Integrand size = 29, antiderivative size = 92 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^{12}} \, dx=-\frac {(b c-a d)^3}{8 b^4 (a+b x)^8}-\frac {3 d (b c-a d)^2}{7 b^4 (a+b x)^7}-\frac {d^2 (b c-a d)}{2 b^4 (a+b x)^6}-\frac {d^3}{5 b^4 (a+b x)^5} \] Output:
-1/8*(-a*d+b*c)^3/b^4/(b*x+a)^8-3/7*d*(-a*d+b*c)^2/b^4/(b*x+a)^7-1/2*d^2*( -a*d+b*c)/b^4/(b*x+a)^6-1/5*d^3/b^4/(b*x+a)^5
Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^{12}} \, dx=-\frac {a^3 d^3+a^2 b d^2 (5 c+8 d x)+a b^2 d \left (15 c^2+40 c d x+28 d^2 x^2\right )+b^3 \left (35 c^3+120 c^2 d x+140 c d^2 x^2+56 d^3 x^3\right )}{280 b^4 (a+b x)^8} \] Input:
Integrate[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^12,x]
Output:
-1/280*(a^3*d^3 + a^2*b*d^2*(5*c + 8*d*x) + a*b^2*d*(15*c^2 + 40*c*d*x + 2 8*d^2*x^2) + b^3*(35*c^3 + 120*c^2*d*x + 140*c*d^2*x^2 + 56*d^3*x^3))/(b^4 *(a + b*x)^8)
Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1121, 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.
\(\displaystyle \int \frac {\left (x (a d+b c)+a c+b d x^2\right )^3}{(a+b x)^{12}} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \left (\frac {3 d^2 (b c-a d)}{b^3 (a+b x)^7}+\frac {3 d (b c-a d)^2}{b^3 (a+b x)^8}+\frac {(b c-a d)^3}{b^3 (a+b x)^9}+\frac {d^3}{b^3 (a+b x)^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d^2 (b c-a d)}{2 b^4 (a+b x)^6}-\frac {3 d (b c-a d)^2}{7 b^4 (a+b x)^7}-\frac {(b c-a d)^3}{8 b^4 (a+b x)^8}-\frac {d^3}{5 b^4 (a+b x)^5}\) |
Input:
Int[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^12,x]
Output:
-1/8*(b*c - a*d)^3/(b^4*(a + b*x)^8) - (3*d*(b*c - a*d)^2)/(7*b^4*(a + b*x )^7) - (d^2*(b*c - a*d))/(2*b^4*(a + b*x)^6) - d^3/(5*b^4*(a + b*x)^5)
Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
Time = 1.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.20
method | result | size |
risch | \(\frac {-\frac {x^{3} d^{3}}{5 b}-\frac {d^{2} \left (a d +5 b c \right ) x^{2}}{10 b^{2}}-\frac {d \left (a^{2} d^{2}+5 a b c d +15 b^{2} c^{2}\right ) x}{35 b^{3}}-\frac {a^{3} d^{3}+5 a^{2} b c \,d^{2}+15 a \,b^{2} c^{2} d +35 b^{3} c^{3}}{280 b^{4}}}{\left (b x +a \right )^{8}}\) | \(110\) |
gosper | \(-\frac {56 x^{3} b^{3} d^{3}+28 a \,b^{2} d^{3} x^{2}+140 x^{2} b^{3} c \,d^{2}+8 a^{2} b \,d^{3} x +40 x a \,b^{2} c \,d^{2}+120 x \,b^{3} c^{2} d +a^{3} d^{3}+5 a^{2} b c \,d^{2}+15 a \,b^{2} c^{2} d +35 b^{3} c^{3}}{280 b^{4} \left (b x +a \right )^{8}}\) | \(115\) |
default | \(-\frac {-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}}{8 b^{4} \left (b x +a \right )^{8}}-\frac {d^{3}}{5 b^{4} \left (b x +a \right )^{5}}-\frac {3 d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{7 b^{4} \left (b x +a \right )^{7}}+\frac {d^{2} \left (a d -b c \right )}{2 b^{4} \left (b x +a \right )^{6}}\) | \(122\) |
parallelrisch | \(\frac {-56 d^{3} x^{3} b^{7}-28 a \,b^{6} d^{3} x^{2}-140 b^{7} c \,d^{2} x^{2}-8 a^{2} b^{5} d^{3} x -40 a \,b^{6} c \,d^{2} x -120 b^{7} c^{2} d x -a^{3} b^{4} d^{3}-5 a^{2} b^{5} c \,d^{2}-15 a \,c^{2} d \,b^{6}-35 c^{3} b^{7}}{280 b^{8} \left (b x +a \right )^{8}}\) | \(123\) |
orering | \(-\frac {\left (56 x^{3} b^{3} d^{3}+28 a \,b^{2} d^{3} x^{2}+140 x^{2} b^{3} c \,d^{2}+8 a^{2} b \,d^{3} x +40 x a \,b^{2} c \,d^{2}+120 x \,b^{3} c^{2} d +a^{3} d^{3}+5 a^{2} b c \,d^{2}+15 a \,b^{2} c^{2} d +35 b^{3} c^{3}\right ) \left (a c +\left (a d +b c \right ) x +b d \,x^{2}\right )^{3}}{280 b^{4} \left (b x +a \right )^{11} \left (d x +c \right )^{3}}\) | \(143\) |
norman | \(\frac {\frac {a^{3} \left (-a^{3} b^{7} d^{3}-5 a^{2} b^{8} c \,d^{2}-15 a \,b^{9} c^{2} d -35 b^{10} c^{3}\right )}{280 b^{11}}-\frac {b^{2} d^{3} x^{6}}{5}+\frac {\left (-7 a \,d^{3} b^{7}-5 c \,d^{2} b^{8}\right ) x^{5}}{10 b^{6}}+\frac {\left (-13 a^{2} b^{7} d^{3}-23 a \,d^{2} c \,b^{8}-6 c^{2} d \,b^{9}\right ) x^{4}}{14 b^{7}}+\frac {\left (-33 a^{3} b^{7} d^{3}-109 a^{2} b^{8} c \,d^{2}-75 a \,b^{9} c^{2} d -7 b^{10} c^{3}\right ) x^{3}}{56 b^{8}}+\frac {a \left (-11 a^{3} b^{7} d^{3}-55 a^{2} b^{8} c \,d^{2}-81 a \,b^{9} c^{2} d -21 b^{10} c^{3}\right ) x^{2}}{56 b^{9}}+\frac {a^{2} \left (-11 a^{3} b^{7} d^{3}-55 a^{2} b^{8} c \,d^{2}-165 a \,b^{9} c^{2} d -105 b^{10} c^{3}\right ) x}{280 b^{10}}}{\left (b x +a \right )^{11}}\) | \(289\) |
Input:
int((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^12,x,method=_RETURNVERBOSE)
Output:
(-1/5/b*x^3*d^3-1/10/b^2*d^2*(a*d+5*b*c)*x^2-1/35/b^3*d*(a^2*d^2+5*a*b*c*d +15*b^2*c^2)*x-1/280/b^4*(a^3*d^3+5*a^2*b*c*d^2+15*a*b^2*c^2*d+35*b^3*c^3) )/(b*x+a)^8
Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (84) = 168\).
Time = 0.07 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.10 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^{12}} \, dx=-\frac {56 \, b^{3} d^{3} x^{3} + 35 \, b^{3} c^{3} + 15 \, a b^{2} c^{2} d + 5 \, a^{2} b c d^{2} + a^{3} d^{3} + 28 \, {\left (5 \, b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{2} + 8 \, {\left (15 \, b^{3} c^{2} d + 5 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}{280 \, {\left (b^{12} x^{8} + 8 \, a b^{11} x^{7} + 28 \, a^{2} b^{10} x^{6} + 56 \, a^{3} b^{9} x^{5} + 70 \, a^{4} b^{8} x^{4} + 56 \, a^{5} b^{7} x^{3} + 28 \, a^{6} b^{6} x^{2} + 8 \, a^{7} b^{5} x + a^{8} b^{4}\right )}} \] Input:
integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^12,x, algorithm="fricas")
Output:
-1/280*(56*b^3*d^3*x^3 + 35*b^3*c^3 + 15*a*b^2*c^2*d + 5*a^2*b*c*d^2 + a^3 *d^3 + 28*(5*b^3*c*d^2 + a*b^2*d^3)*x^2 + 8*(15*b^3*c^2*d + 5*a*b^2*c*d^2 + a^2*b*d^3)*x)/(b^12*x^8 + 8*a*b^11*x^7 + 28*a^2*b^10*x^6 + 56*a^3*b^9*x^ 5 + 70*a^4*b^8*x^4 + 56*a^5*b^7*x^3 + 28*a^6*b^6*x^2 + 8*a^7*b^5*x + a^8*b ^4)
Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (82) = 164\).
Time = 11.37 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.25 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^{12}} \, dx=\frac {- a^{3} d^{3} - 5 a^{2} b c d^{2} - 15 a b^{2} c^{2} d - 35 b^{3} c^{3} - 56 b^{3} d^{3} x^{3} + x^{2} \left (- 28 a b^{2} d^{3} - 140 b^{3} c d^{2}\right ) + x \left (- 8 a^{2} b d^{3} - 40 a b^{2} c d^{2} - 120 b^{3} c^{2} d\right )}{280 a^{8} b^{4} + 2240 a^{7} b^{5} x + 7840 a^{6} b^{6} x^{2} + 15680 a^{5} b^{7} x^{3} + 19600 a^{4} b^{8} x^{4} + 15680 a^{3} b^{9} x^{5} + 7840 a^{2} b^{10} x^{6} + 2240 a b^{11} x^{7} + 280 b^{12} x^{8}} \] Input:
integrate((a*c+(a*d+b*c)*x+b*d*x**2)**3/(b*x+a)**12,x)
Output:
(-a**3*d**3 - 5*a**2*b*c*d**2 - 15*a*b**2*c**2*d - 35*b**3*c**3 - 56*b**3* d**3*x**3 + x**2*(-28*a*b**2*d**3 - 140*b**3*c*d**2) + x*(-8*a**2*b*d**3 - 40*a*b**2*c*d**2 - 120*b**3*c**2*d))/(280*a**8*b**4 + 2240*a**7*b**5*x + 7840*a**6*b**6*x**2 + 15680*a**5*b**7*x**3 + 19600*a**4*b**8*x**4 + 15680* a**3*b**9*x**5 + 7840*a**2*b**10*x**6 + 2240*a*b**11*x**7 + 280*b**12*x**8 )
Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (84) = 168\).
Time = 0.04 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.10 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^{12}} \, dx=-\frac {56 \, b^{3} d^{3} x^{3} + 35 \, b^{3} c^{3} + 15 \, a b^{2} c^{2} d + 5 \, a^{2} b c d^{2} + a^{3} d^{3} + 28 \, {\left (5 \, b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{2} + 8 \, {\left (15 \, b^{3} c^{2} d + 5 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}{280 \, {\left (b^{12} x^{8} + 8 \, a b^{11} x^{7} + 28 \, a^{2} b^{10} x^{6} + 56 \, a^{3} b^{9} x^{5} + 70 \, a^{4} b^{8} x^{4} + 56 \, a^{5} b^{7} x^{3} + 28 \, a^{6} b^{6} x^{2} + 8 \, a^{7} b^{5} x + a^{8} b^{4}\right )}} \] Input:
integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^12,x, algorithm="maxima")
Output:
-1/280*(56*b^3*d^3*x^3 + 35*b^3*c^3 + 15*a*b^2*c^2*d + 5*a^2*b*c*d^2 + a^3 *d^3 + 28*(5*b^3*c*d^2 + a*b^2*d^3)*x^2 + 8*(15*b^3*c^2*d + 5*a*b^2*c*d^2 + a^2*b*d^3)*x)/(b^12*x^8 + 8*a*b^11*x^7 + 28*a^2*b^10*x^6 + 56*a^3*b^9*x^ 5 + 70*a^4*b^8*x^4 + 56*a^5*b^7*x^3 + 28*a^6*b^6*x^2 + 8*a^7*b^5*x + a^8*b ^4)
Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^{12}} \, dx=-\frac {56 \, b^{3} d^{3} x^{3} + 140 \, b^{3} c d^{2} x^{2} + 28 \, a b^{2} d^{3} x^{2} + 120 \, b^{3} c^{2} d x + 40 \, a b^{2} c d^{2} x + 8 \, a^{2} b d^{3} x + 35 \, b^{3} c^{3} + 15 \, a b^{2} c^{2} d + 5 \, a^{2} b c d^{2} + a^{3} d^{3}}{280 \, {\left (b x + a\right )}^{8} b^{4}} \] Input:
integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^12,x, algorithm="giac")
Output:
-1/280*(56*b^3*d^3*x^3 + 140*b^3*c*d^2*x^2 + 28*a*b^2*d^3*x^2 + 120*b^3*c^ 2*d*x + 40*a*b^2*c*d^2*x + 8*a^2*b*d^3*x + 35*b^3*c^3 + 15*a*b^2*c^2*d + 5 *a^2*b*c*d^2 + a^3*d^3)/((b*x + a)^8*b^4)
Time = 0.06 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.03 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^{12}} \, dx=-\frac {\frac {a^3\,d^3+5\,a^2\,b\,c\,d^2+15\,a\,b^2\,c^2\,d+35\,b^3\,c^3}{280\,b^4}+\frac {d^3\,x^3}{5\,b}+\frac {d\,x\,\left (a^2\,d^2+5\,a\,b\,c\,d+15\,b^2\,c^2\right )}{35\,b^3}+\frac {d^2\,x^2\,\left (a\,d+5\,b\,c\right )}{10\,b^2}}{a^8+8\,a^7\,b\,x+28\,a^6\,b^2\,x^2+56\,a^5\,b^3\,x^3+70\,a^4\,b^4\,x^4+56\,a^3\,b^5\,x^5+28\,a^2\,b^6\,x^6+8\,a\,b^7\,x^7+b^8\,x^8} \] Input:
int((a*c + x*(a*d + b*c) + b*d*x^2)^3/(a + b*x)^12,x)
Output:
-((a^3*d^3 + 35*b^3*c^3 + 15*a*b^2*c^2*d + 5*a^2*b*c*d^2)/(280*b^4) + (d^3 *x^3)/(5*b) + (d*x*(a^2*d^2 + 15*b^2*c^2 + 5*a*b*c*d))/(35*b^3) + (d^2*x^2 *(a*d + 5*b*c))/(10*b^2))/(a^8 + b^8*x^8 + 8*a*b^7*x^7 + 28*a^6*b^2*x^2 + 56*a^5*b^3*x^3 + 70*a^4*b^4*x^4 + 56*a^3*b^5*x^5 + 28*a^2*b^6*x^6 + 8*a^7* b*x)
Time = 0.22 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.09 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^3}{(a+b x)^{12}} \, dx=\frac {-56 b^{3} d^{3} x^{3}-28 a \,b^{2} d^{3} x^{2}-140 b^{3} c \,d^{2} x^{2}-8 a^{2} b \,d^{3} x -40 a \,b^{2} c \,d^{2} x -120 b^{3} c^{2} d x -a^{3} d^{3}-5 a^{2} b c \,d^{2}-15 a \,b^{2} c^{2} d -35 b^{3} c^{3}}{280 b^{4} \left (b^{8} x^{8}+8 a \,b^{7} x^{7}+28 a^{2} b^{6} x^{6}+56 a^{3} b^{5} x^{5}+70 a^{4} b^{4} x^{4}+56 a^{5} b^{3} x^{3}+28 a^{6} b^{2} x^{2}+8 a^{7} b x +a^{8}\right )} \] Input:
int((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^12,x)
Output:
( - a**3*d**3 - 5*a**2*b*c*d**2 - 8*a**2*b*d**3*x - 15*a*b**2*c**2*d - 40* a*b**2*c*d**2*x - 28*a*b**2*d**3*x**2 - 35*b**3*c**3 - 120*b**3*c**2*d*x - 140*b**3*c*d**2*x**2 - 56*b**3*d**3*x**3)/(280*b**4*(a**8 + 8*a**7*b*x + 28*a**6*b**2*x**2 + 56*a**5*b**3*x**3 + 70*a**4*b**4*x**4 + 56*a**3*b**5*x **5 + 28*a**2*b**6*x**6 + 8*a*b**7*x**7 + b**8*x**8))