Integrand size = 24, antiderivative size = 94 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx=\frac {3 \left (b^2-4 a c\right )^2 x}{64 c^3 d^2}+\frac {\left (b^2-4 a c\right )^3}{128 c^4 d^2 (b+2 c x)}-\frac {\left (b^2-4 a c\right ) (b+2 c x)^3}{128 c^4 d^2}+\frac {(b+2 c x)^5}{640 c^4 d^2} \] Output:
3/64*(-4*a*c+b^2)^2*x/c^3/d^2+1/128*(-4*a*c+b^2)^3/c^4/d^2/(2*c*x+b)-1/128 *(-4*a*c+b^2)*(2*c*x+b)^3/c^4/d^2+1/640*(2*c*x+b)^5/c^4/d^2
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx=\frac {\frac {10 \left (b^4-12 a b^2 c+48 a^2 c^2\right ) x}{c^3}-\frac {20 b \left (b^2-12 a c\right ) x^2}{c^2}+\frac {40 \left (b^2+4 a c\right ) x^3}{c}+80 b x^4+32 c x^5+\frac {5 \left (b^2-4 a c\right )^3}{c^4 (b+2 c x)}}{640 d^2} \] Input:
Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^2,x]
Output:
((10*(b^4 - 12*a*b^2*c + 48*a^2*c^2)*x)/c^3 - (20*b*(b^2 - 12*a*c)*x^2)/c^ 2 + (40*(b^2 + 4*a*c)*x^3)/c + 80*b*x^4 + 32*c*x^5 + (5*(b^2 - 4*a*c)^3)/( c^4*(b + 2*c*x)))/(640*d^2)
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1107, 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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {3 \left (4 a c-b^2\right ) (b d+2 c d x)^2}{64 c^3 d^4}+\frac {3 \left (4 a c-b^2\right )^2}{64 c^3 d^2}+\frac {\left (4 a c-b^2\right )^3}{64 c^3 (b d+2 c d x)^2}+\frac {(b d+2 c d x)^4}{64 c^3 d^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (b^2-4 a c\right ) (b+2 c x)^3}{128 c^4 d^2}+\frac {\left (b^2-4 a c\right )^3}{128 c^4 d^2 (b+2 c x)}+\frac {3 x \left (b^2-4 a c\right )^2}{64 c^3 d^2}+\frac {(b+2 c x)^5}{640 c^4 d^2}\) |
Input:
Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^2,x]
Output:
(3*(b^2 - 4*a*c)^2*x)/(64*c^3*d^2) + (b^2 - 4*a*c)^3/(128*c^4*d^2*(b + 2*c *x)) - ((b^2 - 4*a*c)*(b + 2*c*x)^3)/(128*c^4*d^2) + (b + 2*c*x)^5/(640*c^ 4*d^2)
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; F reeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b*e, 0] && IGtQ[p, 0] && !(EqQ[ m, 3] && NeQ[p, 1])
Time = 0.83 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80
method | result | size |
gosper | \(-\frac {-2 c^{3} x^{6}-6 b \,c^{2} x^{5}-10 a \,c^{2} x^{4}-5 b^{2} c \,x^{4}-20 a b c \,x^{3}-30 a^{2} c \,x^{2}+10 a^{3}}{20 \left (2 c x +b \right ) c \,d^{2}}\) | \(75\) |
parallelrisch | \(\frac {2 c^{3} x^{6}+6 b \,c^{2} x^{5}+10 a \,c^{2} x^{4}+5 b^{2} c \,x^{4}+20 a b c \,x^{3}+30 a^{2} c \,x^{2}-10 a^{3}}{20 c \,d^{2} \left (2 c x +b \right )}\) | \(75\) |
orering | \(-\frac {\left (-2 c^{3} x^{6}-6 b \,c^{2} x^{5}-10 a \,c^{2} x^{4}-5 b^{2} c \,x^{4}-20 a b c \,x^{3}-30 a^{2} c \,x^{2}+10 a^{3}\right ) \left (2 c x +b \right )}{20 c \left (2 c d x +b d \right )^{2}}\) | \(81\) |
norman | \(\frac {\frac {a b \,x^{3}}{d}+\frac {a^{3} x}{b d}+\frac {c^{2} x^{6}}{10 d}+\frac {3 a^{2} x^{2}}{2 d}+\frac {\left (2 a c +b^{2}\right ) x^{4}}{4 d}+\frac {3 b c \,x^{5}}{10 d}}{d \left (2 c x +b \right )}\) | \(82\) |
default | \(\frac {\frac {\frac {16}{5} c^{4} x^{5}+8 b \,c^{3} x^{4}+16 a \,c^{3} x^{3}+4 b^{2} c^{2} x^{3}+24 a b \,c^{2} x^{2}-2 b^{3} c \,x^{2}+48 a^{2} c^{2} x -12 a \,b^{2} c x +b^{4} x}{64 c^{3}}-\frac {64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}}{128 c^{4} \left (2 c x +b \right )}}{d^{2}}\) | \(135\) |
risch | \(\frac {c \,x^{5}}{20 d^{2}}+\frac {b \,x^{4}}{8 d^{2}}+\frac {a \,x^{3}}{4 d^{2}}+\frac {b^{2} x^{3}}{16 c \,d^{2}}+\frac {3 a b \,x^{2}}{8 c \,d^{2}}-\frac {b^{3} x^{2}}{32 c^{2} d^{2}}+\frac {3 a^{2} x}{4 c \,d^{2}}-\frac {3 a \,b^{2} x}{16 c^{2} d^{2}}+\frac {b^{4} x}{64 c^{3} d^{2}}-\frac {a^{3}}{2 c \,d^{2} \left (2 c x +b \right )}+\frac {3 a^{2} b^{2}}{8 c^{2} d^{2} \left (2 c x +b \right )}-\frac {3 a \,b^{4}}{32 c^{3} d^{2} \left (2 c x +b \right )}+\frac {b^{6}}{128 c^{4} d^{2} \left (2 c x +b \right )}\) | \(187\) |
Input:
int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x,method=_RETURNVERBOSE)
Output:
-1/20*(-2*c^3*x^6-6*b*c^2*x^5-10*a*c^2*x^4-5*b^2*c*x^4-20*a*b*c*x^3-30*a^2 *c*x^2+10*a^3)/(2*c*x+b)/c/d^2
Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx=\frac {64 \, c^{6} x^{6} + 192 \, b c^{5} x^{5} + 640 \, a b c^{4} x^{3} + 960 \, a^{2} c^{4} x^{2} + 5 \, b^{6} - 60 \, a b^{4} c + 240 \, a^{2} b^{2} c^{2} - 320 \, a^{3} c^{3} + 160 \, {\left (b^{2} c^{4} + 2 \, a c^{5}\right )} x^{4} + 10 \, {\left (b^{5} c - 12 \, a b^{3} c^{2} + 48 \, a^{2} b c^{3}\right )} x}{640 \, {\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x, algorithm="fricas")
Output:
1/640*(64*c^6*x^6 + 192*b*c^5*x^5 + 640*a*b*c^4*x^3 + 960*a^2*c^4*x^2 + 5* b^6 - 60*a*b^4*c + 240*a^2*b^2*c^2 - 320*a^3*c^3 + 160*(b^2*c^4 + 2*a*c^5) *x^4 + 10*(b^5*c - 12*a*b^3*c^2 + 48*a^2*b*c^3)*x)/(2*c^5*d^2*x + b*c^4*d^ 2)
Time = 0.32 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.70 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx=\frac {b x^{4}}{8 d^{2}} + \frac {c x^{5}}{20 d^{2}} + x^{3} \left (\frac {a}{4 d^{2}} + \frac {b^{2}}{16 c d^{2}}\right ) + x^{2} \cdot \left (\frac {3 a b}{8 c d^{2}} - \frac {b^{3}}{32 c^{2} d^{2}}\right ) + x \left (\frac {3 a^{2}}{4 c d^{2}} - \frac {3 a b^{2}}{16 c^{2} d^{2}} + \frac {b^{4}}{64 c^{3} d^{2}}\right ) + \frac {- 64 a^{3} c^{3} + 48 a^{2} b^{2} c^{2} - 12 a b^{4} c + b^{6}}{128 b c^{4} d^{2} + 256 c^{5} d^{2} x} \] Input:
integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**2,x)
Output:
b*x**4/(8*d**2) + c*x**5/(20*d**2) + x**3*(a/(4*d**2) + b**2/(16*c*d**2)) + x**2*(3*a*b/(8*c*d**2) - b**3/(32*c**2*d**2)) + x*(3*a**2/(4*c*d**2) - 3 *a*b**2/(16*c**2*d**2) + b**4/(64*c**3*d**2)) + (-64*a**3*c**3 + 48*a**2*b **2*c**2 - 12*a*b**4*c + b**6)/(128*b*c**4*d**2 + 256*c**5*d**2*x)
Time = 0.03 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx=\frac {b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}{128 \, {\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}} + \frac {16 \, c^{4} x^{5} + 40 \, b c^{3} x^{4} + 20 \, {\left (b^{2} c^{2} + 4 \, a c^{3}\right )} x^{3} - 10 \, {\left (b^{3} c - 12 \, a b c^{2}\right )} x^{2} + 5 \, {\left (b^{4} - 12 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} x}{320 \, c^{3} d^{2}} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x, algorithm="maxima")
Output:
1/128*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)/(2*c^5*d^2*x + b*c^ 4*d^2) + 1/320*(16*c^4*x^5 + 40*b*c^3*x^4 + 20*(b^2*c^2 + 4*a*c^3)*x^3 - 1 0*(b^3*c - 12*a*b*c^2)*x^2 + 5*(b^4 - 12*a*b^2*c + 48*a^2*c^2)*x)/(c^3*d^2 )
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (86) = 172\).
Time = 0.14 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.35 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx=\frac {{\left (\frac {15 \, b^{4} d^{4}}{{\left (2 \, c d x + b d\right )}^{4}} - \frac {120 \, a b^{2} c d^{4}}{{\left (2 \, c d x + b d\right )}^{4}} + \frac {240 \, a^{2} c^{2} d^{4}}{{\left (2 \, c d x + b d\right )}^{4}} - \frac {5 \, b^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {20 \, a c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + 1\right )} {\left (2 \, c d x + b d\right )}^{5}}{640 \, c^{4} d^{7}} + \frac {\frac {b^{6} c^{5} d^{11}}{2 \, c d x + b d} - \frac {12 \, a b^{4} c^{6} d^{11}}{2 \, c d x + b d} + \frac {48 \, a^{2} b^{2} c^{7} d^{11}}{2 \, c d x + b d} - \frac {64 \, a^{3} c^{8} d^{11}}{2 \, c d x + b d}}{128 \, c^{9} d^{12}} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x, algorithm="giac")
Output:
1/640*(15*b^4*d^4/(2*c*d*x + b*d)^4 - 120*a*b^2*c*d^4/(2*c*d*x + b*d)^4 + 240*a^2*c^2*d^4/(2*c*d*x + b*d)^4 - 5*b^2*d^2/(2*c*d*x + b*d)^2 + 20*a*c*d ^2/(2*c*d*x + b*d)^2 + 1)*(2*c*d*x + b*d)^5/(c^4*d^7) + 1/128*(b^6*c^5*d^1 1/(2*c*d*x + b*d) - 12*a*b^4*c^6*d^11/(2*c*d*x + b*d) + 48*a^2*b^2*c^7*d^1 1/(2*c*d*x + b*d) - 64*a^3*c^8*d^11/(2*c*d*x + b*d))/(c^9*d^12)
Time = 5.14 (sec) , antiderivative size = 293, normalized size of antiderivative = 3.12 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx=x^3\,\left (\frac {b^2+a\,c}{4\,c\,d^2}-\frac {3\,b^2}{16\,c\,d^2}\right )-x^2\,\left (\frac {b^3}{16\,c^2\,d^2}-\frac {b^3+6\,a\,c\,b}{8\,c^2\,d^2}+\frac {b\,\left (\frac {3\,\left (b^2+a\,c\right )}{4\,c\,d^2}-\frac {9\,b^2}{16\,c\,d^2}\right )}{2\,c}\right )+x\,\left (\frac {b\,\left (\frac {b^3}{8\,c^2\,d^2}-\frac {b^3+6\,a\,c\,b}{4\,c^2\,d^2}+\frac {b\,\left (\frac {3\,\left (b^2+a\,c\right )}{4\,c\,d^2}-\frac {9\,b^2}{16\,c\,d^2}\right )}{c}\right )}{c}-\frac {b^2\,\left (\frac {3\,\left (b^2+a\,c\right )}{4\,c\,d^2}-\frac {9\,b^2}{16\,c\,d^2}\right )}{4\,c^2}+\frac {3\,a\,\left (b^2+a\,c\right )}{4\,c^2\,d^2}\right )+\frac {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}{2\,c\,\left (128\,x\,c^4\,d^2+64\,b\,c^3\,d^2\right )}+\frac {b\,x^4}{8\,d^2}+\frac {c\,x^5}{20\,d^2} \] Input:
int((a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^2,x)
Output:
x^3*((a*c + b^2)/(4*c*d^2) - (3*b^2)/(16*c*d^2)) - x^2*(b^3/(16*c^2*d^2) - (b^3 + 6*a*b*c)/(8*c^2*d^2) + (b*((3*(a*c + b^2))/(4*c*d^2) - (9*b^2)/(16 *c*d^2)))/(2*c)) + x*((b*(b^3/(8*c^2*d^2) - (b^3 + 6*a*b*c)/(4*c^2*d^2) + (b*((3*(a*c + b^2))/(4*c*d^2) - (9*b^2)/(16*c*d^2)))/c))/c - (b^2*((3*(a*c + b^2))/(4*c*d^2) - (9*b^2)/(16*c*d^2)))/(4*c^2) + (3*a*(a*c + b^2))/(4*c ^2*d^2)) + (b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)/(2*c*(64*b*c^3 *d^2 + 128*c^4*d^2*x)) + (b*x^4)/(8*d^2) + (c*x^5)/(20*d^2)
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx=\frac {x \left (2 b \,c^{2} x^{5}+6 b^{2} c \,x^{4}+10 a b c \,x^{3}+5 b^{3} x^{3}+20 a \,b^{2} x^{2}+30 a^{2} b x +20 a^{3}\right )}{20 b \,d^{2} \left (2 c x +b \right )} \] Input:
int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x)
Output:
(x*(20*a**3 + 30*a**2*b*x + 20*a*b**2*x**2 + 10*a*b*c*x**3 + 5*b**3*x**3 + 6*b**2*c*x**4 + 2*b*c**2*x**5))/(20*b*d**2*(b + 2*c*x))