Integrand size = 24, antiderivative size = 100 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx=\frac {\left (b^2-4 a c\right )^3}{256 c^4 d^3 (b+2 c x)^2}-\frac {3 \left (b^2-4 a c\right ) (b+2 c x)^2}{256 c^4 d^3}+\frac {(b+2 c x)^4}{512 c^4 d^3}+\frac {3 \left (b^2-4 a c\right )^2 \log (b+2 c x)}{128 c^4 d^3} \] Output:
1/256*(-4*a*c+b^2)^3/c^4/d^3/(2*c*x+b)^2-3/256*(-4*a*c+b^2)*(2*c*x+b)^2/c^ 4/d^3+1/512*(2*c*x+b)^4/c^4/d^3+3/128*(-4*a*c+b^2)^2*ln(2*c*x+b)/c^4/d^3
Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx=\frac {-\frac {8 b \left (b^2-6 a c\right ) x}{c^3}+\frac {48 a x^2}{c}+\frac {16 b x^3}{c}+8 x^4+\frac {\left (b^2-4 a c\right )^3}{c^4 (b+2 c x)^2}+\frac {6 \left (b^2-4 a c\right )^2 \log (b+2 c x)}{c^4}}{256 d^3} \] Input:
Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^3,x]
Output:
((-8*b*(b^2 - 6*a*c)*x)/c^3 + (48*a*x^2)/c + (16*b*x^3)/c + 8*x^4 + (b^2 - 4*a*c)^3/(c^4*(b + 2*c*x)^2) + (6*(b^2 - 4*a*c)^2*Log[b + 2*c*x])/c^4)/(2 56*d^3)
Time = 0.31 (sec) , antiderivative size = 100, 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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle \int \left (\frac {3 \left (4 a c-b^2\right ) (b d+2 c d x)}{64 c^3 d^4}+\frac {3 \left (4 a c-b^2\right )^2}{64 c^3 d^2 (b d+2 c d x)}+\frac {\left (4 a c-b^2\right )^3}{64 c^3 (b d+2 c d x)^3}+\frac {(b d+2 c d x)^3}{64 c^3 d^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 \left (b^2-4 a c\right ) (b+2 c x)^2}{256 c^4 d^3}+\frac {\left (b^2-4 a c\right )^3}{256 c^4 d^3 (b+2 c x)^2}+\frac {3 \left (b^2-4 a c\right )^2 \log (b+2 c x)}{128 c^4 d^3}+\frac {(b+2 c x)^4}{512 c^4 d^3}\) |
Input:
Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^3,x]
Output:
(b^2 - 4*a*c)^3/(256*c^4*d^3*(b + 2*c*x)^2) - (3*(b^2 - 4*a*c)*(b + 2*c*x) ^2)/(256*c^4*d^3) + (b + 2*c*x)^4/(512*c^4*d^3) + (3*(b^2 - 4*a*c)^2*Log[b + 2*c*x])/(128*c^4*d^3)
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 = 0.84 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {\frac {\left (2 c^{2} x^{2}+2 c b x +6 a c -b^{2}\right )^{2}}{128 c^{4}}+\frac {\left (48 a^{2} c^{2}-24 c a \,b^{2}+3 b^{4}\right ) \ln \left (2 c x +b \right )}{128 c^{4}}-\frac {64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}}{256 c^{4} \left (2 c x +b \right )^{2}}}{d^{3}}\) | \(114\) |
norman | \(\frac {\frac {c^{2} x^{6}}{8 d}+\frac {3 \left (8 a c +3 b^{2}\right ) x^{4}}{32 d}-\frac {64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+72 a \,b^{4} c -9 b^{6}}{256 c^{4} d}+\frac {3 b c \,x^{5}}{8 d}+\frac {b \left (24 a c -b^{2}\right ) x^{3}}{16 c d}-\frac {b \left (24 c a \,b^{2}-3 b^{4}\right ) x}{32 d \,c^{3}}}{d^{2} \left (2 c x +b \right )^{2}}+\frac {3 \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \ln \left (2 c x +b \right )}{128 c^{4} d^{3}}\) | \(173\) |
risch | \(\frac {x^{4}}{32 d^{3}}+\frac {b \,x^{3}}{16 c \,d^{3}}+\frac {3 a \,x^{2}}{16 c \,d^{3}}+\frac {3 a b x}{16 c^{2} d^{3}}-\frac {b^{3} x}{32 c^{3} d^{3}}+\frac {9 a^{2}}{32 c^{2} d^{3}}-\frac {3 a \,b^{2}}{32 c^{3} d^{3}}+\frac {b^{4}}{128 c^{4} d^{3}}-\frac {a^{3}}{4 c \,d^{3} \left (2 c x +b \right )^{2}}+\frac {3 a^{2} b^{2}}{16 c^{2} d^{3} \left (2 c x +b \right )^{2}}-\frac {3 a \,b^{4}}{64 c^{3} d^{3} \left (2 c x +b \right )^{2}}+\frac {b^{6}}{256 c^{4} d^{3} \left (2 c x +b \right )^{2}}+\frac {3 \ln \left (2 c x +b \right ) a^{2}}{8 c^{2} d^{3}}-\frac {3 \ln \left (2 c x +b \right ) a \,b^{2}}{16 c^{3} d^{3}}+\frac {3 \ln \left (2 c x +b \right ) b^{4}}{128 c^{4} d^{3}}\) | \(226\) |
parallelrisch | \(\frac {384 a b \,c^{4} x^{3}+9 b^{6}-72 a \,b^{4} c +96 \ln \left (\frac {b}{2}+c x \right ) a^{2} b^{2} c^{2}-48 \ln \left (\frac {b}{2}+c x \right ) a \,b^{4} c -192 x a \,b^{3} c^{2}+72 x^{4} b^{2} c^{4}-64 a^{3} c^{3}-16 b^{3} c^{3} x^{3}+192 a \,c^{5} x^{4}+32 x^{6} c^{6}+24 \ln \left (\frac {b}{2}+c x \right ) x^{2} b^{4} c^{2}+24 \ln \left (\frac {b}{2}+c x \right ) x \,b^{5} c +384 \ln \left (\frac {b}{2}+c x \right ) x^{2} a^{2} c^{4}-192 \ln \left (\frac {b}{2}+c x \right ) x^{2} a \,b^{2} c^{3}+384 \ln \left (\frac {b}{2}+c x \right ) x \,a^{2} b \,c^{3}-192 \ln \left (\frac {b}{2}+c x \right ) x a \,b^{3} c^{2}+96 x^{5} b \,c^{5}+24 x c \,b^{5}+6 \ln \left (\frac {b}{2}+c x \right ) b^{6}+48 a^{2} b^{2} c^{2}}{256 c^{4} d^{3} \left (2 c x +b \right )^{2}}\) | \(280\) |
Input:
int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^3,x,method=_RETURNVERBOSE)
Output:
1/d^3*(1/128/c^4*(2*c^2*x^2+2*b*c*x+6*a*c-b^2)^2+1/128*(48*a^2*c^2-24*a*b^ 2*c+3*b^4)/c^4*ln(2*c*x+b)-1/256*(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6 )/c^4/(2*c*x+b)^2)
Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (92) = 184\).
Time = 0.08 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.52 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx=\frac {32 \, c^{6} x^{6} + 96 \, b c^{5} x^{5} + b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3} + 24 \, {\left (3 \, b^{2} c^{4} + 8 \, a c^{5}\right )} x^{4} - 16 \, {\left (b^{3} c^{3} - 24 \, a b c^{4}\right )} x^{3} - 16 \, {\left (2 \, b^{4} c^{2} - 15 \, a b^{2} c^{3}\right )} x^{2} - 8 \, {\left (b^{5} c - 6 \, a b^{3} c^{2}\right )} x + 6 \, {\left (b^{6} - 8 \, a b^{4} c + 16 \, a^{2} b^{2} c^{2} + 4 \, {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{2} + 4 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x\right )} \log \left (2 \, c x + b\right )}{256 \, {\left (4 \, c^{6} d^{3} x^{2} + 4 \, b c^{5} d^{3} x + b^{2} c^{4} d^{3}\right )}} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^3,x, algorithm="fricas")
Output:
1/256*(32*c^6*x^6 + 96*b*c^5*x^5 + b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64* a^3*c^3 + 24*(3*b^2*c^4 + 8*a*c^5)*x^4 - 16*(b^3*c^3 - 24*a*b*c^4)*x^3 - 1 6*(2*b^4*c^2 - 15*a*b^2*c^3)*x^2 - 8*(b^5*c - 6*a*b^3*c^2)*x + 6*(b^6 - 8* a*b^4*c + 16*a^2*b^2*c^2 + 4*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^2 + 4* (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x)*log(2*c*x + b))/(4*c^6*d^3*x^2 + 4 *b*c^5*d^3*x + b^2*c^4*d^3)
Time = 0.64 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.56 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx=\frac {3 a x^{2}}{16 c d^{3}} + \frac {b x^{3}}{16 c d^{3}} + x \left (\frac {3 a b}{16 c^{2} d^{3}} - \frac {b^{3}}{32 c^{3} d^{3}}\right ) + \frac {- 64 a^{3} c^{3} + 48 a^{2} b^{2} c^{2} - 12 a b^{4} c + b^{6}}{256 b^{2} c^{4} d^{3} + 1024 b c^{5} d^{3} x + 1024 c^{6} d^{3} x^{2}} + \frac {x^{4}}{32 d^{3}} + \frac {3 \left (4 a c - b^{2}\right )^{2} \log {\left (b + 2 c x \right )}}{128 c^{4} d^{3}} \] Input:
integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**3,x)
Output:
3*a*x**2/(16*c*d**3) + b*x**3/(16*c*d**3) + x*(3*a*b/(16*c**2*d**3) - b**3 /(32*c**3*d**3)) + (-64*a**3*c**3 + 48*a**2*b**2*c**2 - 12*a*b**4*c + b**6 )/(256*b**2*c**4*d**3 + 1024*b*c**5*d**3*x + 1024*c**6*d**3*x**2) + x**4/( 32*d**3) + 3*(4*a*c - b**2)**2*log(b + 2*c*x)/(128*c**4*d**3)
Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx=\frac {b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}{256 \, {\left (4 \, c^{6} d^{3} x^{2} + 4 \, b c^{5} d^{3} x + b^{2} c^{4} d^{3}\right )}} + \frac {c^{3} x^{4} + 2 \, b c^{2} x^{3} + 6 \, a c^{2} x^{2} - {\left (b^{3} - 6 \, a b c\right )} x}{32 \, c^{3} d^{3}} + \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left (2 \, c x + b\right )}{128 \, c^{4} d^{3}} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^3,x, algorithm="maxima")
Output:
1/256*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)/(4*c^6*d^3*x^2 + 4* b*c^5*d^3*x + b^2*c^4*d^3) + 1/32*(c^3*x^4 + 2*b*c^2*x^3 + 6*a*c^2*x^2 - ( b^3 - 6*a*b*c)*x)/(c^3*d^3) + 3/128*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*log(2*c *x + b)/(c^4*d^3)
Time = 0.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.48 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx=\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, c x + b \right |}\right )}{128 \, c^{4} d^{3}} + \frac {b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}{256 \, {\left (2 \, c x + b\right )}^{2} c^{4} d^{3}} + \frac {c^{12} d^{9} x^{4} + 2 \, b c^{11} d^{9} x^{3} + 6 \, a c^{11} d^{9} x^{2} - b^{3} c^{9} d^{9} x + 6 \, a b c^{10} d^{9} x}{32 \, c^{12} d^{12}} \] Input:
integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^3,x, algorithm="giac")
Output:
3/128*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*log(abs(2*c*x + b))/(c^4*d^3) + 1/256 *(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)/((2*c*x + b)^2*c^4*d^3) + 1/32*(c^12*d^9*x^4 + 2*b*c^11*d^9*x^3 + 6*a*c^11*d^9*x^2 - b^3*c^9*d^9*x + 6*a*b*c^10*d^9*x)/(c^12*d^12)
Time = 5.15 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.23 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx=x^2\,\left (\frac {3\,\left (b^2+a\,c\right )}{16\,c^2\,d^3}-\frac {3\,b^2}{16\,c^2\,d^3}\right )-x\,\left (\frac {5\,b^3}{32\,c^3\,d^3}-\frac {b^3+6\,a\,c\,b}{8\,c^3\,d^3}+\frac {3\,b\,\left (\frac {3\,\left (b^2+a\,c\right )}{8\,c^2\,d^3}-\frac {3\,b^2}{8\,c^2\,d^3}\right )}{2\,c}\right )+\frac {x^4}{32\,d^3}+\frac {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}{8\,c\,\left (32\,b^2\,c^3\,d^3+128\,b\,c^4\,d^3\,x+128\,c^5\,d^3\,x^2\right )}+\frac {b\,x^3}{16\,c\,d^3}+\frac {\ln \left (b+2\,c\,x\right )\,\left (48\,a^2\,c^2-24\,a\,b^2\,c+3\,b^4\right )}{128\,c^4\,d^3} \] Input:
int((a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^3,x)
Output:
x^2*((3*(a*c + b^2))/(16*c^2*d^3) - (3*b^2)/(16*c^2*d^3)) - x*((5*b^3)/(32 *c^3*d^3) - (b^3 + 6*a*b*c)/(8*c^3*d^3) + (3*b*((3*(a*c + b^2))/(8*c^2*d^3 ) - (3*b^2)/(8*c^2*d^3)))/(2*c)) + x^4/(32*d^3) + (b^6 - 64*a^3*c^3 + 48*a ^2*b^2*c^2 - 12*a*b^4*c)/(8*c*(32*b^2*c^3*d^3 + 128*c^5*d^3*x^2 + 128*b*c^ 4*d^3*x)) + (b*x^3)/(16*c*d^3) + (log(b + 2*c*x)*(3*b^4 + 48*a^2*c^2 - 24* a*b^2*c))/(128*c^4*d^3)
Time = 0.19 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.87 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^3} \, dx=\frac {96 \,\mathrm {log}\left (2 c x +b \right ) a^{2} b^{2} c^{2}+384 \,\mathrm {log}\left (2 c x +b \right ) a^{2} b \,c^{3} x +384 \,\mathrm {log}\left (2 c x +b \right ) a^{2} c^{4} x^{2}-48 \,\mathrm {log}\left (2 c x +b \right ) a \,b^{4} c -192 \,\mathrm {log}\left (2 c x +b \right ) a \,b^{3} c^{2} x -192 \,\mathrm {log}\left (2 c x +b \right ) a \,b^{2} c^{3} x^{2}+6 \,\mathrm {log}\left (2 c x +b \right ) b^{6}+24 \,\mathrm {log}\left (2 c x +b \right ) b^{5} c x +24 \,\mathrm {log}\left (2 c x +b \right ) b^{4} c^{2} x^{2}-64 a^{3} c^{3}+48 a^{2} b^{2} c^{2}-24 a \,b^{4} c +192 a \,b^{2} c^{3} x^{2}+384 a b \,c^{4} x^{3}+192 a \,c^{5} x^{4}+3 b^{6}-24 b^{4} c^{2} x^{2}-16 b^{3} c^{3} x^{3}+72 b^{2} c^{4} x^{4}+96 b \,c^{5} x^{5}+32 c^{6} x^{6}}{256 c^{4} d^{3} \left (4 c^{2} x^{2}+4 b c x +b^{2}\right )} \] Input:
int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^3,x)
Output:
(96*log(b + 2*c*x)*a**2*b**2*c**2 + 384*log(b + 2*c*x)*a**2*b*c**3*x + 384 *log(b + 2*c*x)*a**2*c**4*x**2 - 48*log(b + 2*c*x)*a*b**4*c - 192*log(b + 2*c*x)*a*b**3*c**2*x - 192*log(b + 2*c*x)*a*b**2*c**3*x**2 + 6*log(b + 2*c *x)*b**6 + 24*log(b + 2*c*x)*b**5*c*x + 24*log(b + 2*c*x)*b**4*c**2*x**2 - 64*a**3*c**3 + 48*a**2*b**2*c**2 - 24*a*b**4*c + 192*a*b**2*c**3*x**2 + 3 84*a*b*c**4*x**3 + 192*a*c**5*x**4 + 3*b**6 - 24*b**4*c**2*x**2 - 16*b**3* c**3*x**3 + 72*b**2*c**4*x**4 + 96*b*c**5*x**5 + 32*c**6*x**6)/(256*c**4*d **3*(b**2 + 4*b*c*x + 4*c**2*x**2))