Integrand size = 20, antiderivative size = 138 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx=-\frac {c (3 c d-2 b e) x}{e^4}+\frac {c^2 x^2}{2 e^3}-\frac {\left (c d^2-b d e+a e^2\right )^2}{2 e^5 (d+e x)^2}+\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{e^5 (d+e x)}+\frac {\left (6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)\right ) \log (d+e x)}{e^5} \] Output:
-c*(-2*b*e+3*c*d)*x/e^4+1/2*c^2*x^2/e^3-1/2*(a*e^2-b*d*e+c*d^2)^2/e^5/(e*x +d)^2+2*(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)/e^5/(e*x+d)+(6*c^2*d^2+b^2*e^2-2* c*e*(-a*e+3*b*d))*ln(e*x+d)/e^5
Time = 0.06 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {e^2 (b d-a e) (3 b d+a e+4 b e x)+c^2 \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )+2 c e \left (a d e (3 d+4 e x)+b \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )\right )+2 \left (6 c^2 d^2+b^2 e^2+2 c e (-3 b d+a e)\right ) (d+e x)^2 \log (d+e x)}{2 e^5 (d+e x)^2} \] Input:
Integrate[(a + b*x + c*x^2)^2/(d + e*x)^3,x]
Output:
(e^2*(b*d - a*e)*(3*b*d + a*e + 4*b*e*x) + c^2*(7*d^4 + 2*d^3*e*x - 11*d^2 *e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4) + 2*c*e*(a*d*e*(3*d + 4*e*x) + b*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + 2*(6*c^2*d^2 + b^2*e^2 + 2*c*e* (-3*b*d + a*e))*(d + e*x)^2*Log[d + e*x])/(2*e^5*(d + e*x)^2)
Time = 0.34 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1140, 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.
| \(\displaystyle \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle \int \left (\frac {-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{e^4 (d+e x)}+\frac {2 (b e-2 c d) \left (a e^2-b d e+c d^2\right )}{e^4 (d+e x)^2}+\frac {\left (a e^2-b d e+c d^2\right )^2}{e^4 (d+e x)^3}-\frac {c (3 c d-2 b e)}{e^4}+\frac {c^2 x}{e^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\log (d+e x) \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^5}-\frac {\left (a e^2-b d e+c d^2\right )^2}{2 e^5 (d+e x)^2}+\frac {2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^5 (d+e x)}-\frac {c x (3 c d-2 b e)}{e^4}+\frac {c^2 x^2}{2 e^3}\) |
Input:
Int[(a + b*x + c*x^2)^2/(d + e*x)^3,x]
Output:
-((c*(3*c*d - 2*b*e)*x)/e^4) + (c^2*x^2)/(2*e^3) - (c*d^2 - b*d*e + a*e^2) ^2/(2*e^5*(d + e*x)^2) + (2*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(e^5*(d + e*x)) + ((6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))*Log[d + e*x])/e^5
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] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Time = 0.92 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.30
| method | result | size |
| default | \(\frac {c \left (\frac {1}{2} c e \,x^{2}+2 b e x -3 c d x \right )}{e^{4}}+\frac {\left (2 a c \,e^{2}+b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{5}}-\frac {2 a \,e^{3} b -4 a d \,e^{2} c -2 d \,e^{2} b^{2}+6 d^{2} e b c -4 c^{2} d^{3}}{e^{5} \left (e x +d \right )}-\frac {a^{2} e^{4}-2 d \,e^{3} a b +2 a c \,d^{2} e^{2}+d^{2} e^{2} b^{2}-2 b c \,d^{3} e +c^{2} d^{4}}{2 e^{5} \left (e x +d \right )^{2}}\) | \(180\) |
| norman | \(\frac {-\frac {a^{2} e^{4}+2 d \,e^{3} a b -6 a c \,d^{2} e^{2}-3 d^{2} e^{2} b^{2}+18 b c \,d^{3} e -18 c^{2} d^{4}}{2 e^{5}}+\frac {c^{2} x^{4}}{2 e}-\frac {2 \left (a \,e^{3} b -2 a d \,e^{2} c -d \,e^{2} b^{2}+6 d^{2} e b c -6 c^{2} d^{3}\right ) x}{e^{4}}+\frac {2 c \left (b e -c d \right ) x^{3}}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {\left (2 a c \,e^{2}+b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{5}}\) | \(182\) |
| risch | \(\frac {c^{2} x^{2}}{2 e^{3}}+\frac {2 c b x}{e^{3}}-\frac {3 c^{2} d x}{e^{4}}+\frac {\left (-2 a \,e^{3} b +4 a d \,e^{2} c +2 d \,e^{2} b^{2}-6 d^{2} e b c +4 c^{2} d^{3}\right ) x -\frac {a^{2} e^{4}+2 d \,e^{3} a b -6 a c \,d^{2} e^{2}-3 d^{2} e^{2} b^{2}+10 b c \,d^{3} e -7 c^{2} d^{4}}{2 e}}{e^{4} \left (e x +d \right )^{2}}+\frac {2 \ln \left (e x +d \right ) a c}{e^{3}}+\frac {b^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {6 \ln \left (e x +d \right ) b c d}{e^{4}}+\frac {6 \ln \left (e x +d \right ) c^{2} d^{2}}{e^{5}}\) | \(201\) |
| parallelrisch | \(\frac {-4 d \,c^{2} x^{3} e^{3}+24 x \,c^{2} d^{3} e -2 d \,e^{3} a b +18 c^{2} d^{4}+4 \ln \left (e x +d \right ) x^{2} a c \,e^{4}+12 \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{2}+c^{2} x^{4} e^{4}-24 \ln \left (e x +d \right ) x b c \,d^{2} e^{2}-12 \ln \left (e x +d \right ) x^{2} b c d \,e^{3}+6 a c \,d^{2} e^{2}-18 b c \,d^{3} e +8 \ln \left (e x +d \right ) x a c d \,e^{3}+8 x a c d \,e^{3}-a^{2} e^{4}+4 \ln \left (e x +d \right ) a c \,d^{2} e^{2}+4 \ln \left (e x +d \right ) x \,b^{2} d \,e^{3}+24 \ln \left (e x +d \right ) x \,c^{2} d^{3} e -12 \ln \left (e x +d \right ) b c \,d^{3} e -24 x b c \,d^{2} e^{2}+4 x^{3} b c \,e^{4}+2 \ln \left (e x +d \right ) b^{2} d^{2} e^{2}-4 x a b \,e^{4}+4 x \,b^{2} d \,e^{3}+2 \ln \left (e x +d \right ) x^{2} b^{2} e^{4}+3 d^{2} e^{2} b^{2}+12 \ln \left (e x +d \right ) c^{2} d^{4}}{2 e^{5} \left (e x +d \right )^{2}}\) | \(342\) |
Input:
int((c*x^2+b*x+a)^2/(e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
c/e^4*(1/2*c*e*x^2+2*b*e*x-3*c*d*x)+1/e^5*(2*a*c*e^2+b^2*e^2-6*b*c*d*e+6*c ^2*d^2)*ln(e*x+d)-(2*a*b*e^3-4*a*c*d*e^2-2*b^2*d*e^2+6*b*c*d^2*e-4*c^2*d^3 )/e^5/(e*x+d)-1/2*(a^2*e^4-2*a*b*d*e^3+2*a*c*d^2*e^2+b^2*d^2*e^2-2*b*c*d^3 *e+c^2*d^4)/e^5/(e*x+d)^2
Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (134) = 268\).
Time = 0.08 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.07 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {c^{2} e^{4} x^{4} + 7 \, c^{2} d^{4} - 10 \, b c d^{3} e - 2 \, a b d e^{3} - a^{2} e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} - 4 \, {\left (c^{2} d e^{3} - b c e^{4}\right )} x^{3} - {\left (11 \, c^{2} d^{2} e^{2} - 8 \, b c d e^{3}\right )} x^{2} + 2 \, {\left (c^{2} d^{3} e - 4 \, b c d^{2} e^{2} - 2 \, a b e^{4} + 2 \, {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x + 2 \, {\left (6 \, c^{2} d^{4} - 6 \, b c d^{3} e + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + {\left (6 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} + {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 2 \, {\left (6 \, c^{2} d^{3} e - 6 \, b c d^{2} e^{2} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \] Input:
integrate((c*x^2+b*x+a)^2/(e*x+d)^3,x, algorithm="fricas")
Output:
1/2*(c^2*e^4*x^4 + 7*c^2*d^4 - 10*b*c*d^3*e - 2*a*b*d*e^3 - a^2*e^4 + 3*(b ^2 + 2*a*c)*d^2*e^2 - 4*(c^2*d*e^3 - b*c*e^4)*x^3 - (11*c^2*d^2*e^2 - 8*b* c*d*e^3)*x^2 + 2*(c^2*d^3*e - 4*b*c*d^2*e^2 - 2*a*b*e^4 + 2*(b^2 + 2*a*c)* d*e^3)*x + 2*(6*c^2*d^4 - 6*b*c*d^3*e + (b^2 + 2*a*c)*d^2*e^2 + (6*c^2*d^2 *e^2 - 6*b*c*d*e^3 + (b^2 + 2*a*c)*e^4)*x^2 + 2*(6*c^2*d^3*e - 6*b*c*d^2*e ^2 + (b^2 + 2*a*c)*d*e^3)*x)*log(e*x + d))/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)
Time = 1.41 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {c^{2} x^{2}}{2 e^{3}} + x \left (\frac {2 b c}{e^{3}} - \frac {3 c^{2} d}{e^{4}}\right ) + \frac {- a^{2} e^{4} - 2 a b d e^{3} + 6 a c d^{2} e^{2} + 3 b^{2} d^{2} e^{2} - 10 b c d^{3} e + 7 c^{2} d^{4} + x \left (- 4 a b e^{4} + 8 a c d e^{3} + 4 b^{2} d e^{3} - 12 b c d^{2} e^{2} + 8 c^{2} d^{3} e\right )}{2 d^{2} e^{5} + 4 d e^{6} x + 2 e^{7} x^{2}} + \frac {\left (2 a c e^{2} + b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right ) \log {\left (d + e x \right )}}{e^{5}} \] Input:
integrate((c*x**2+b*x+a)**2/(e*x+d)**3,x)
Output:
c**2*x**2/(2*e**3) + x*(2*b*c/e**3 - 3*c**2*d/e**4) + (-a**2*e**4 - 2*a*b* d*e**3 + 6*a*c*d**2*e**2 + 3*b**2*d**2*e**2 - 10*b*c*d**3*e + 7*c**2*d**4 + x*(-4*a*b*e**4 + 8*a*c*d*e**3 + 4*b**2*d*e**3 - 12*b*c*d**2*e**2 + 8*c** 2*d**3*e))/(2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) + (2*a*c*e**2 + b**2*e **2 - 6*b*c*d*e + 6*c**2*d**2)*log(d + e*x)/e**5
Time = 0.04 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {7 \, c^{2} d^{4} - 10 \, b c d^{3} e - 2 \, a b d e^{3} - a^{2} e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 4 \, {\left (2 \, c^{2} d^{3} e - 3 \, b c d^{2} e^{2} - a b e^{4} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} + \frac {c^{2} e x^{2} - 2 \, {\left (3 \, c^{2} d - 2 \, b c e\right )} x}{2 \, e^{4}} + \frac {{\left (6 \, c^{2} d^{2} - 6 \, b c d e + {\left (b^{2} + 2 \, a c\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \] Input:
integrate((c*x^2+b*x+a)^2/(e*x+d)^3,x, algorithm="maxima")
Output:
1/2*(7*c^2*d^4 - 10*b*c*d^3*e - 2*a*b*d*e^3 - a^2*e^4 + 3*(b^2 + 2*a*c)*d^ 2*e^2 + 4*(2*c^2*d^3*e - 3*b*c*d^2*e^2 - a*b*e^4 + (b^2 + 2*a*c)*d*e^3)*x) /(e^7*x^2 + 2*d*e^6*x + d^2*e^5) + 1/2*(c^2*e*x^2 - 2*(3*c^2*d - 2*b*c*e)* x)/e^4 + (6*c^2*d^2 - 6*b*c*d*e + (b^2 + 2*a*c)*e^2)*log(e*x + d)/e^5
Time = 0.33 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {{\left (6 \, c^{2} d^{2} - 6 \, b c d e + b^{2} e^{2} + 2 \, a c e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{5}} + \frac {c^{2} e^{3} x^{2} - 6 \, c^{2} d e^{2} x + 4 \, b c e^{3} x}{2 \, e^{6}} + \frac {7 \, c^{2} d^{4} - 10 \, b c d^{3} e + 3 \, b^{2} d^{2} e^{2} + 6 \, a c d^{2} e^{2} - 2 \, a b d e^{3} - a^{2} e^{4} + 4 \, {\left (2 \, c^{2} d^{3} e - 3 \, b c d^{2} e^{2} + b^{2} d e^{3} + 2 \, a c d e^{3} - a b e^{4}\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{5}} \] Input:
integrate((c*x^2+b*x+a)^2/(e*x+d)^3,x, algorithm="giac")
Output:
(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2 + 2*a*c*e^2)*log(abs(e*x + d))/e^5 + 1/2* (c^2*e^3*x^2 - 6*c^2*d*e^2*x + 4*b*c*e^3*x)/e^6 + 1/2*(7*c^2*d^4 - 10*b*c* d^3*e + 3*b^2*d^2*e^2 + 6*a*c*d^2*e^2 - 2*a*b*d*e^3 - a^2*e^4 + 4*(2*c^2*d ^3*e - 3*b*c*d^2*e^2 + b^2*d*e^3 + 2*a*c*d*e^3 - a*b*e^4)*x)/((e*x + d)^2* e^5)
Time = 5.48 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (b^2\,e^2-6\,b\,c\,d\,e+6\,c^2\,d^2+2\,a\,c\,e^2\right )}{e^5}-\frac {\frac {a^2\,e^4+2\,a\,b\,d\,e^3-6\,a\,c\,d^2\,e^2-3\,b^2\,d^2\,e^2+10\,b\,c\,d^3\,e-7\,c^2\,d^4}{2\,e}-x\,\left (2\,b^2\,d\,e^2-6\,b\,c\,d^2\,e-2\,a\,b\,e^3+4\,c^2\,d^3+4\,a\,c\,d\,e^2\right )}{d^2\,e^4+2\,d\,e^5\,x+e^6\,x^2}-x\,\left (\frac {3\,c^2\,d}{e^4}-\frac {2\,b\,c}{e^3}\right )+\frac {c^2\,x^2}{2\,e^3} \] Input:
int((a + b*x + c*x^2)^2/(d + e*x)^3,x)
Output:
(log(d + e*x)*(b^2*e^2 + 6*c^2*d^2 + 2*a*c*e^2 - 6*b*c*d*e))/e^5 - ((a^2*e ^4 - 7*c^2*d^4 - 3*b^2*d^2*e^2 + 2*a*b*d*e^3 + 10*b*c*d^3*e - 6*a*c*d^2*e^ 2)/(2*e) - x*(4*c^2*d^3 + 2*b^2*d*e^2 - 2*a*b*e^3 + 4*a*c*d*e^2 - 6*b*c*d^ 2*e))/(d^2*e^4 + e^6*x^2 + 2*d*e^5*x) - x*((3*c^2*d)/e^4 - (2*b*c)/e^3) + (c^2*x^2)/(2*e^3)
Time = 0.22 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.69 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^3} \, dx=\frac {2 \,\mathrm {log}\left (e x +d \right ) b^{2} d \,e^{4} x^{2}+12 \,\mathrm {log}\left (e x +d \right ) c^{2} d^{3} e^{2} x^{2}+12 \,\mathrm {log}\left (e x +d \right ) c^{2} d^{5}+24 \,\mathrm {log}\left (e x +d \right ) c^{2} d^{4} e x -4 a c d \,e^{4} x^{2}+12 b c \,d^{2} e^{3} x^{2}+4 b c d \,e^{4} x^{3}+4 \,\mathrm {log}\left (e x +d \right ) a c \,d^{3} e^{2}+4 \,\mathrm {log}\left (e x +d \right ) b^{2} d^{2} e^{3} x -12 \,\mathrm {log}\left (e x +d \right ) b c \,d^{4} e -a^{2} d \,e^{4}+b^{2} d^{3} e^{2}+2 a b \,e^{5} x^{2}+2 a c \,d^{3} e^{2}+4 \,\mathrm {log}\left (e x +d \right ) a c d \,e^{4} x^{2}-12 \,\mathrm {log}\left (e x +d \right ) b c \,d^{2} e^{3} x^{2}+c^{2} d \,e^{4} x^{4}+2 \,\mathrm {log}\left (e x +d \right ) b^{2} d^{3} e^{2}-2 b^{2} d \,e^{4} x^{2}-12 c^{2} d^{3} e^{2} x^{2}-4 c^{2} d^{2} e^{3} x^{3}+6 c^{2} d^{5}+8 \,\mathrm {log}\left (e x +d \right ) a c \,d^{2} e^{3} x -24 \,\mathrm {log}\left (e x +d \right ) b c \,d^{3} e^{2} x -6 b c \,d^{4} e}{2 d \,e^{5} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((c*x^2+b*x+a)^2/(e*x+d)^3,x)
Output:
(4*log(d + e*x)*a*c*d**3*e**2 + 8*log(d + e*x)*a*c*d**2*e**3*x + 4*log(d + e*x)*a*c*d*e**4*x**2 + 2*log(d + e*x)*b**2*d**3*e**2 + 4*log(d + e*x)*b** 2*d**2*e**3*x + 2*log(d + e*x)*b**2*d*e**4*x**2 - 12*log(d + e*x)*b*c*d**4 *e - 24*log(d + e*x)*b*c*d**3*e**2*x - 12*log(d + e*x)*b*c*d**2*e**3*x**2 + 12*log(d + e*x)*c**2*d**5 + 24*log(d + e*x)*c**2*d**4*e*x + 12*log(d + e *x)*c**2*d**3*e**2*x**2 - a**2*d*e**4 + 2*a*b*e**5*x**2 + 2*a*c*d**3*e**2 - 4*a*c*d*e**4*x**2 + b**2*d**3*e**2 - 2*b**2*d*e**4*x**2 - 6*b*c*d**4*e + 12*b*c*d**2*e**3*x**2 + 4*b*c*d*e**4*x**3 + 6*c**2*d**5 - 12*c**2*d**3*e* *2*x**2 - 4*c**2*d**2*e**3*x**3 + c**2*d*e**4*x**4)/(2*d*e**5*(d**2 + 2*d* e*x + e**2*x**2))