Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^3}{x^4} \, dx=\frac {108 \,\mathrm {log}\left (x \right ) a^{2} b c \,x^{3}+48 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{3}-4 a^{4}-24 a^{3} b x -36 a^{3} c \,x^{2}-72 a^{2} b^{2} x^{2}+36 a^{2} c^{2} x^{4}+108 a \,b^{2} c \,x^{4}+36 a b \,c^{2} x^{5}+4 a \,c^{3} x^{6}+12 b^{4} x^{4}+18 b^{3} c \,x^{5}+12 b^{2} c^{2} x^{6}+3 b \,c^{3} x^{7}}{12 x^{3}} \] Input:
int((B*x+A)*(c*x^2+b*x+a)^3/x^4,x)
Output:
(108*log(x)*a**2*b*c*x**3 + 48*log(x)*a*b**3*x**3 - 4*a**4 - 24*a**3*b*x - 36*a**3*c*x**2 - 72*a**2*b**2*x**2 + 36*a**2*c**2*x**4 + 108*a*b**2*c*x** 4 + 36*a*b*c**2*x**5 + 4*a*c**3*x**6 + 12*b**4*x**4 + 18*b**3*c*x**5 + 12* b**2*c**2*x**6 + 3*b*c**3*x**7)/(12*x**3)